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Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume

For better preparation and quality education, our team has shared the Big Ideas Math Book Geometry Answer Key Chapter 11 Circumference, Area, and Volume for high school students. Provided Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume are aligned topic-wise as per the latest common core curriculum. Students can easily understand the concepts of Chapter 11 Circumference, Area, and Volume. Learn how to solve the problems using the Circumference, Area, and Volume formulas in detail from here.

Big Ideas Math Book Geometry Answer Key Chapter 11 Circumference, Area, and Volume

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Circumference, Area, and Volume Maintaining Mathematical Proficiency

Circumference, area, and volume monitoring progress, 11.1 circumference and arc length, lesson 11.1 circumference and arc length, exercise 11.1 circumference and arc length, 11.2 areas of circles and sectors, lesson 11.2 areas of circles and sectors, exercise 11.2 areas of circles and sectors, 11.3 areas of polygons, lesson 11.3 areas of polygons, exercise 11.3 areas of polygons, 11.4 three-dimensional figures, lesson 11.4 three-dimensional figures, exercise 11.4 three-dimensional figures, 11.1 – 11.4 quiz, 11.5 volumes of prisms and cylinders, lesson 11.5 volumes of prisms and cylinders, exercise 11.5 volumes of prisms and cylinders, 11.6 volumes of pyramids, lesson 11.6 volumes of pyramids, exercise 11.6 volumes of pyramids, 11.7 surface areas and volumes of cones, lesson 11.7 surface areas and volumes of cones, exercise 11.7 surface areas and volumes of cones, 11.8 surface areas and volumes of spheres, lesson 11.8 surface areas and volumes of spheres, exercise 11.8 surface areas and volumes of spheres, circumference, area, and volume review, circumference, area, and volume test, circumference, area, and volume cumulative assessment.

Find the surface area of the prism.

Answer: The surface area of the prism = 158.

Explanation: In the above-given question, given that, l = 5 ft, w = 8 ft, and h = 3 ft. where l = length, w = width, and h = height. The surface area of the rectangular prism = 2(lw + lh + wh). surface area = 2(5×8 + 5×3 + 8×3). surface area = 2(40 + 15 + 24). surface area = 2(79). surface area = 158.

Answer: The surface area of the triangular prism = 68m.

Explanation: In the above-given question, given that, l = 10 m, p = 4 m, and h = 10. the surface area of the triangular prism = 2B + ph. b = base, p = perimeter, and h = height. surface area = 2(6 + 8) + 4(10). surface area = 2(14) + 40. surface area = 28 + 40. surface area = 68m.

Answer: The surface area of the triangular prism = 42m.

Explanation: In the above-given question, given that, w = 10 cm, p = 4 cm, and h = 5 cm, l = 6 cm. the surface area of the triangular prism = 2B + ph. b = base, p = perimeter, and h = height. surface area = 2(6 + 5) + 4(5). surface area = 2(11) + 20. surface area = 22 + 20. surface area = 42cm.

Find the missing dimension.

Question 4. A rectangle has a perimeter 0f 28 inches and a width of 5 inches. What is the length of the rectangle?

Answer: The length of the rectangle = 9 in.

Explanation: In the above-given question, given that, A rectangle has a perimeter of 28 inches and a width of 5 inches. length of the rectangle = p/2 – w. length = 28/2 – 5. where perimeter = 28 in, and w = 5 in. length = 14 – 5. length = 9. so the length of the rectangle = 9 in.

Question 5. A triangle has an area of 12 square centimeters and a height of 12 centimeters. What is the base of the triangle?

Answer: The base of the triangle = 2 cm.

Explanation: In the above-given question, given that, A triangle has an area of 12 sq cm and a height of 12 cm. The base of the triangle = 2(A)/h. base = 2(12)/12. base = 24/12. base = 2cm. so the base of the triangle = 2 cm.

Question 6. A rectangle has an area of 84 square feet and a width of 7 feet. What is the length of the rectangle?

Answer: The length of the rectangle = 12 ft.

Explanation: In the above-given question, given that, A rectangle has an area of 84 sq ft and a width of 7 feet. area of the rectangle = l x b. 84 = l x 7. l = 84/7. l = 12. so the length of the rectangle = 12 ft.

Question 7. ABSTRACT REASONING Write an equation for the surface area of a Prism with a length, width, and height of x inches. What solid figure does the prism represent?

Answer: The surface area of a prism = 2(lw + wh + lh).

Explanation: In the above-given question, given that, length = l, width = w, and height = x inches. the surface area of the prism = 2(lw + wh + lh). the solid figure does the prism represent the rectangular prism.

Draw a net of the three-dimensional figure. Label the dimensions.

Answer: The surface area of the prism = 64 cm.

Explanation: In the above-given question, given that, l = 2 ft, w = 4 ft, and h = 4 ft. where l = length, w = width, and h = height. The surface area of the rectanguler prism = 2(lw + lh + wh). surface area = 2(2×4 + 4×4 + 4×2). surface area = 2(8 + 16 + 8). surface area = 2(32). surface area = 64.

Answer: The surface area of the prism = 392 m.

Explanation: In the above-given question, given that, l =8 m, w = 12 m, and h = 5 m. where l = length, w = width, and h = height. The surface area of the rectanguler prism = 2(lw + lh + wh). surface area = 2(8×12 + 12×5 + 5×8). surface area = 2(96 + 60 + 40). surface area = 2(196). surface area = 392.

Answer: The surface area of the triangular prism = 170 in.

Explanation: In the above-given question, given that, B = 10 in, p = 15 in, and h = 15 in, l = 10 in. the surface area of the triangular prism = 2B + ph. b = base, p = perimeter, and h = height. surface area = 2(10) + 15(10). surface area = 2(10) + 150. surface area = 20 + 150. surface area = 170 in.

Exploration 1

Finding the Length of a Circular Arc

Work with a partner: Find the length of each red circular arc.

Answer: The Formula for the Arc Length is 2r(θ/360) r = 4 θ = 260 degrees Arc length = 2(4)(360/360) = 8(1) = 8 Therefore the Arc length is 8.

Exploration 2

Using Arc Length

Answer: C = πd C = 25π or 78.54 inch Half of revolution = 1/2 (78.54) = 39.27 inch No, the time has gone past the box by 3.27 inch

Communicate Your Answer

Question 3. How can you find the length of a circular arc? Answer: The length of a circular arc = 2 LOOKING FOR REGULARITY IN REPEATED REASONING To be proficient in math, you need to notice if calculations are repeated and look both for general methods and for shortcuts.

Question 4. A motorcycle tire has a diameter of 24 inches. Approximately how many inches does the motorcycle travel when its front tire makes three-fourths of a revolution? Answer: The diameter of the motorcycle is 24 inches. One revelation = 360 degrees. Three-fourths of one revelation = 270 degrees. The motorcycles travel = 2r(θ/360) Radius r = d/2 = 24/2 = 12 θ = 270 degrees. = 2(12)(270/360) = 24(270/360) = 18 Therefore the motorcycle travels 18 cm.

Monitoring Progress

Question 1. Find the circumference of a circle with a diameter of 5 inches.

Answer: Circumference C = πd C = 3.14x 5 = 15.7 in

Question 2. Find the diameter of a circle with a circumference of 17 feet.

Answer: Diameter d = C/π d = 17/π = 5.41 ft

Find the indicated measure.

Answer: arc length of \(\widehat{P Q}\) is 5.887

Explanation: \(\widehat{P Q}\) = \(\frac { 75 }{ 360 } \) . π(9) = 5.887

Answer: arc length of LM/C = LM/360 61.26/C = 270/360 C = 81.68

Answer: arc length of EF = \(\frac { 60 }{ 360 } \) • 2πr 10.5 = \(\frac { 1 }{ 6 } \) • 2πr r = 10.02

Question 6. A car tire has a diameter of 28 inches. How many revolutions does the tire make while traveling 500 feet?

Answer: The car tire have to make 69 revolutions to travel 500 ft.

Explanation: Circumference C = 2πr = πd C = 28π Distance travelled = number of revolutions x C 500 x 12 = number of revolutions x 28π number of revolutions = 68.2

Question 7. In Example 4. the radius of the arc for a runner on the blue path is 44.02 meters, as shown in the diagram. About how far does this runner travel to go once around the track? Round to the nearest tenth of a meter.

Answer: Given that, The arc radius for a runner on the blue path is 44.02 meters. The diameter of the track is 2r = 2(44.02) = 88.04 The circumference of the track is πd = π(88.04) = 276.44 The runner travels around the track 276.44 cm.

Question 8. Convert 15° to radians.

Answer: 15° = 15 . \(\frac { π radians }{ 180° } \) = \(\frac { π }{ 12 } \) radians

Question 9. Convert \(\frac{4 \pi}{3}\) radians to degrees.

Answer: \(\frac{4 \pi}{3}\) radians = \(\frac{4 \pi}{3}\) radians . \(\frac { 180° }{ π radians } \) = 240 degrees

Vocabulary and Core Concept Check

Question 2. WRITING Describe the difference between an arc measure and an arc length.

Answer: An arc measure is measured in degrees while an arc length is the distance along an arc measured in linear units.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 10, find the indicated measure.

Question 4. diameter of a circle with a circumference of 63 feet

Answer: C = 63 ft πd = 63 d = 20.05

Question 6. exact circumference of a circle with a diameter of 5 inches

Answer: C = πd C = 5π = 15.707

Answer: \(\frac { arc length of DE }{ 2πr } \) = \(\frac { DE }{ 360 } \) \(\frac { 8.73 }{ 2π(10) } \) = \(\frac { DE }{ 360 } \) DE = 50.01°

Answer: \(\frac { arc length of GH }{ 2πr } \) = \(\frac { m GH }{ 360 } \) \(\widehat{G H}\). = \(\frac { 5 }{ 24 } \) . 2π(10) = 13.08

Answer: Circumference of the front wheel = 2π(32.5) = 65π cm Distance covered = 40 m = 40 x 100 = 4000 cm Number of revolutions = \(\frac { 4000 }{ 65π } \) = 19.58

In Exercises 15-18 find the perimeter of the shaded region.

In Exercises 19 – 22, convert the angle measure.

Question 20. Convert 300° to radians.

Answer: 300 • (\(\frac { π }{ 180 } \)) = \(\frac { 5π }{ 3 } \) radian

Question 22. Convert \(\frac{\pi}{8}\) radian to degrees.

Answer: \(\frac { π }{ 8 } \) • \(\frac { 180 }{ π } \)) = 22.5°

Answer: C = 38 ft 2πr = 38 r = 6.04

In Exercises 25 and 26, find the circumference of the circle with the given equation. Write the circumference in terms of π

Question 26. (x + 2) 2 + (y – 3) 2 = 9

Answer: The radius of circle (x + 2)² + (y – 3)² = 9 is 3 C = 2πr = 2π(3) = 6π The circumference of the circle is 6π units.

Question 28. REASONING \(\widehat{E F}\) is an arc on a circle with radius r. Let x° be the measure of \(\widehat{E F}\). Describe the effect on the length of \(\widehat{E F}\) if you (a) double the radius of the circle, and (b) double the measure of \(\widehat{E F}\). Answer: Given x° is the measure of \(\widehat{E F}\) arc length of a circle = x/360 × 2πr arc length of \(\widehat{E F}\) = x/360 × 2πr a. double the radius of the circle x/360 × 2π(2r) = 2x/360 × 2πr x/360 × 2π(2r) = 2arc length of \(\widehat{E F}\) (b) double the measure of \(\widehat{E F}\) 2x/360 × 2πr = 2 × x/360 × 2πr 2x/360 × 2πr = 2arc length \(\widehat{E F}\)

Question 32. ANALYZING RELATIONSHIPS A 45° arc in ⊙C and a 30° arc in ⊙P have the same length. What is the ratio of the radius r 1 of ⊙C to the radius r 2 of ⊙P? Explain your reasoning. Answer: Given, A 45° arc in ⊙C and a 30° arc in ⊙P have the same length. r1/r2 = 45/30 = 3/2 So, the ratio of the radius r 1 of ⊙C to the radius r 2 of ⊙P is 3/2.

Question 38. MODELING WITH MATHEMATICS What is the measure (in radians) of the angle formed by the hands of a clock at each time? Explain your reasoning. a. 1 : 30 P.M.

Answer: 3π/4

b. 3:15 P.M.

Answer: π/24

Question 40. THOUGHT PROVOKING Is π a rational number? Compare the rational number \(\frac{355}{113}\) to π. Find a different rational number that is even closer π.

Answer: π is not a rational number as it can not be represented as an equivalent fraction. π = 3.14 and 355/113 = 3.14. This fraction resembles that value of π. Therefore a more accurate fraction will be starting by the value of 7 decimal places of π, therefore 3.1415926 x x = a.

b. What would the sum of the arc lengths be if \(\overline{A B}\) was divided into 8 congruent segments? 16 congruent segments? n congruent segments? Explain your reasoning. Answer: 360/2 = 180 degrees Then the arc length of 1 semicircle is 180/360 × 2π(r/2) 1/2 × πr = πr/2 Therefore the arc length of 8 semicircles will be 8 × πr/2 = 4πr

Maintaining Mathematical Proficiency

Find the area of the polygon with the given vertices.

Question 44. L(- 3, 1), M(4, 1), N(4, – 5), P(- 3, – 5)

Finding the Area of a Sector of a Circle

Work with a partner: A sector of a circle is the region bounded by two radii of the circle and their intercepted arc. Find the area of each shaded circle or sector of a circle.

Finding the Area of a Circular Sector

Question 3. How can you find the area of a sector of a circle? Answer: The formula for sector area is simple, just multiply the central angle by the radius squared, and divide by 2 Area of a sector = θ/360 × πr²

Question 4. In Exploration 2, find the area of the sector that is irrigated in 2 hours. Answer:

Monitoring progress

Question 1. Find the area of a circle with a radius of 4.5 meters.

Answer: Circle area = πr² A = π(4.5)² = 20.25π

Question 2. Find the radius of a circle with an area of 176.7 square feet.

Answer: Circle area = πr² 176.7 = πr² r² = 56.24 r = 7.499

Question 3. About 58,000 people live in a region with a 2-mile radius. Find the population density in people per square mile.

Answer: The population density is about 4615.49 people per square mile.

Explanation: A = πr² = π • 2² = 4π Population density = \(\frac { number of people }{ area of land } \) = \(\frac { 58000 }{ 4π } \) = 4615.49

Question 4. A region with a 3-mile radius has a population density of about 1000 people per square mile. Find the number of people who live in the region.

Answer: The number of people who live in the region are 28274.

Explanation: A = πr² = π • 3² = 9π Population density = \(\frac { number of people }{ area of land } \) Number of people = 1000 x 9π = 28274

Find the indicated measure

Question 5. area of red sector

Answer: The area of red sector = 205.25

Explanation: m∠FDE = 120°, FE = 120° and FGE = 360° – 120° = 240° Area of red sector = \(\frac { FE }{ 360° } \) • πr² = \(\frac { 120 }{ 360° } \) • π(14²) = 205.25

Question 6. area of blue sector

Answer: Area of blue sector = 410.5

Explanation: Area of blue sector = \(\frac { FGE }{ 360° } \) • πr² = \(\frac { 240 }{ 360° } \) • π(14²) = 410.5

Question 7. Find the area of ⊙H.

Answer: Area of ⊙H = 907.92 sq cm

Explanation: Area of sector FHG =\(\frac { FG }{ 360° } \) • Area of ⊙H 214.37 = \(\frac { 85 }{ 360° } \) • Area of ⊙H Area of ⊙H = 907.92 sq cm

Question 8. Find the area of the figure.

Answer: Area of triangle = \(\frac { 1 }{ 2 } \) • 7 • 7 = 24.5 sq m Area of semi circle = πr²/2 = π(3.5)²/2 = 19.242255 Area of the figure = 24.5 + 19.24 = 43.74 sq m

Question 9. If you know the area and radius of a sector of a circle, can you find the measure of the intercepted arc? Explain.

Answer: Yes, we can find the intercepted arc when we the area and the radius of the sector of the circle. Because the intercepted arc is the arc inside the inscribed angles and whose endpoints are on the angle.

Question 2. WRITING The arc measure of a sector in a given circle is doubled. will the area of the sector also be doubled? Explain our reasoning.

Answer: Yes

Explanation: Area of sector with arc measure x and radius r is s = π/180(xr) If x becomes doube, then s1 = π/180(2xr) = 2s This means that if the arc measure doubles, area of the sector also doubles.

In Exercise 3 – 10, find the indicated measure,

Answer: Area A = πr² A = π(10)² = 100π sq in

Question 6. area of a circle with a diameter of 16 feet

Answer: d = 2r Circle area = πr² = (π/4)d² = (π/4)16² = 64π

Question 8. radius of a circle with an area of 380 square inches

Answer: A = πr² 380 = πr² r = 10.99

Question 10. diameter of a circle with an area of 676π square centimeters

Answer: Area A = 676π square centimeters (π/4)d² = 676π d² = 2704 d = 52

In Exercises 11 – 14, find the indicated measure.

Question 12. About 650,000 people live in a region with a 6-mile radius. Find the population density in people per square mile.

Answer: The population density is about 5747 people per square mile.

Explanation: Area of region = π(6)² = 36π Population density = \(\frac { Number of people }{ area of land } \) = \(\frac { 650,000 }{ 36π } \) = 5747.2

Question 14. About 79,000 people live in a circular region with a population density of about 513 people per square mile. Find the radius of the region.

Answer: The radius of the region is 7

Explanation: Population density = \(\frac { Number of people }{ area of land } \) 513 = \(\frac { 79,000 }{ πr² } \) πr² = 153.99 r = 7

In Exercises 15-18 find the areas of the sectors formed by∠DFE.

Answer: Area of sector = \(\frac { 104° }{ 360° } \) • π(14)² = 177.88 Area of red region is 177.88 sq cm Area of blue region = \(\frac { 256° }{ 360° } \) • π(14)² = 437.86 sq cm

Answer: Area of red region is 10.471 sq ft Area of the blue region is 39.79 sq ft

Explanation: Area of sector = \(\frac { 75° }{ 360° } \) • π(4)² = 10.471 Area of red region is 10.471 sq ft Area of blue region = \(\frac { 285° }{ 360° } \) • π(4)² = 39.79 sq ft

Answer: Area of ⊙Z is 255 square feet πr² = 255 r = 9 Area of sector XZY = \(\frac { 115 }{ 360 } \) • 255 n = 81.458 sq ft

In Exercises 21 and 22, the area of the shaded sector is show. Find the indicated measure.

Answer: radius of ⊙M = 3.98

Explanation: Area of region = \(\frac { 89 }{ 360 } \) . Area of ⊙M 12.36 = \(\frac { 89 }{ 360 } \) . Area of ⊙M Area of ⊙M = 49.99 πr² = 49.99 r = 3.98

In Exercises 23 – 28, find the area of the shaded region.

Answer: The area of the shaded region is 85.840 sq in.

Explanation: Area of square = 20² = 400 Diameter of one circle = 10 radius of one circle = 5 in Area of one circle = π(5)² = 78.53 Areas of four circle = 314.159 Area of shaded region = 400 – 314.159 = 85.840

Answer: The area of shaded region is 301.59

Explanation: The radius of smaller circle is 8 cm The radius of bigger circle is 16 cm Area of smaller semicircle = \(\frac { 1 }{ 2 } \)(π(8)²) = 100.53 Area of lager semicircle = \(\frac { 1 }{ 2 } \)(π(16)²) = 402.123 Area of shaded region = 402.123 – 100.53 = 301.59

Explanation: c² = 3² + 4² = 25 c = 5 Radius = 2.5 Circle area = π(2.5)² = 19.63 Area of triangle = (3 x 4)/2 = 6 Area of shaded region = 19.63 – 12 = 7.63

Question 30. MAKING AN ARGUMENT Your friend claims that if the radius of a circle is doubled, then its area doubles. Is your friend correct? Explain your reasoning.

Answer: The friend is not correct. doubling the radius quadruples the area.

Explanation: Area of circle with radius r = πr² Area of circle with radius 2r = π(2r)² = 4πr²

b. What is the area of land that can be covered by the light from the lighthouse? Answer: Area = \(\frac { 245 }{ 360 } \) x π(18)² = 692.72 sq mi

Question 34. ANALYZING RELATIONSHIPS A square is inscribed in a circle. The same square is also circumscribed about a smaller circle. Draw a diagram that represents this situation. Then find the ratio of the area of the larger circle to the area of the smaller circle.

Answer: We start by assigning a variable to the radius of the inner circle. It is r, therefore the area of the circle is πr² It can be seen that the side length of square is twice this radius. Therefore it can be said that the side length of this square is 2r. Next, it can be seen that the diagonal of the square is diameter of outer circle. Therefore, length of the diagonal of the circle d = 2r√2. outer circle radius = r√2 Area of outer circle 2πr² The ratio of the area of larger circle to the smaller circle = 2.

Answer: Ellipse area = πab

Maintaining Mathematical proficiency

Find the area of the figure.

Answer: Area = \(\frac { 1 }{ 2 } \)(base x height) Area = \(\frac { 1 }{ 2 } \)(18 x 6) = 54 sq in

Finding the Area of a Regular Polygon

Writing a Formula for Area

Work with a partner: Generalize the steps you used in Exploration 1 to develop a formula for the area of a regular polygon. REASONING ABSTRACTLY To be proficient in math, you need to know and flexibly use different properties of operations and objects. Answer:

Question 3. How can you find the area of a regular polygon? Answer: You can find the area of the regular pentagon using the formulas. They are, The formula for the regular pentagon if only the side is known is A = 1/4 x square root of 5(5 + 2 square root(5) x (a)². The formula for the area of the regular pentagon is 1/2 x p x a. Where a = apothem P = perimeter

Question 4. Regular pentagon ABCDE has side lengths of 6 meters and an apothem of approximately 4.13 meters. Find the area of ABCDE. Answer: Given that, The side length of the regular pentagon ABCDE is 6 meters. The apothem of the regular pentagon is 4.13 meters. The formula for the area of the regular pentagon is 1/2 x p x a. Where a = apothem P = perimeter The formula for the perimeter of a regular pentagon is = 5a = 5(6) = 30 = 1/2 x 30 x 4.13 = 1/2 x 123.9 = 61.95 square cm

Question 1. Find the area of a rhombus with diagonals d 1 = 4 feet and d 2 = 5 feet.

Answer: Area of rhombus = \(\frac { 1 }{ 4 } \)(d₁d₂) = \(\frac { 1 }{ 4 } \)(4 x 5) = 5 sq ft

Question 2. Find the area of a kite with diagonals d 1 = 12 inches and d 1 = 9 inches.

Answer: Area of kite = \(\frac { 1 }{ 4 } \)(d₁d₂) = \(\frac { 1 }{ 4 } \)(12 x 9) = 27 sq in

In the diagram. WXYZ is a square inscribed in ⊙P.

Question 3. Identify the center, a radius, an apothem, and a central angle of the polygon.

Answer: P is the center, PY or PX is the radius, PQ is apothem, ∠XPY is the central angle.

Question 4. Find m∠XPY, m∠XPQ, and m∠PXQ.

Answer: m∠XPY = \(\frac { 360 }{ 4 } \) = 90 m∠XPQ = 90/2 = 45 m∠PXQ = 180 – (90 + 45) = 45

Find the area of regular polygon

Answer: c = √(8² + 6.5²) = 10.3 a = 20.61 Area = 0.25(√5(5+2√5) a² Area = 0.25(√5(5+2√5) 20.61² = 730.8

Answer: EF = radius = 6.8

Find the apothem of polygon ABCDE.

Answer: GF = apothem = 5.5

Answer: AF = √4² + 5.5² AF = 6.8

Find the radius of polygon ABCDE. Answer: AF = radius = 6.8

In Exercises 3 – 6, find the area of the kite or rhombus.

Answer: d₁ = 6 + 6 = 12 d₂ = 2 + 10 = 12 area A = \(\frac { 1 }{ 4 } \)(d₁d₂) = \(\frac { 1 }{ 4 } \)(12 x 12) = 36

In Exercises 7 – 10, use the diagram

Question 8. Identify a central angle of polygon JKLMN.

Answer: ∠NPM is the central angle of polygon JKLMN

Question 10. What is the apothem of polygon JKLMN? Answer: QP is the apothem of polygon JKLMN

In Exercises 11 – 14, find the measure of a central angle of a regular polygon with the given number of sides. Round answers to the nearest tenth of a degree, if necessary.

Question 12. 18 sides

Answer: The measure of central angle = \(\frac { 360 }{ 18 } \) = 20

Question 14. 7 sides

Answer: The measure of central angle = \(\frac { 360 }{ 7 } \) = 51.42

In Exercises 15 – 18, find the given angle measure for regular octagon ABCDEFGH.

Question 16. m∠GJK

Answer: m∠GJK = m∠GJH/2 m∠GJK = 22.5

Question 18. m∠EJH Answer: m∠EJH = 3(45) = 135

In Exercises 19 – 24, find the area of the regular polygon.

Question 24. a pentagon with an apothem of 5 units

Answer: A = 90.75

We know apothem a = and it divides pentagon into triangles, the central angle is divided into 360/5 = 72 After that, we halved this angle and got 2 right triangles with x = 44 and y = 36. Since we know one side and all three angles of the triangle, we can calculate p with the tangent function. tan y = p/a tan 36 = p/5 p = 3.63 Since p is just half of the length of the side, we have to multiply it by 2 2 . p = 2 . 3.63 = 7.26 = s Area = \(\frac { a . s. n }{ 2 } \) A = \(\frac { 15 x 7.26 x 5 }{ 2 } \) A = 90.75

Answer: s = √15² – 13² = 7.48 Area = \(\frac { 1 }{ 2 } \)(a . ns) A = \(\frac { 1 }{ 2 } \)(13 x 6 x 7.48) A = 291.72

In Exercises 27 – 30, find the area of the shaded region.

Answer: Area of the shaded region = 223.75

Explanation: Square side = diagonal/√2 = 28/√2 = 19.79 Area of square = 19.79² = 392 Circle area = π(14)² = 615.75 Area of the shaded region = 615.75 – 392 = 223.75

CRITICAL THINKING In Exercises 33 – 35, tell whether the statement is true or false. Explain your reasoning

Question 34. The apothem of a regular polygon is always less than the radius.

Answer: true, the radius always reaches the end of the circle but the apothem never does

Explanation: (A) area = π(6.5)² = 132.73 (B) area = 139.25 (C) area = \(\frac { 1 }{ 2 } \)(18 x 15) = 135

Question 38. REASONING What happens to the area of a kite if you double the length of one of the diagonals? if you double the length of both diagonals? Justify your answer.

Answer: Area of a kite = \(\frac { 1 }{ 4 } \)(d₁d₂) If you double the length of one diagonal, then d₁ = 2d₁ Area of kite = \(\frac { 1 }{ 2 } \)(d₁d₂) If you double length of both diagonals Area = \(\frac { 1 }{ 4 } \)(2d₁2d₂) = d₁d₂ If you double the length of one diagonal, then the area becomes halve. If you double length of both diagonals, then area becomes 4 times.

MATHEMATICAL CONNECTIONS In Exercises 39 and 40, write and solve an equation to find the indicated lengths. Round decimal answers to the nearest tenth.

Question 40. One diagonal of a rhombus is four times the length of the other diagonal. The area of the rhombus is 98 square feet. Find the length of each diagonal.

Answer: The length of each diagonal is 9.89, 2.47.

Explanation: One diagonal of a rhombus is four times the length of the other diagonal. d₁ = 4d₂ Area = \(\frac { 1 }{ 4 } \)(d₁d₂) 98 = \(\frac { 1 }{ 4 } \)(d₁(4d₁)) d₁ = 9.89 d₂ = 2.47

Question 42. MAKING AN ARGUMENT Your friend claims that it is possible to find the area of any rhombus if you only know the perimeter of the rhombus. Is your friend correct? Explain your reasoning.

Answer: No; A rhombus is not a regular polygon.

Answer: Given that, The hexagon has 6 sides. The hexagon is divided into 6 equilateral triangles. The area of the equilateral triangle is (square root of 3)/4 x a² a = side length = (square root of 3)/4 x (6)² = (square root of 3)/4 x 36 = 15.588 square cm.

Question 48. CRITICAL THINKING The area of a dodecagon, or 12-gon, is 140 square inches. Find the apothem of the polygon.

Answer: Let the side length of dodecagon be 2x. The measure of each interior angle of a regular decagon is 150. This implies that the base angle C and A of the resulting isosceles triangle formed by the red sides is equal to 150/2 = 75. The adjacent to this angle is the length 2x/2 = x inches, while the opposite to it is the blue apothem in the right triangle BDC formed. Therefore a = x tan 75. Therefore, area of dodecagon is 140 = 1/2 (x tan75)(12 . 2x) 140 = 44.785 x² x² = 3.126 x = 1.768

Question 52. USING STRUCTURE Two regular polygons both have n sides. One of the polygons is inscribed in, and the other is circumscribed about, a circle of radius r. Find the area between the two polygons in terms of n and r.

Answer: The radius of the smaller polygon is equal to the apothem of the larger polygon. The central angle is 360/n, therefore the apothem makes an angle of 180/n. Use sine and cosine to find the apothem and side length of the smaller polygon. a small = r sin\(\frac { 180 }{ n } \) s small = 2r cos\(\frac { 180 }{ n } \) Use tangent to find the side length of the large polygon. S large = 2r tan\(\frac { 180 }{ n } \) Use the formula to find the area of the smaller polygon. A small = 1/2 . a small . n . s small A small = 1/2 . r sin\(\frac { 180 }{ n } \) . n . 2r cos\(\frac { 180 }{ n } \) A small = nr² sin \(\frac { 180 }{ n } \) cos\(\frac { 180 }{ n } \) Use the formula to find the area of the larger polygon. A Large = 1/2 . a large . n . slarge = nr² tan\(\frac { 180 }{ n } \) The area between the polygons is equal to the area of the larger polygon minus the area of the smaller polygon. Use some trig identities to simplify the expression. A = A large – A small A = nr² tan\(\frac { 180 }{ n } sin²[latex]\frac { 180 }{ n }

Determine whether the figure has line symmetry, rotational symmetry, both, or neither. If the figure has line symmetry. determine the number of lines of symmetry. It the figure has rotational symmetry, describe any rotations that map the figure onto itself.

Analyzing a Property of Polyhedra

Question 2. What is the relationship between the numbers of vertices V, edges E, and faces F of a polyhedron? (Note: Swiss mathematician Leonhard Euler (1707 – 1783) discovered a formula that relates these quantities.) CONSTRUCTING VIABLE ARGUMENTS To be proficient in math, you need to reason inductively about data. Answer: The relationship between the vertices, edges, and faces of a polyhedron according to Euler’s formula is F + V = E + 2. Where F = number of faces. V = number of vertices. E = number of edges.

now we count vertices, edges, and faces in each solids A triangular prism has vertices 6, edges 9 and faces 5 6 – 9 + 5 = 2 A Pentagonal prism has vertices 10, edges 15 and faces 7 10 – 15 + 7 = 2 A triangular pyramid has vertices 4, edges6 and faces 4 – 6 + 2 = 2

Tell whether the solid is a polyhedron. If it is, name the polyhedron.

Answer: The solid is formed by polygons, so it is a polyhedron. The base is a square, it is a square pyramid.

Describe the shape formed by the intersection of the plane and the solid.

Sketch the solid produced by rotating the figure around the given axis. Then identify and describe the solid.

Answer: Cone does not belong with the other three as it has a curved surface and others not.

In Exercises 3 – 6, match the polyhedron with its name.

In Exercises 7 – 10, tell whether the solid is a polyhedron. If it is, name the polyhedron.

In Exercises 11 – 14, describe the cross section formed by the intersection of the plane and the solid.

In Exercises 15 – 18, sketch the solid produced by rotating the figure around the given axis. Then identify and describe the solid.

In Exercises 21 – 26, sketch the polyhedron.

Question 22. rectangular prism

Question 26. pentagonal pyramid

b. What is the perimeter of the cross section? Answer: The perimeter is 2(l + b)

c. What is the area of the cross section? Answer: Area is lb.

REASONING In Exercises 29 – 34, tell whether it is possible for a cross section of a cube to have the given shape. If it is, describe or sketch how the plane could intersect the cube.

Question 30. pentagon Answer: yes, cross-section of the cube can be a pentagon.

Question 32. isosceles triangle Answer: Yes, the cross-section can be an isosceles triangle.

Question 34. scalene triangle Answer: Yes, the cross-section can be scalene triangle.

Question 36. THOUGHT-PROVOKING Describe how Plato might have argued that there are precisely five Platonic Solids (see page 617). (Hint: Consider the angles that meet at a vertex.) Answer:

Decide whether enough information is given to prove that the triangles are congruent. It so, state the theorem you would use.

Answer: ∆JLK ≅ ∆JLM by SAS congruence theorem.

Question 4. Convert 26° to radians and \(\frac{5 \pi}{9}\) radians to degrees.

Answer: 26° = 26 . \(\frac { π }{ 180 } \) = \(\frac { 13π }{ 90 } \) radians \(\frac{5 \pi}{9}\) = \(\frac{5 \pi}{9}\) . \(\frac { 180 }{ π } \) = 100°

Use the figure to find the indicated measure.

Answer: area of red sector = \(\frac { 100 }{ 360 } \) . π(12)² = 125.66

Answer: area of blue sector = \(\frac { 260 }{ 360 } \) . π(12)² = 326.72

In the diagram, RSTUVWXY is a reuIar octagon inscribed in ⊙C.

Question 7. Identify the center, a radius, an apothem, and a central angle of the polygon.

Answer: C is center, CY is radius, CZ is apothem, ∠YCR is central angle of the polygon

Question 8. Find m∠RCY, m∠RCZ, and m∠ZRC.

Answer: m∠RCY = 360/8 = 45 m∠RCZ = 45/2 = 22.5 m∠ZRC = 180 – (22.5 + 90) = 67.5

Question 9. The radius of the circle is 8 units. Find the area of the octagon.

Answer: Area of octagon = 0.5 x 8 x 8 sin 45 = 22.62

Answer: Area of yellow tile = \(\frac { 1 }{ 4 } \)(15.7 x 11.4) = 44.745 area of red tile = \(\frac { 1 }{ 4 } \)(18.5 x 6) = 27.75 Area of pattern = 32(44.745) + 23(27.75) = 2070.09 sq mm

Finding volume

Work with a partner: Consider a stack of square papers that is in the form of a right prism.

a. What is the volume the prism? Answer: The volume of the prism is B = 1/2 x h(b1 + b2) h = 8 b1 = 2 b2 = 2 = 1/2 x 8(2 + 2) = 1/2 x 8(4) = 1/2 x 32 = 16 cube inches. Therefore the volume of the prism is 16 cu. inches.

b. When you twist the slack of papers, as shown at the right, do you change the volume? Explain your reasoning. Answer: The volume of the prism and the twist of the slack of paper volume are the same. Because the different shapes of the prism have the same volume.

c. Write a carefully worded conjecture that describes the conclusion you reached in part (b). ATTENDING TO PRECISION To be proficient in math, you need to communicate precisely to others. Answer: The conjecture is that the different shapes of the prism have the same volume but are different in surface area.

d. Use your conjecture to find the volume of the twisted stack of papers. Answer: The volume of the twist of the slack of paper is B = 1/2 x h(b1 + b2) h = 8 b1 = 2 b2 = 2 = 1/2 x 8(2 + 2) = 1/2 x 8(4) = 1/2 x 32 = 16 cu. inches. Therefore the volume of the twist and the slack of the paper is 16 cu. inches. It is the same as the volume of the prism.

Work with a partner: Use the conjecture you wrote in Exploration I to find the volume of the cylinder.

Question 3. How can you find the volume of a prism or cylinder that is not a right prism or right cylinder? Answer: Using π we can find the volume of the prism or cylinder that is not a right prism of the right cylinder. The cylinder and the prism have the same cross-sectional area of πr². At every level and same height. Both the cylinder and prism have the same volume it is V = πr²h.

Question 4. In Exploration 1, would the conjecture you wrote change if the papers in each stack were not squares? Explain your reasoning. Answer:

Find the volume of the solid.

Answer: Area of circle = πr² = π(8)² = 64π Volume = 64π x 14 = 2814.86 cubic ft

Question 4. WHAT IF? In Example 4, you want the length to be 5 meters, the width to be 3 meters. and the volume to be 60 cubic meters. What should the height be? Answer: volume = lbh 60 = 5 x 3 x h h = 4 m

Question 5. WHAT IF? In Example 5, you want the height to be 5 meters and the volume to be 75 cubic meters. What should the area of the base be? Give a possible length and width. Answer: volume V = base x height 75 = base x 5 Base = 15 sq m

Question 2. COMPLETE THE SENTENCE Density is the amount of _______ that an object has in a given unit of __________ .

Answer: Density is the mass of the object divided by its volume.

In Exercises 3 – 6, find the volume of the prism.

In Exercises 7 – 10. find the volume of the cylinder.

In Exercises 11 and 12. make a sketch of the solid and find its volume. Round your answer to the nearest hundredth.

Question 12. A pentagonal prism has a height of 9 feet and each base edge is 3 feet.

Answer: volume is 139.32 ft³

explanation: Pentagon area = 15.48 Height h = 9 ft Volume V = area x height = 15.48 x 9 = 139.32

Answer: Density = mass / volume Density = \(\frac { 24 }{ 28.3 } \) Density = 0.8480

In Exercises 17 – 22, find the missing dimension of the prism or cylinder.

Answer: Volume = 2700 yd³ 12 x 5 x v = 2700 v = 15 yd

In Exercises 23 and 24, find the area of the base of the rectangular prism with the given volume and height. Then give a possible length and width.

Question 24. V = 27 m 3 ,h = 3m

Answer: V = Bh 27 = B x 3 B = 9

In Exercises 25 and 26, the solids are similar. Find the volume of solid B.

In Exercises 27 and 28, the solids are similar. Find the indicated measure.

Explanation: \(\frac { 7π }{ 5 } \) = \(\frac { 56π }{ h } \) h = 40

In Exercises 29 – 32. find the volume of the composite solid.

Explanation: Volume of square = 4³ = 64 Volume of semicircle = π(2)² x 4 = 8π

Answer: The volume of composite solid is 35 cubic ft

Explanation: Volume of larger prism = 4 x 2 x 5 = 40 Volume of the smaller prism = 1 x 1 x 5 = 5 Volume of larger prism – volume of the smaller prism = 40 – 5 = 35 cubic ft

Question 34. COMPARING METHODS The Volume Addition Postulate states that the volume of a solid is the sum of the volumes of all its non overlapping parts. Use this postulate to find the volume of the block of concrete in Example 7 by subtracting the volume of each hole from the volume of the large rectangular prism. Which method do you prefer? Explain your reasoning. Answer:

REASONING In Exercises 35 and 36, you are melting a rectangular block of wax to make candles. how many candles of the given shape can be made using a block that measures 10 centimeters by 9 centimeters by 20 centimeters?

Answer: 7 triangular prism candles with the given measures can be made.

Explanation: Volume of block = 1800 The volume of triangular prism = 4 x 6 x 10 = 240 1800/240 = 7.5

Question 38. PROBLEM SOLVING You drop an irregular piece of metal into a container partially filled with water and measure that the waler level rises 4.8 centimeters. The square base of the container has a side length of 8 centimeters. You measure the mass of the metal to be 450 grams. What is the density of the metal?

Answer: The density of metal is 1.4648

Explanation: Density = \(\frac { Mass }{ Volume } \) Volume V = 4.8 x 64 = 307.2 Density = \(\frac { 450 }{ 307.2 } \) = 1.4648

Answer: Volume = 2.5 x 3.5 x 6 Volume = 52.5

Answer: First one gives more cerel for your money.

Explanation: Bigger one volume = 16 x 4 x 10 = 640 Smaller one volume = 2 x 8 x 10 = 160 6 – 640 means 1 – 106.66 2 – 160 means 1 – 80

Question 46. CRITICAL THINKING The height of cylinder X is twice the height of cylinder Y. The radius of cylinder X is half the radius of cylinder Y. Compare the volumes of cylinder X and cylinder Y. Justify your answer.

Answer: Let the height of cylinder X be h, radius be r and its volume is πr²h So, the height of cylinder Y is h/2 and radius is 2r, then the volume is 2πr²h From both expressions, it can be seen that the volume of cylinder y is twice that of cylinder X.

Question 48. MATHEMATICAL CONNECTIONS You drill a circular hole of radius r through the base of a cylinder of radius R. Assume the hole is drilled completely through to the other base. You want the volume of the hole to be half the volume of the cylinder. Express r as a function of R.

Answer: r = √R²/2

Explanation: The radius of a solid cylinder without a hole is R. So its volume is πR²h As per the given condition, the volume of the hole must be half of that of the solid cylinder, hole volume is πR²h/2 Volume of cylinder V = πr²h πR²h/2 = πr²h R²/2 = r² r = √R²/2 r = \(\frac { R√2 }{ 2 } \)

Question 50. ANALYZING RELATIONSHIPS How can you change the edge length of a cube so that the volume is reduced by 40%?

Answer: Write the equation of volume of rectangular prism which can be used to evaluate the cube volume Volume = s x s x s The above equation shows that the volume of a cube is directly proportional to one of its side length, therefore, if the volume is to be reduced by 40%, then its the length of one of its side must be reduced by 40%, without changing the 2 other of its sides.

Find the surface area of the regular pyramid.

Finding the Volume of a Pyramid

Question 3. How can you find the volume of a pyramid? Answer: The volume of a pyramid is found using the formula V = (1/3) Bh, where ‘B’ is the base area and ‘h’ is the height of the pyramid. As we know the base of a pyramid is any polygon, we can apply the area of polygons formulas to find ‘B’.

Question 4. In Section 11 .7, you will study volumes of cones. How do you think you could use a method similar to the one presented in Exploration 1 to write a formula for the volume of a cone? Explain your reasoning. Answer:

Find the volume of the pyramid.

Explanation: Volume V = \(\frac { 1 }{ 3 } \)Bh V = \(\frac { 1 }{ 3 } \)(10 x 10 x 12) V = 400

Answer: The volume of the pyramid is 2494.13 cm³

Explanation: Volume V = \(\frac { 1 }{ 3 } \)Bh V = \(\frac { 1 }{ 3 } \)(374.12 x 20) V = 2494.13

Question 3. The volume of a square pyramid is 75 cubic meters and the height is 9 meters. Find the side length of the square base.

Answer: The side length of the square base is 5 m

Explanation: Volume V = \(\frac { 1 }{ 3 } \)Bh = 75 \(\frac { 1 }{ 3 } \)B(9) = 75 B = 25 s = 5

Answer: The height of the triangular pyramid is 8 m

Explanation: V = 24 \(\frac { 1 }{ 3 } \)Bh = 24 B = \(\frac { 1 }{ 2 } \)(3 x 6) = 9 \(\frac { 1 }{ 3 } \)(9)h = 24 h = 8

Answer: The volume of pyramid D is 12 m³

Explanation: \(\frac { volume of pyramid C }{ volume of pyramid D } \) = (\(\frac { pyramid C base }{ pyramid D base } \))³ \(\frac { 324 }{ V } \) = (\(\frac { 9 }{ 3 } \))³ V = 12

Answer: the volume of solid = 96

Explanation: Volume of prism = Bh B = 8 x 2 = 16 V = 16 x 5 = 80 \(\frac { 1 }{ 3 } \)Bh = \(\frac { 1 }{ 3 } \)(16 x 3) = 16 the volume of solid = 16 + 80 = 96

Question 2. REASONING A square pyramid and a cube have the same base and height. Compare the volume of the square pyramid to the volume of the cube.

Answer: Square pyramid = 1/3 Bh Cube = BH So, the volume of the square pyramid is 1/3 of the volume of the cube.

In Exercises 3 and 4, find the volume of the pyramid.

Answer: V = 6 in³

Explanation: V =  \(\frac { 1 }{ 3 } \)Bh B = 2 x 3 = 6 V = \(\frac { 1 }{ 3 } \)(6 x 3)

In Exercises 5 – 8, find the indicated measure.

Question 6. A pyramid with a square base has a volume of 912 cubic feet and a height of 19 feet. Find the side length of the square base.

Answer: The side length of the square base is 12 ft

Explanation: A pyramid with a square base has a volume of 912 cubic feet h = 19 \(\frac { 1 }{ 3 } \)Bh = 912 \(\frac { 1 }{ 3 } \)B(19) = 912 B = 144 s = 12

Question 8. A pyramid with a rectangular base has a volume of 105 cubic centimeters and a height of 15 centimeters. The length of the rectangular base is 7 centimeters. Find the width of the rectangular base.

Answer: The width of the rectangular base is 3 cm

Explanation: A pyramid with a rectangular base has a volume of 105 cubic centimeters h = 15 l = 7 \(\frac { 1 }{ 3 } \)Bh = 105 \(\frac { 1 }{ 3 } \)lbh = 105 \(\frac { 1 }{ 3 } \)(7 x 15 x b) = 105 b = 3

Question 10. OPEN-ENDED Give an example of a pyramid and a prism that have the same base and the same volume. Explain your reasoning. Answer: Let the rectangular prism have the base dimensions 4 x 2 nad a height of 5 so its volume is 4 x 2 x 5 = 40 cubic units Therefore the base of the rectangular prism also have the dimensions of 4 x 2 and a height of 5 x 3 = 15 units so its volume V = 1/3 x 4 x 2 x 15 = 40 cubic units

In Exercises 11 – 14, find the height of the pyramid.

Answer: The height of the pyramid is 10.5 in

Explanation: Volume = 224 \(\frac { 1 }{ 3 } \)Bh = 224 B = 8² = 64 \(\frac { 1 }{ 3 } \)(64)h = 224 h = 10.5

Answer: The height of the pyramid is 12 cm

Explanation: Volume = 392 \(\frac { 1 }{ 3 } \)Bh = 392 B = 14 x 7 = 98 \(\frac { 1 }{ 3 } \)(98)h = 392 h = 12

In Exercises 15 and 16, the pyramids are similar. Find the volume of pyramid B.

Answer: Volume of A = 80

Explanation: \(\frac { Volume of B }{ Volume of A } \) = (\(\frac { Side of B }{ side of A } \))³ \(\frac { V }{ 10 } \) = (\(\frac { 6 }{ 3 } \))³ V = 8 x 10 Volume of A = 80

In Exercises 17 – 20, find the volume of the composite solid.

Answer: Composite solid volume = 306

Explanation: Base area = \(\frac { 1 }{ 2 } \)bh = \(\frac { 1 }{ 2 } \)(12 x 9) = 54 Bottom solid volume V = \(\frac { 1 }{ 3 } \)Bh = \(\frac { 1 }{ 3 } \)(54 x 10) V = 180 Top solid volume v = \(\frac { 1 }{ 3 } \)(54 x 7) = 126 Composite solid volume = 180 + 126 = 306

Answer: Composite solid volume = 1152

Explanation: Volume of Box = 12 x 12 x 12 = 1728 Square pyramid volume = \(\frac { 1 }{ 3 } \)Bh = \(\frac { 1 }{ 3 } \)(144 x 12) = 576 Composite solid volume = 1728 – 576 = 1152

Question 21. ABSTRACT REASONING A pyramid has a height of 8 feet and a square base with a side length of 6 feet.

Find the value of X. Round your answer to the nearest tenth.

Finding the Surface Area of a Cone

b. What is the area of the original circle? What is the area with one sector missing? Answer:

c. Describe the surface area of the cone, including the base. Use your description to find the surface area. Answer:

Finding the Volume of a Cone

Question 3. How can you find the surface area and the volume of a cone? Answer:

Question 4. In Exploration 1, cut another sector from the circle and make a cone. Find the radius of the base and the surface area of the cone. Repeat this three times, recording your results in a table. Describe the pattern. Answer:

Answer: The surface area of the right cone is 436.17 m²

Explanation: r = 7.8 l = 10 S = πr² + πrl S = π(7.8)² + π(7.8 x 10) S = 436.17

Find the volume of the cone.

Answer: The volume of the cone is 2206.44 in³

Explanation: r = 7, h = 13 l = √13² – 7²= 10.95 S = πr² + πrl S = π7² + π(7 x 10.95) S = 394.74 Volume V = \(\frac { 1 }{ 3 } \)(πr²h) V = \(\frac { 1 }{ 3 } \)(π x 7² x 13) V = 2206.44

Answer: The volume of the cone is 163.4 m³

Explanation: h = √8² – 5² = 6.24 Volume V = \(\frac { 1 }{ 3 } \)(πr²h) V = \(\frac { 1 }{ 3 } \)(π x 5² x 6.24) V = 163.4

Answer: Volume of cone D = 18.84 cm³

Explanation: \(\frac { Volumeof cone C }{ Volume of cone D } \) = (\(\frac { height of C }{ height of D } \))³ \(\frac { 384π }{ Volume of cone D } \)= (\(\frac { 8 }{ 2 } \))³ Volume of cone D = 18.84

Answer: Composite solid volume = 329.86 cm³

Explanation: Volume of cylinder = πr²h = π(3)² x 10 = 90π Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h) = \(\frac { 1 }{ 3 } \)(π x 3² x 5) = 15π Composite solid volume = 15π + 90π = 105π

Question 2. COMPLETE THE SENTENCE The volume of a cone with radius r and height h is \(\frac{1}{3}\) the volume of a(n) __________ with radius r and height h. Answer:

In Exercises 3 – 6, find the surface area of the right cone.

Answer: The surface area of cone is 219.44 sq cm.

Explanation: S = πr² + πrl S = π(5.5)² + π(5.5 x 7.2) S = 219.44

Question 6. A right cone has a diameter of 11.2 feet and a height of 19.2 feet.

Answer: The surface area is 421.52 sq ft.

Explanation: r = 5.6 h = 19.2 l = √19.2² – 5.6² = 18.36 Surface area S = πr² + πrl S = π(5.6)² + π(5.6 x 18.36) S = 421.52

In Exercises 7 – 10, find the volume of the cone.

Answer: The volume is 2.09 cubic meter

Explanation: Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h) V = \(\frac { 1 }{ 3 } \)(π(1)² x 2) V = 2.09

Question 10. A right cone has a radius of 3 feet and a slant height of 6 feet.

Answer: The volume is 56.54 cubic ft

Explanation: Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h) V = \(\frac { 1 }{ 3 } \)(π x 3² x 6) V = 56.54

In Exercises 11 and 12, find the missing dimension(s).

Question 12. Volume = 216π in. 3

Answer: The radius is 6.13 in

Explanation: Volume = 216π in. 3 \(\frac { 1 }{ 3 } \)(πr²h) = 216 \(\frac { 1 }{ 3 } \)(πr² x 18) = 216 r = 6.13

In Exercises 13 and 14, the cones are similar. Find the volume of cone B.

Answer: Volume of cone B = 24.127

Explanation: \(\frac { Volumeof cone A }{ Volume of cone B } \) = (\(\frac { height of A }{ height of B } \))³ \(\frac { 120π }{ Volume of cone B } \) = (\(\frac { 10 }{ 4 } \))³ Volume of cone B = 24.127

In Exercises 15 and 16, find the volume of the composite solid.

Answer: Volume of the composite solid = 97.93 cubic m

Explanation: Volume of box = lbh V = 51 x 5.1 x 5.1 = 132.651 Cone volume = \(\frac { 1 }{ 3 } \)(πr²h) v = \(\frac { 1 }{ 3 } \)(π x 2.55² x 5.1) v = 34.72 Volume of the composite solid = 132.651 – 34.72 = 97.93

Answer: Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h) V = \(\frac { 1 }{ 3 } \)(π x 3² x 8) = 75.39 Volume of cylinder = πr²h = π x 3² x 8 = 226.19 Volume of cylinder / Volume of cone = \(\frac { 226.19 }{ 75.39 } \) = 3 You have to buy 3 small containers of popcorn to equal the amount of popcorn in a large container.

b. Which container gives you more popcorn for your money? Explain. Answer: $1.25 -> 75.39 i.e $1 = 60.312 $2.50 -> 226.19 i.e $1 = 90.47 So, large containers gives you more popcorn for your money

In Exercises 19 and 20. find the volume of the right cone.

Answer: Volume of cone is 575.62 cubic yd

Explanation: tan 32 = \(\frac { 7 }{ h } \) h = 11.21 Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h) V = \(\frac { 1 }{ 3 } \)(π x 7² x 11.21) V = 575.62

Question 22. MODELING WITH MATHEMATICS During a chemistry lab, you use a funnel to pour a solvent into a flask. The radius of the funnel is 5 centimeters and its height is 10 centimeters. You pour the solvent into the funnel at a rate of 80 milliliters per second and the solvent flows out of the funnel at a rate of 65 milliliters per second. How long will it be before the funnel overflows? (1 mL = 1 cm 3 ) Answer: 17.45 seconds

Explanation: Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h) V = \(\frac { 1 }{ 3 } \)(π x 5² x 10) V = 261.8 \(\frac { 261.8 }{ 15 } \) = 17.45

Question 28. area of a circle with a diameter of 22 centimeters

Answer: d = 11 A = πr² A = 121π

Question 30. radius of a circle with an area of 529 π square inches Answer: A = πr² 529π = πr² r = 23

Finding the Surface Area of a Sphere

Finding the volume of a sphere

Question 3. How can you find the surface area and the volume of a sphere? Answer:

Question 4. Use the results of Explorations 1 and 2 to find the surface area and the volume of a sphere with a radius of(a) 3 inches and (b) 2 centimeters. Answer:

Find the surface area of the sphere.

Answer: The surface area of the sphere is 5026.54 ft²

Explanation: D = 40 r = 20 The surface area of the sphere = 4πr² S = 4 x π x (20)² S = 5026.54 ft²

Answer: The surface area of the sphere is 113.09 ft²

Explanation: Circumference C = 6π 2πr = 6π r = 3 The surface area of the sphere = 4πr² S = 4π x 3² S = 113.09

Answer: The radius of the sphere is 2.73 m

Explanation: The surface area of the sphere = 4πr² 30π = 4πr² r = 2.73

Question 4. The radius of a sphere is 5 yards. Find the volume of the sphere.

Answer: The volume of the sphere is 523.59 yards³

Explanation: r = 5 The volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³ V = \(\frac { 4 }{ 3 } \)π x 5³ V = 523.59 yards³

Question 5. The diameter of a sphere is 36 inches. Find the volume of the sphere.

Answer: The volume of the sphere is 24429.02 in³

Explanation: D = 36 r = 18 The volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³ V = \(\frac { 4 }{ 3 } \)π x 18³ V = 24429.02

Question 6. The surface area of a sphere is 576π square centimeters. Find the volume of the sphere.

Answer: The volume of the sphere is 2304π cm³

Explanation: The surface area of the sphere = 4πr² 576π = 4πr² r = 12 The volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³ V = \(\frac { 4 }{ 3 } \)π x 12³ V = 2304π

Answer: The volume of the composite solid is 7.324 m³

Explanation: The volume of cone = πr²\(\frac { h }{ 3 } \) = π x 1² x \(\frac { 5 }{ 3 } \) = 5.23 The volume of sphere = \(\frac { 4 }{ 3 } \)πr³ = \(\frac { 4 }{ 3 } \)π x 1³ = 4.188 The volume of the composite solid = The volume of cone + The volume of sphere/2 = 5.23 + 4.188/2 = 7.324 m³

Question 2. WRITING Explain the difference between a sphere and a hemisphere.

Answer: Hemisphere is a related term of the sphere. Sphere and hemisphere are three-dimensional solids. The volume of sphere is \(\frac { 4 }{ 3 } \)πr³ and hemisphere volume is \(\frac { 2 }{ 3 } \)πr³. The surface area of the sphere is 4πr² and hemisphere surface area is 3πr².

Monitoring progress and Modeling with Mathematics

In Exercises 3 – 6, find the surface area of the sphere.

Answer: The surface area of the sphere is 225π cm²

Explanation: The surface area of the sphere = 4πr² S = 4π x 7.5² S = 225π

Answer: The surface area of the sphere is 8π ft²

Explanation: C = 4π 2πr = 4π r = 2 The surface area of the sphere = 4πr² S = 4π x 2² S = 8π

In Exercises 7 – 10. find the indicated measure.

Question 8. Find the radius of a sphere with a surface area of 1024π square inches.

Answer: The radius of a sphere is 16 in

Explanation: The surface area of the sphere = 1024π 4πr² = 1024π r = 16

Question 10. Find the diameter of a sphere with a surface area of 196π square centimeters.

Answer: The diameter of a sphere is 14 cm

Explanation: The surface area of the sphere = 196π 4πr² = 196π r = 7 D = 2(7) = 14

In Exercises 11 and 12, find the surface area of the hemisphere.

Answer: The surface area of the hemisphere is 108π in²

Explanation: D = 12, r = 6 The surface area of the sphere = 3πr² S = 3π x 6² S = 108π

In Exercises 13 – 18. find the volume of the sphere.

Answer: The volume of the sphere is 268.08 ft³

Explanation: r = 4 ft Volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³ V = \(\frac { 4 }{ 3 } \)π x 4³ V = 268.08 ft

Answer: The volume of the sphere is 1436.75 ft³

Explanation: D = 14 ft r = 7 ft Volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³ V = \(\frac { 4 }{ 3 } \)π x 7³ V = 1436.75 ft

Answer: The volume of the sphere is 179.89 in³

Explanation: C = 7π 2πr = 7π r = 3.5 Volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³ V = \(\frac { 4 }{ 3 } \)π x 3.5³ V = 179.89 in

In Exercises 19 and 20, find the volume of the sphere with the given surface area.

Question 20. Surface area = 484π cm 2

Answer: The volume of the sphere is 5575.27 cm³

Explanation: Surface area = 484π 4πr² = 484π r = 11 Volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³ V = \(\frac { 4 }{ 3 } \)π x 11³ V = 5575.27

Answer: Diameter = 3 radius = 1.5 Volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³ V = \(\frac { 4 }{ 3 } \)π x (1.5)³ V = 14.137 cubic in

In Exercises 23 – 26, find the volume of the composite solid.

Answer: Volume is 288π ft³

Explanation: Volume of hemipshere = \(\frac { 2 }{ 3 } \)πr³ = \(\frac { 2 }{ 3 } \)π x 6³ = 144π volume of the cone = πr²\(\frac { h }{ 3 } \) = π x 6² x \(\frac { 12 }{ 3 } \) = 144π Area of circle = πr² = π x 6² = 36π Volume of hemipshere + volume of the cone = 144π + 144π = 288π

Answer: The volume of solid is 296π m³

Explanation: Volume of hemipshere = \(\frac { 2 }{ 3 } \)πr³ = \(\frac { 2 }{ 3 } \)π x 6³ = 144π Volume of cylinder = πr²h = π x 6² x 14 = 504π Volume of solid = 504π – 2(144π) = 296π

In Exercises 27 – 32, find the surface area and volume of the ball.

Answer: The surface area is 277 in², volume is 43212.27 in³

Explanation: C = 29.5 2πr = 29.5 r = 4.69 Surface area = 4πr² S = 4π x 4.69² = 277 Volume V = \(\frac { 4 }{ 3 } \)πr³ V = \(\frac { 4 }{ 3 } \)π x 4.69³ V = 43212.27

Answer: The surface area is 9.07 in², volume is 2.57 in³

Explanation: d = 1.7 r = 0.85 Surface area = 4πr² S = 4π x 0.85² = 9.07 Volume V = \(\frac { 4 }{ 3 } \)πr³ V = \(\frac { 4 }{ 3 } \)π x 0.85³ V = 2.57

Answer: The surface area is 25.78 in², volume is 12.24 in³

Explanation: C = 9 2πr = 9 r = 1.43 Surface area = 4πr² S = 4π x 1.43² S = 25.78 Volume V = \(\frac { 4 }{ 3 } \)πr³ V = \(\frac { 4 }{ 3 } \)π x 1.43³ V = 12.24

Question 34. REASONING A semicircle with a diameter of 18 inches is rotated about its diameter. Find the surface area and the volume of the solid formed.

Answer: The surface area is 1018 in², volume is 3054.02 in³

Explanation: Diameter = 18 radius r = 9 Volume V = \(\frac { 4 }{ 3 } \)πr³ V = \(\frac { 4 }{ 3 } \)π x 9³ V = 3054.02 Surface area = 4πr² S = 4π x 9² S = 1018

Answer: C = 8 in 2πr = 8 r = 1.27 Volume V = \(\frac { 4 }{ 3 } \)πr³ V = \(\frac { 4 }{ 3 } \)π x 1.27³ = 8.64 The volume of tennis ball = 8.64 in³

b. Find the amount of space within the cylinder not taken up by the tennis balls. Answer: The surface area of tennis ball S = 4πr² S = 4π x 1.27² = 20.26 Area of cylinder s = 2πrh+2πr² s = 2π x 1.43 x 8+2π x 1.43² s = 84.72 Remaining space = 84.72 – 20.26 = 64.46 in²

Question 38. MATHEMATICAL CONNECTIONS A sphere has a diameter of 4(x + 3) centimeters and a surface area of 784 π square centimeters. Find the value of x.

Answer: x =11

Explanation: Surface area = 4πr² 784π = πr² r = 28 2r = diameter = 4(x + 3) r = 2(x + 3) 28 = 2(x + 3) x = 11

b. A meteorite is equally likely to hit anywhere on Earth. Estimate the probability that a meteorite will land in the Torrid Zone. Answer: Probability of meteorites hitting the torrid zone = 80875080/197086348.8 = 0.4104

Explanation: Volume of hemisphere v = \(\frac { 2 }{ 3 } \)πr³ Volume of cone V = \(\frac { 1 }{ 3 } \)πr²h If r = h Volume of cone V = \(\frac { 1 }{ 3 } \)πr² x r = \(\frac { 1 }{ 3 } \)πr³ So, the hemisphere has the highest volume.

b. r = 34 cm, a = 30 cm Answer: The formula for the volume of the spherical cap is V = πh/6 x (3a² +h²). Where a is the radius h is the height of the cap. Where a = r = 34cm h = 30cm V = π(30)/6 x (3(34)² +(30)²) = π(30)/6 x (3(1156) + (900)) = π(30)/6 x 6,168 = 94.2/6 x 6,168 Therefore the volume of the spherical cap is 96,837.6 cu. cm.

c. r = 13 m, h = 8 m Answer: The formula for the volume of the spherical cap is V = πh/6 x (3a² +h²). Where a is the radius h is the height of the cap. Where a = r = 13cm h = 8cm V = π(8)/6 x (3(13)² +(8)²) = 8π/6 x (3(169 + 64)) = 8π/6 x (699) = 25.12/6 x 699 Therefore the volume of the spherical cap is 2,926.48 cu. cm

d. r=75 in., h = 54in. Answer: The formula for the volume of the spherical cap is V = πh/6 x (3a² +h²). Where a is the radius h is the height of the cap. Where r = 75in h = 54in V = π(54)/6 x (3(75)² +(54)²) = 54π/6 x (3(5,625 + 2,916) = 54π/6 x (25,623) = 169.56π/6 x 25,623 = 532.4184 x 25,623 Therefore the volume of the spherical cap is 13,642,156.66 cu. in.

Solve the triangle. Round decimal answers to the nearest tenth.

Question 48. A = 26°, C = 35°, b = 13

Answer: B = 119°, a = 7.16, c = 9.5

Explanation: B = 180 – (26 + 35) = 119 \(\frac { sin A }{ a } \) = \(\frac { sin B }{ b } \) \(\frac { sin 26 }{ a } \) = \(\frac { sin 119 }{ 13 } \) a = 7.16 \(\frac { sin C }{ c } \) = \(\frac { sin B }{ b } \) \(\frac { sin 35 }{ c } \) = \(\frac { sin 119 }{ 13 } \) c = 9.5

Question 50. a = 23, b = 24, c = 20

Answer: A = 62.2, B = 65.5, C = 49.4

Explanation: a² = b² + c² – 2bc cos A 23² = 24²+ 20² – 2(24 x 20) cos A A = 62.2 \(\frac { sin 62.2 }{ 23 } \) = \(\frac { sin B }{ 24 } \) B = 65.5 \(\frac { sin 62.2 }{ 23 } \) = \(\frac { sin C }{ 20 } \) C = 49.4

Answer: diameter of ⊙P is 29.99

Explanation: Circumference = 94.24 πd = 94.24 d = 29.99

Answer: circumference of ⊙F = 56.57

Explanation: 5.5 = \(\frac { 35 }{ 360 } \) . C C = 56.57

Answer: arc length of \(\widehat{A B}\) = 26.09

Explanation: arc length of \(\widehat{A B}\) = \(\frac { 115 }{ 360 } \) . 2π(13) = 26.09

Question 4. A mountain bike tire has a diameter of 26 inches. To the nearest foot, how far does the tire travel when it makes 32 revolutions?

Answer: The tire travels 2613.80 inches.

Explanation: D = 26 in r = 13 in Circumference C = 2π(13) = 81.68 32 revolutions = 32 x 81.68 = 2613.80

Find the area of the blue shaded region.

Answer: Area = \(\frac { 240 }{ 360 } \) . π(9)² = 169.64

Answer: Area of shaded region = 11.43

Explanation: Area of rectangle = 6 x 4 = 24 Area of semicircle = π(2)² = 4π Area of shaded region = 24 – 4π = 11.43

Answer: Area of shaded region = 173.13

Explanation: Area of small region = 27.93 = \(\frac { 50 }{ 360 } \) . πr² πr² = 201.096 r = 8 Area of shaded region = \(\frac { 310 }{ 360 } \) . π(8)² = 173.13

Find the area of the kite or rhombus.

Explanation: Area = \(\frac { 1 }{ 4 } \)(d₁d₂) A = \(\frac { 1 }{ 4 } \)(13 x 20) A = 65

Explanation: Area = \(\frac { 1 }{ 4 } \)(d₁d₂) A = \(\frac { 1 }{ 4 } \)(16 x 12) A = 48

Explanation: Area = \(\frac { 1 }{ 4 } \)(d₁d₂) A = \(\frac { 1 }{ 4 } \)(14 x 15) A = 52.5

Find the area of the regular polygon.

Question 14. A platter is in the shape of a regular octagon with an apothem of 6 inches. Find the area of the platter.

Answer: Area = \(\frac { 1 }{ 2 } \)(n . a. s) A = \(\frac { 1 }{ 2 } \)(8 . 6 . sin 45) A = 16.97

Describe the cross section formed by the intersection of the plane and the solid.

Answer: Volume = lbh V = 3.6 x 2.1 x 1.5 = 113.4 m³

Answer: Volume = πr²h V = π(2)² x 8 = 100.53 mm³

Answer: Pentagon area = 6.88 Volume = Area x height V = 6.88 x 4 = 27.52 yd³

Answer: Volume V = Base area x height/3 Base Area = 9² = 81 V = 81 x 7/3 = 189 ft³

Answer: Volume V = Base area x height/3 base area = 18 x 10 = 180 V = 180 x 5/3 = 300 m³

Question 27. The volume of a square pyramid is 60 cubic inches and the height is 15 inches. Find the side length of the square base.

Answer: The side length of the square base is 3.46 in

Explanation: The volume of a square pyramid is 60 cubic inches V = 60 s²h/3 = 60 s² x 15/3 = 60 s² = 12 s = 3.46

Question 28. The volume of a square pyramid is 1024 cubic inches. The base has a side length of 16 inches. Find the height of the pyramid Answer: The volume of a square pyramid is 1024 cubic inches s²h/3 = 1024 16²h = 3072 h = 12

Find the surface area and the volume of the cone.

Explanation: Surface area of cone S = πr² + πrl S = π x 9² + π x 9 x 15 S = 678.58 Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h) V = \(\frac { 1 }{ 3 } \)(π x 9² x 12) V = 1017.87

Answer: Surface area is 2513.27 cm² volume is 8042.47 cm³

Explanation: Surface area of cone S = πr² + πrl S = π x 16² + π x 16 x 34 S = 2513.27 Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h) V = \(\frac { 1 }{ 3 } \)(π x 16² x 30) V = 8042.47

Answer: Surface area is 439.82 m² volume is 562.102 m³

Explanation: Surface area of cone S = πr² + πrl S = π x 7² + π x 7 x 13 S = 439.82 h = √13² – 7² = 10.95 Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h) V = \(\frac { 1 }{ 3 } \)(π x 7² x 10.95) V = 562.102

Question 32. A cone with a diameter of 16 centimeters has a volume of 320π cubic centimeters. Find the height of the cone.

Answer: The height of the cone = 15 cm.

Explanation: r = 8 Volume V = 320π \(\frac { 1 }{ 3 } \)(πr²h) = 320π \(\frac { 1 }{ 3 } \)(π x 8² x h) = 320π h = 15

Find the surface area and the volume of the sphere.

Answer: The surface area is 615.75 in², volume is 1436.75 in³

Explanation: Surface area S = 4πr² S = 4π x 7² S = 615.75 Volume V = \(\frac { 4 }{ 3 } \)πr³ V = \(\frac { 4 }{ 3 } \)π x 7³ V = 1436.75

Answer: The surface area is 907.92 ft², volume is 2572.44 ft³

Explanation: d = 17 r = 8.5 Surface area S = 4πr² S = 4π x 8.5² S = 907.92 Volume V = \(\frac { 4 }{ 3 } \)πr³ V = \(\frac { 4 }{ 3 } \)π x 7³ V = 2572.44

Answer: The surface area is 2827.43 ft², volume is 14137.16 ft³

Explanation: C = 30π 2πr = 30π r = 15 Surface area S = 4πr² S = 4π x 15² S = 2827.43 Volume V = \(\frac { 4 }{ 3 } \)πr³ V = \(\frac { 4 }{ 3 } \)π x 15³ V = 14137.16

Question 36. The shape of Mercury can be approximated by a sphere with a diameter of 4880 kilometers. Find the surface area and the volume of Mercury.

Answer: The surface area and the volume of Mercury is 23814400π, 19369045330π

Explanation: d = 4880 r = 2440 Surface area S = 4πr² S = 4π x 2440² = 23814400π Volume V = \(\frac { 4 }{ 3 } \)πr³ V = \(\frac { 4 }{ 3 } \)π x 2440³ V = 19369045330π

Question 37. A solid is composed of a cube with a side length of 6 meters and a hemisphere with a diameter of 6 meters. Find the volume of the composite solid.

Answer: Volume of the composite solid = 272.52

Explanation: Volume of cube = a³ = 6³ = 216 Volume of hemisphere = \(\frac { 4 }{ 6 } \)πr³ = \(\frac { 4 }{ 6 } \)π x 3³ = 18π Volume of the composite solid = 216 + 18π = 272.52

Answer: Volume = 2577.29 m³

Explanation: Volume = \(\frac { 3√3 }{ 2 } \)a²h = \(\frac { 3√3 }{ 2 } \) x 8² x 15.5 = 2577.29

Answer: Volume is 17.157 ft³

Explanation: d = 3.2 r = 1.6 Volume V = \(\frac { 4 }{ 3 } \)πr³ V = \(\frac { 4 }{ 3 } \)π x 1.6³ V = 17.157

Answer: Volume of sloid = 402.11 m³

Explanation: Volume of cone = \(\frac { 1 }{ 3 } \)πr²h = \(\frac { 1 }{ 3 } \)π x 4² x 3 = 50.26 Volume of cylinder = πr²h = π x 4² x 6 = 301.59 Volume of sloid = 2(50.26) + 301.59 = 402.11

Answer: Volume of solid = 106.66

Explanation: Volume of rectangular box = 5 x 2 x 8 = 80 Volume of pyramid = 80/3 = 26.66 Volume of solid = 80 + 26.66 = 106.66

Answer: circumference of ⊙F is 109.71 in

Explanation: 64 = \(\frac { 210 }{ 360 } \) • C C = 109.7

Answer: m\(\widehat{G H}\) = 74.27

Explanation: 35 = \(\frac { x }{ 360 } \) • 2π x 27 x = 74.27

Answer: Area is 142.41 in²

Explanation: Area = \(\frac { 360 – 105 }{ 360 } \) • π x 8² Area = 142.41

Question 9. Find the surface area of a right cone with a diameter of 10 feet and a height of 12 feet.

Answer: The surface area is 486.7 sq ft

Explanation: l² = r² + h² l² = 5² + 12² l = 13 Surface area S = πr² + 2πrl S = π x 5² + 2π x 5 x 13 S = 486.7

Answer: Volume = \(\frac { 1 }{ 3 } \)πr²h = \(\frac { 1 }{ 3 } \)π x 6² x 10 = 376.99

b. You use the funnel to put oil in a ear. Oil flows out of the funnel at a rate of 45 milliliters per second. How long will it take to empty the funnel when it is full of oil? (1 mL = 1 cm 3 ) Answer: T = 376.8 ml/45 ml per sec T = 8.373 sec

c. How long would it take to empty a funnel with a radius of 10 centimeters and a height of 6 centimeters if oil flows out of the funnel at a rate of 45 milliliters per second? Answer: V = 1/3 πr²h = 1/3 (3.14 × 10² × 6) = 1/3(1884) = 628 cu. cm T = 628 ml/45 ml per sec T = 13.95 sec

d. Explain why you can claim that the time calculated in part (c) is greater than the time calculated in part (b) without doing any calculations. Answer:

• In cone type shaped object if the radius is large then the volume of the cone increases.
• In the b part of the funnel, the radius is smaller than the height and in the c part, the radius is larger than the height of the cone.
• V = 1/3 πr²h
• It means if radius increase volume also increases with the same rate. So, the time is taken by the c part (13.95 sec) is larger than the b part.

Question 11. A water bottle in the shape of a cylinder has a volume of 500 cubic centimeters. The diameter of a base is 7.5 centimeters. What is the height of the bottle? Justify your answer.

Answer: The height of the bottle is 11.3 cm

Explanation: Volume of cylinder = 500 πr²h = 500 π(3.75)²h = 500 h = 11.3 cm

Question 12. Find the area of a dodecagon (12 sides) with a side length of 9 inches.

Answer: Area is 237.31

Explanation: Area = \(\frac { 1 }{ 4 } \)πa²cot(π/n) = \(\frac { 1 }{ 4 } \)π x 9² x cot(π/12) = 237.31

Answer: Area of sector for fan of radius 9 cm = 120/360 × 3.14 × 9 × 9 = 1/3 × 254.57 = 84.85 sq. cm Area of sector of fan of radius 6 cm = 150/360 × 3.14 × 6 × 6 = 5/12 × 113.14 = 47.14 sq. cm

Question 1. Identify the shape of the cross section formed by the intersection of the plane and the solid.

b. Find the amount of space within the crayon box not taken up by the crayons. Answer: Volume of box = 94 x 28 x 71 = 186872 The volume of a crayon = 4650.21 Remaining space = 186872 – 24 x 4650.21 = 75266.96

Question 4. What is the equation ol the line passing through the point (2, 5) that is parallel to the line x + \(\frac{1}{2}\)y = – 1? (A) y = – 2x + 9 (B) y = 2x + 1 (C) y = \(\frac{1}{2}\)x + 4 (D) y = –\(\frac{1}{2}\)x + 6 Answer: (A) y = – 2x + 9

Explanation: x + \(\frac{1}{2}\)y = – 1 y = -2 – 2x The slope of the line is -2 The euation of line is y – 5 = -2(x – 2) y – 5 = -2x + 4 y = -2x + 9

Answer: (A) 22,019.63 ft 3

Explanation: Volume = a²\(\frac { h }{ 3 } \) = 34.5² x \(\frac { 55.5 }{ 3 } \) = 22019.62

Question 6. Prove or disprove that the point (1, √3 ) lies on the circle centered at the origin and containing the point (0, 2). Answer: We consider the circle centered at the origin and containing the point (0, 2). Therefore, we canconclude that rdaius is 2 and points be (0, 0), (1, √3) distance = √(1 – 0)² + (√3 – 0)² = 2 As radius and distance are same. The point B(1, √3) lies on the circle.

Explanation: Volume of square pyramid = a²\(\frac { h }{ 3 } \) Square diagonal = √2a radius = √2a/2 a = 2r/√2 Volume of cone = \(\frac { 1 }{ 3 } \)πr²h

Answer: The number of people per square mile is 247

Explanation: S = πr² = π x 5² = 78.5 Number of people per square mile = 19400/78.5 = 247

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Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume

Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume contains 6th Standard Go Math solutions which will make students understand the concepts easily help the students to score well in the exams. This Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume. And in this, each and every question was explained intimately . The answers in this chapter are explained in a simple way that anyone can understand easily.

This chapter 11 contains Three-Dimensional Figures and Nets, Explore Surface Area Using Nets, Surface Area of Prisms, etc. are explained clearly which makes the scholars learn quickly. Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume. questions are explained in a basic way that students will never feel any difficulty in learning. By this, students can gain good knowledge and this is helpful in finish student’s assignments also.

Lesson 1: Three-Dimensional Figures and Nets

Share and Show – Page No. 599

Problem solving + applications – page no. 600, three-dimensional figures and nets – page no. 601, lesson check – page no. 602.

Lesson 2: Investigate • Explore Surface Area Using Nets

Share and Show – Page No. 605

What’s the error – page no. 606, explore surface area using nets – page no. 607, lesson check – page no. 608.

Lesson 3: Algebra • Surface Area of Prisms

Share and Show – Page No. 611

Unlock the problem – page no. 612, surface area of prisms – page no. 613, lesson check – page no. 614.

Lesson 4: Algebra • Surface Area of Pyramids

Share and Show – Page No. 617

Problem solving + applications – page no. 618, surface area of pyramids – page no. 619, lesson check – page no. 620.

 Mid-Chapter Checkpoint

Mid-Chapter Checkpoint – Vocabulary – Page No. 621

Page no. 622.

Lesson 5: Investigate • Fractions and Volume

Share and Show – Page No. 625

Problem solving + applications – page no. 626, fractions and volume – page no. 627, lesson check – page no. 628.

Lesson 6: Algebra • Volume of Rectangular Prisms

Share and Show – Page No. 631

Aquariums – page no. 632, volume of rectangular prisms – page no. 633, lesson check – page no. 634.

Lesson 7: Problem Solving • Geometric Measurements

Share and Show – Page No. 637

On your own – page no. 638, problem solving geometric measurements – page no. 639, lesson check – page no. 640.

Chapter 11 Review/Test

Chapter 11 Review/Test – Page No. 641

Page no. 642, page no. 643, page no. 644, page no. 645, page no. 646.

Identify and draw a net for the solid figure.

Answer: The base Square or Rectangle, and lateral faces are Triangle and the figure is a Square pyramid or Rectangular pyramid.

Explanation: The base is a square or rectangle and lateral faces are squares are rectangle. The figure is a Cube or Rectangular prism.

Identify and sketch the solid figure that could be formed by the net.

Answer: Triangular pyramid.

Explanation: The net has four triangles, so it is a triangular pyramid.

On Your Own

Answer: Triangular prism.

Explanation: The base is a rectangle and the lateral faces are triangles and rectangles, so it is a triangular prism.

Answer:  Rectangular Prism.

Explanation: The base is a rectangle and the lateral faces are squares and rectangles. And it is a Rectangular prism.

Explanation: The shape of the galena is Cube.

Answer: No, she cannot fold her net into a cube. Rhianon’s net has seven squares but there are only six squares in the net of a cube.

Answer: We can see that Diamond crystal consists of two square pyramids with congruent bases and the pyramids are reversed and placed base to base.

Explore Surface Area Using Nets Question 10. Sasha makes a triangular prism from paper. The bases are _____. The lateral faces are _____.

Answer: The bases are Triangle The lateral faces are Rectangle

Answer: Rectangular Prism

Answer: Cube, Rectangular prism

Answer: Square Pyramid

Answer: Triangular Prism

Problem Solving

Question 5. Hobie’s Candies are sold in triangular pyramid-shaped boxes. How many triangles are needed to make one box?

Question 6. Nina used plastic rectangles to make 6 rectangular prisms. How many rectangles did she use?

Question 7. Describe how you could draw more than one net to represent the same three-dimensional figure. Give examples.

Explanation:

Question 1. How many vertices does a square pyramid have?

Question 2. Each box of Fred’s Fudge is constructed from 2 triangles and 3 rectangles. What is the shape of each box?

Spiral Review

Question 3. Bryan jogged the same distance each day for 7 days. He ran a total of 22.4 miles. The equation 7d = 22.4 can be used to find the distance d in miles he jogged each day. How far did Bryan jog each day?

Answer: 3.2 miles

Explanation: As given in equation 7d= 22.4, d= 22.4÷7 = 3.2 miles.

Question 4. A hot-air balloon is at an altitude of 240 feet. The balloon descends 30 feet per minute. What equation gives the altitude y, in feet, of the hot-air balloon after x minutes?

Answer: Y= 240- 30X.

Explanation: Given altitude Y, and the ballon was descended 30 feet per minute. So the equation is Y= 240- 30X.

Go Math Grade 6 Chapter 11 Answer Key Pdf Question 5. A regular heptagon has sides measuring 26 mm and is divided into 7 congruent triangles. Each triangle has a height of 27 mm. What is the area of the heptagon?

Answer: 351 mm 2

Explanation: Area of heptagon= 1/2 b×h = 1/2 (26)×(27) = 13×27 = 351 mm 2

Question 6. Alexis draws quadrilateral STUV with vertices S(1, 3), T(2, 2), U(2, –3), and V(1, –2). What name best classifies the quadrilateral?

Answer: Parallelogram

Use the net to find the surface area of the prism.

Explanation: First we must find the area of each face A= 4×3= 12 B= 4×3= 12 C= 5×4= 20 D= 5×4= 20 E= 5×3= 15 F= 5×3= 15 So, the surface area is 12+12+20+20+15+15= 94 cm 2

Find the surface area of the rectangular prism.

Answer: 222 cm 2

Explanation: Area of a rectangular prism is 2(wl+hl+hw) = 2(7×9+ 3×9+ 3×7) = 2(63+27+21) = 2(111) = 222 cm 2

Explanation: Area of a rectangular prism is 2(wl+hl+hw) = 2(10×10+ 10×10+ 10×10) = 2(100+100+100) = 2(300) = 600 cm 2

Answer: 350 cm 2

Explanation: Area of a rectangular prism is 2(wl+hl+hw) = 2(15×5+ 5×5+ 15×5) = 2(75+25+75) = 2(175) = 350 cm 2

Problem Solving + Applications

Question 5. A cereal box is shaped like a rectangular prism. The box is 20 cm long by 5 cm wide by 30 cm high. What is the surface area of the cereal box?

Answer: 1700 cm 2

Explanation: The length of the box is 20 cm, the wide is 5 cm and the height is 30 cm. So surface area of the cereal box is 2(wl+hl+hw)= 2(20×5+30×20+30×5) = 2(100+600+150) = 2(850) = 1700 cm 2

Question 6. Darren is painting a wooden block as part of his art project. The block is a rectangular prism that is 12 cm long by 9 cm wide by 5 cm high. Describe the rectangles that make up the net for the prism.

Question 7. In Exercise 6, what is the surface area, in square meters, that Darren has to paint?

Answer: 416 cm 2

Explanation: Surface area = 2(wl+hl+hw) = 2(9×12+5×12+ 5×9) = 2(108+60+45) = 2(213) = 416 cm 2

Answer: Emilio drew the net incorrectly Face D and F should have been 2 cm by 5 cm, not 3 cm by 5 cm

Explanation: Face A: 3×2= 6 cm 2 Face B: 3×5= 15 cm 2 Face C: 3×2= 6 cm 2 Face D: 2×5= 10 cm 2 Face E: 3×5= 15 cm 2 Face F: 2×5= 10 cm 2 So, the surface area of the prism area is 6+15+6+10+15+10= 62 cm 2.

9a. The area of face A is 10 cm 2 .

Answer: True

Explanation: The area of face A is 2×5= 10 cm 2 .

9b. The area of face B is 10 cm 2 .

Answer: False

Explanation: The area of face B is 2×8= 16  cm 2 .

9c. The area of face C is 40 cm 2 .

Answer: The area of face C is 8×5= 40 cm 2 .

9d. The surface area of the prism is 66 cm 2 .

Answer: 160 cm 2 .

Explanation: The surface area of the prism is = 2×10+2×10+2×40 = 20+20+80 = 160 cm 2 .

Use the net to find the surface area of the rectangular prism.

Answer: 52 square units.

Explanation: The area of face A is 6 squares. The area of face B is 8 squares. The area of face C is 6 squares. The area of face D is 12 squares. The area of face E is 8 squares. The area of face F is 12 squares. The surface area is 6+8+6+12+8+12= 52 square units.

Answer: 112 square units.

Explanation: The area of face A is 16 squares. The area of face B is 8 squares. The area of face C is 32 squares. The area of face D is 16 squares. The area of face E is 32 squares. The area of face F is 8 squares. The surface area is 112 square units.

Answer: 102 mm 2

Explanation: Area= 2(wl+hl+hw) = 2(3×7+3×7+3×3) = 2(21+21+9) = 2(51) = 102 mm 2

Answer: 58 in. 2

Explanation: Area= 2(wl+hl+hw) = 2(5×1+ 4×1+ 4×5) = 2(5+4+20) = 2(29) = 58 in. 2

Answer: 77 ft 2

Explanation: Area= 2(wl+hl+hw) = 2(6.5×2+3×2+3×6.5) = 2(13+6+19.5) = 2(38.5) = 77 ft 2

Question 6. Jeremiah is covering a cereal box with fabric for a school project. If the box is 6 inches long by 2 inches wide by 14 inches high, how much surface area does Jeremiah have to cover? _______ in. 2

Answer: 248 in. 2

Go Math Grade 6 Answer Key Chapter 11 Question 7. Tia is making a case for her calculator. It is a rectangular prism that will be 3.5 inches long by 1 inch wide by 10 inches high. How much material (surface area) will she need to make the case? _______ in. 2

Answer: 97 in. 2

Explanation: Surface Area= 2(wl+hl+hw) = 2(1×3.5+ 10×3.5+ 10×1) = 2(3.5+35+10) = 2(48.5) = 97 in. 2

Question 8. Explain in your own words how to find the surface area of a rectangular prism.

Answer: To find the surface area we must know the width, length, and height of the prism and then we can apply the formula which is Surface area= 2(width ×length)+ 2(length×height)+ 2(height×width) = 2(width ×length+ length×height+ 2(height×width)

Question 1. Gabriela drew a net of a rectangular prism on centimeter grid paper. If the prism is 7 cm long by 10 cm wide by 8 cm high, how many grid squares does the net cover? _______ cm 2

Answer: 412 cm 2.

Explanation: Surface area is 2(wl+hl+hw) = 2(10×7+8×7+8×10) = 2(70+56+80) = 2(206) = 412 cm 2.

Question 2. Ben bought a cell phone that came in a box shaped like a rectangular prism. The box is 5 inches long by 3 inches wide by 2 inches high. What is the surface area of the box? _______ in. 2

Answer: 62 in. 2

Explanation: Surface area is 2(wl+hl+hw) = 2(3×5+2×5+2×3) = 2(15+10+6) = 2(31) = 62 in. 2

Question 3. Katrin wrote the inequality x + 56 < 533. What is the solution of the inequality?

Answer: X<477.

Explanation: X+56<533 = X<533-56 = X<477.

Answer: y= 5x

Explanation: y/x = 10/2= 15/4= 3 y= 5x The pattern is y is x multiplied by 5.

Question 5. A square measuring 9 inches by 9 inches is cut from a corner of a square measuring 15 inches by 15 inches. What is the area of the L-shaped figure that is formed? _______ in. 2

Answer: 144 in. 2

Explanation: The area of a square A= a 2 , so we will find the area of each square. Area= 9 2 = 9×9 = 81 in. 2 And the area of another square is A= 15 2 = 15×15 = 225 in. 2 So the area of L shaped figure is 225-81= 144 in. 2

Question 6. Boxes of Clancy’s Energy Bars are rectangular prisms. How many lateral faces does each box have?

Explanation: As Lateral faces are not included in the bases, so rectangular prism has 4.

Use a net to find the surface area.

Answer: 24 ft 2

Explanation: The area of each face is 2 ft×2 ft= 4 ft and the number of faces is 6, so surface area is 6×4= 24 ft 2

Answer: 432 cm 2

Explanation: The area of face A is 16×6= 96 cm 2 The area of face B is 16×8= 128 cm 2 The area of face C and D is 1/2 × 6×8= 24 cm 2 The area of face E is 16×10= 160 cm 2 The surface 96+128+2×24+160= 432 cm 2

Answer: 155.5 in. 2

The area of each face is 7 1/2 × 7 1/2 = 15/2 × 15/2 = 225/4 in. 2 The number of faces is 6 and the surface area is 6× 225/4 = 675/4 = 337 1/2 in. 2

Go Math 6th Grade Chapter 11 Mid Chapter Checkpoint Answer Key Question 6. Attend to Precision Calculate the surface area of the cube in Exercise 5 using the formula S = 6s 2 . Show your work.

Answer: 337 1/2 in. 2

Explanation: As S= s 2 = 6(7 1/2) 2 = 6(15/2) 2 = 6(225/4) = 675/2 = 337 1/2 in. 2

Question 7. The Vehicle Assembly Building at Kennedy Space Center is a rectangular prism. It is 218 m long, 158 m wide, and 160 m tall. There are four 139 m tall doors in the building, averaging 29 m in width. What is the building’s outside surface area when the doors are open? a. Draw each face of the building, not including the floor.

Question 7. b. What are the dimensions of the 4 walls?

Answer: The 2 walls measure 218 m ×160 m and 2 walls measure by 158 m×160 m.

Question 7. c. What are the dimensions of the roof?

Answer: The dimensions of the roof are 218 m×158 m.

Question 7. d. Find the building’s surface area (not including the floor) when the doors are closed. _______ m 2

Answer: 1,54,764 m 2

Explanation: The area of two walls is 218×160= 34,880 m 2 The area of the other two walls is 158×160= 25,280 m 2 The area of the roof 158×218= 34,444 m 2 The surface area is 2× 34,880+ 2× 25,280+ 34,444 = 69,760+ 50,560+ 34,444 = 1,54,764 m 2

Question 7. e. Find the area of the four doors. _______ m 2

Answer: 16,124 m 2

Explanation: Area of a door is 139×29 = 4031 m 2 And the area of 4 doors is 4×4031= 16,124 m 2

Question 7. f. Find the building’s surface area (not including the floor) when the doors are open. _______ m 2

Answer: 1,38,640 m 2

Explanation: The building’s surface area (not including the floor) when the doors are open is 1,54,764 – 16,124= 1,38,640 m 2

Go Math Lesson 11.3 Surface Area and Volume Question 8. A rectangular prism is 1 \(\frac{1}{2}\) ft long, \(\frac{2}{3}\) ft wide, and \(\frac{5}{6}\) ft high. What is the surface area of the prism in square inches? _______ in. 2

Answer: 808 in. 2

Explanation: The area of two faces is 1 1/2× 5/6 = 3/2 × 5/6 = 5/4 cm 2 The area of two faces is 2/3 × 5/6 = 5/9 ft 2 The area of two faces is 1 1/2× 2/3 = 3/2 × 2/3 = 1 ft 2 The surface area of the prism is 2(wl+hl+hw) = 2(5/4 + 5/9 + 1) = 2( 1.25+0.55+1) = 2.5+1.1+2 = 5.61 ft 2 As 1 square foot = 144 square inches so 5.61×144 = 807.84 = 808 in. 2

Question 9. A gift box is a rectangular prism. The box measures 8 inches by 10 inches by 3 inches. What is its surface area? _______ in. 2

Answer: 268 in. 2

Explanation: The area of face A and Face E is 8×10= 80 in. 2 The area of face B and Face F is 8×3= 24 in. 2 The area of face C and Face D is 10×3= 30 in. 2 The surface area is 2×80+2×24+2×30 = 160+48+60 = 268 in. 2

Answer: 104 cm 2

Explanation: Surface area= 2(wl+hl+hw) = 2(6×5+2×5+2×6) = 2(30+10+12) = 2(52) = 104 cm 2

Answer: 118 in. 2

Explanation: Surface area= 2(wl+hl+hw) = 2(3.5×4+6×4+6×3.5) = 2(59) = 118 in. 2

Answer: 486 ft 2

Explanation: Surface area= 2(wl+hl+hw) = 2(9×9+9×9+9×9) = 2(81+81+81) = 2(243) = 486 ft 2

Answer: 336 cm 2.

Explanation: Area = 1/2 bh = 1/2 (6)(8) = 3×8 = 24. As there are 2 triangles, so 2×24= 48. Surface Area= (wl+hl+hw) = (6×12+8×12+12×10) = 228 Total Surface area = 228+48 = 336 cm 2

Question 5. A shoe box measures 15 in. by 7 in. by 4 \(\frac{1}{2}\) in. What is the surface area of the box? _______ in. 2

Answer: 408 in. 2

Explanation: The area of two faces is 15×7= 105 in. 2 The area of two faces is 15× 4 1/2 = 15 × 9/2 = 15 × 4.5 = 67.5 in. 2 The area of two faces is 7× 4 1/2 = 7× 9/2 = 7× 4.5 = 31.5 in. 2 The surface area is 2×105+ 2×67.5+ 2×31.5 = 210+ 135+ 63 = 408 in. 2

Mathematics Grade 6 Unit 11 Area and Volume Answers Question 6. Vivian is working with a styrofoam cube for art class. The length of one side is 5 inches. How much surface area does Vivian have to work with? _______ in. 2

Answer: 150 in. 2

Explanation: The area of each face is 5×5= 25 in. 2 The number of faces that styrofoam cube has is 6 So the surface area is 6×25= 150 in. 2

Question 7. Explain why a two-dimensional net is useful for finding the surface area of a three-dimensional figure.

Answer: Two-dimensional net is useful because by using a two-dimensional net you can calculate the surface area of each face and add them up to find the surface area of the three-dimensional figure.

Question 1. What is the surface area of a cubic box that contains a baseball that has a diameter of 3 inches? _______ in. 2

Answer: 54 in. 2

Explanation: The area of each face is 3×3= 9 in. 2 The number of faces for a cubic box is 6 in. 2 The surface area of box that contains a baseball is 6×9= 54 in. 2

Question 2. A piece of wood used for construction is 2 inches by 4 inches by 24 inches. What is the surface area of the wood? _______ in. 2

Answer: 304 in. 2

Explanation: The area of two faces is 4×2= 8 in. 2 The area of two faces is 2×24= 48 in. 2 The area of two faces is 24×4= 96 in. 2 So the surface area is 2×8+ 2×48+ 2×96 = 16+96+192 = 304 in. 2

Question 3. Detergent costs $4 per box. Kendra graphs the equation that gives the cost y of buying x boxes of detergent. What is the equation?

Answer: Y= 4X.

Explanation: The total price Y and the price is equal to 4 × X, and X is the number of boxes that Kendra buys. So the equation is Y=4X.

Question 4. A trapezoid with bases that measure 8 inches and 11 inches has a height of 3 inches. What is the area of the trapezoid? _______ in. 2

Answer: 28.5 in. 2

Explanation: Area of a trapezoid is 1/2(b1+b2)h = 1/2(8+11)3 = 1/2(19)3 = 1/2 (57) = 28.5 in. 2

Question 5. City Park is a right triangle with a base of 40 yd and a height of 25 yd. On a map, the park has a base of 40 in. and a height of 25 in. What is the ratio of the area of the triangle on the map to the area of City Park?

Answer: 1296:1.

Explanation: Area= 1/2 bh = 1/2 (40)(25) = (20)(25) = 500 yd 2 So area of city park is 500 yd 2 Area= 1/2 bh = 1/2 (40)(25) = (20)(25) = 500 in 2 So area on the map is 500 in as 1 yd 2 = 1296 in 2 So 500 in 2 = 500×1296 = 648,000 So, the ratio of the area of the triangle on the map to the area of City Park is 648,000:500 = 1296:1.

Explanation: The area of two faces is 18 squares The area of two faces is 6 squares The area of two faces is 12 squares So the surface area is 2×18+ 2×6+ 2×12 = 72 square units.

Answer: 105 cm 2

Question 2. A triangular pyramid has a base with an area of 43 cm 2 and lateral faces with bases of 10 cm and heights of 8.6 cm. What is the surface area of the pyramid? _______ cm 2

Answer: 172 cm 2

Explanation: The area of one face is 1/2×10×8.6 = 5×8.6 = 43 cm 2 The surface area of the pyramid is 43+3×43 = 43+ 129 = 172 cm 2

Surface Area and Volume Test Answer Key Question 3. A square pyramid has a base with a side length of 3 ft and lateral faces with heights of 2 ft. What is the lateral area of the pyramid? _______ ft 2

Answer: 12 ft 2

Explanation: The area of one face is 1/2×3×2= 3 ft 2 The lateral area of the pyramid is 4×3= 12 ft 2

Use a net to find the surface area of the square pyramid.

Answer: 208 ft 2

Answer: 220 cm 2

Answer: 264 in. 2

Question 7. The Pyramid Arena is located in Memphis, Tennessee. It is in the shape of a square pyramid, and the lateral faces are made almost completely of glass. The base has a side length of about 600 ft and the lateral faces have a height of about 440 ft. What is the total area of the glass in the Pyramid Arena? _______ ft 2

Answer: 5,28,000 ft 2

Explanation: The area of one face is 1/2×600×440= 1,32,000 ft 2 The surface of tha lateral faces is 4× 1,32,000= 5,28,000 ft 2 So, the total area of the glass in the arena is 5,28,000 ft 2

Question 8. The Great Pyramids are located near Cairo, Egypt. They are all square pyramids, and their dimensions are shown in the table. What is the lateral area of the Pyramid of Cheops? _______ m 2

Answer: 82,800 m 2

Explanation: The area of one face is 1/2×230×180 = 230×90 = 20,700 m 2 The lateral area of the pyramid of Cheops is 4×20,700= 82,800 m 2

Question 9. What is the difference between the surface areas of the Pyramid of Khafre and the Pyramid of Menkaure? _______ m 2

Answer: 93,338 m 2

Explanation: The area of the base is 215×215= 46,225 The area of one face is 1/2×215×174 = 215× 87 18,705 m 2 The surface area of Pyramid Khafre is 46,225+4×18,705 = 46,225+ 74820 = 121,045 m 2 The area of the base 103×103= 10,609 The area of one face is 1/2×103×83 = 8549÷2 = 4274.4 m 2 The surface area of the Pyramid of Menkaure is 10,609+4×4274.5 = 10,609+ 17,098 = 27,707 m 2

The difference between the surface areas of the Pyramid of Khafre and the Pyramid of Menkaure = 121,405-27,707 = 93,338 m 2

Answer: 6x+9 ft 2.

Explanation: The expression for the surface area of the square pyramid is 6x+9 ft 2.

Question 11. Make Arguments A square pyramid has a base with a side length of 4 cm and triangular faces with a height of 7 cm. Esther calculated the surface area as (4 × 4) + 4(4 × 7) = 128 cm 2 . Explain Esther’s error and find the correct surface area

Answer: 72 cm 2 .

Explanation: Esther didn’t apply the formula correctly, she forgot to include 1/2 in the calculated surface area. The correct surface area is (4×4)+4(1/2 ×4×7) = 16+4(14) = 16+56 = 72 cm 2 .

Answer: 160 in. 2

Explanation: No, I disagree with Jose as he found surface area instead of the lateral area, so the lateral area is 4×1/2×10×8 = 2×10×8 = 160 in. 2

Answer: 95 mm 2

Answer: 612 cm 2

The area of the base is 18×18= 324 cm 2 The area of one face is 1/2×18×8 = 18×4 =  72 cm 2 The surface area is 324+4×72 = 25+4×17.5 = 25+70 = 612 cm 2

Answer: 51.25 yd 2

The area of the base is 2.5×2.5= 6.25  mm 2 The area of one face is 1/2×2.5×9 = 22.5/2 = 11.25 yd 2 The surface area is 25+4×17.5 = 6.25+4×11.25 = 6.25+45 = 51.25 yd 2

Answer: 180 in 2

The area of the base is 10×10= 100 in 2 The area of one face is 1/2×4×10 = 2×10 = 20 in 2 The surface area is 100+4×20 = 100+4×20 = 100+80 = 180 in 2

Question 5. Cho is building a sandcastle in the shape of a triangular pyramid. The area of the base is 7 square feet. Each side of the base has a length of 4 feet and the height of each face is 2 feet. What is the surface area of the pyramid? _______ ft 2

Answer: 19 ft 2

Explanation: The area of one face is 1/2×4×2= 4 ft 2 The surface area of the triangular pyramid is 7+3×4 = 7+12 = 19 ft 2

Question 6. The top of a skyscraper is shaped like a square pyramid. Each side of the base has a length of 60 meters and the height of each triangle is 20 meters. What is the lateral area of the pyramid? _______ m 2

Answer: 2400 m 2

Explanation: The area of one face is 1/2×60×20 = 600 m 2 The lateral area of the pyramid is 4×600= 2400 m 2

Question 7. Write and solve a problem finding the lateral area of an object shaped like a square pyramid.

Answer: Mary has a triangular pyramid with a base of 10cm and a height of 15cm. What is the lateral area of the pyramid?

Explanation: The area of one face is 1/2×10×15 = 5×15 = 75 cm 2 The lateral area of the triangular pyramid is 3×75 = 225 cm 2

Question 1. A square pyramid has a base with a side length of 12 in. Each face has a height of 7 in. What is the surface area of the pyramid? _______ in. 2

Answer: 312 in. 2

Explanation: The area of the base is 12×12= 144 in. 2 The area of one face is 1/2×12×7 = 6×7 = 42 in. 2 The surface area of the square pyramid is 144+4×42 = 144+ 168 = 312 in. 2

Question 2. The faces of a triangular pyramid have a base of 5 cm and a height of 11 cm. What is the lateral area of the pyramid? _______ cm 2

Answer: 82.5 cm 2

Explanation: The area of one face is 1/2×5×11 = 55/2 = 27.5 cm 2 The lateral area of the triangular pyramid is 3×27.5= 82.5 cm 2

Answer: y=x+1.

Explanation: As the figure represents that every y value is 1 more than the corresponding x value, so the linear equation is y=x+1.

Question 4. A regular octagon has sides measuring about 4 cm. If the octagon is divided into 8 congruent triangles, each has a height of 5 cm. What is the area of the octagon? _______ cm 2

Explanation: Area is 1/2bh = 1/2× 4×5 = 2×5 = 10 cm 2 So the area of each triangle is 10 cm 2 and the area of the octagon is 8×10= 80 cm 2

Question 5. Carly draws quadrilateral JKLM with vertices J(−3, 3), K(3, 3), L(2, −1), and M(−2, −1). What is the best way to classify the quadrilateral?

Answer: It is a Trapezoid.

Surface Area and Volume Answer Key Question 6. A rectangular prism has the dimensions of 8 feet by 3 feet by 5 feet. What is the surface area of the prism? _______ ft 2

Answer: 158 ft 2

Explanation: The area of the two faces of the rectangular prism is 8×3= 24 ft 2 The area of the two faces of the rectangular prism is 8×5= 40 ft 2 The area of the two faces of the rectangular prism is 3×5= 15 ft 2 The surface area of the rectangular prism is 2×24+2×40+2×15 = 48+80+30 = 158 ft 2

Question 1. _____ is the sum of the areas of all the faces, or surfaces, of a solid figure.

Answer: Surface area is the sum of the areas of all the faces, or surfaces, of a solid figure.

Question 2. A three-dimensional figure having length, width, and height is called a(n) _____.

Answer: A three-dimensional figure having length, width, and height is called a(n) solid figure.

Question 3. The _____ of a solid figure is the sum of the areas of its lateral faces.

Answer: The lateral area of a solid figure is the sum of the areas of its lateral faces.

Concepts and Skills

Answer: Triangular prism

Answer: 216 in. 2

Answer: 310 cm 2

Question 7. A machine cuts nets from flat pieces of cardboard. The nets can be folded into triangular pyramids and used as pieces in a board game. What shapes appear in the net? How many of each shape are there?

Answer: 4 triangles.

Explanation: There are 4 triangles.

Question 8. Fran’s filing cabinet is 6 feet tall, 1 \(\frac{1}{3}\) feet wide, and 3 feet deep. She plans to paint all sides except the bottom of the cabinet. Find the area of the sides she intends to paint. _______ ft 2

Answer: 56 ft 2

Explanation: The two lateral face area is 6×1 1/3 = 6× 4/3 = 2×4 = 8 ft 2 The area of the other two lateral faces is 6×3= 18 The area of the top and bottom is 3× 1 1/3 = 3× 4/3 = 4 ft 2 The area of the sides she intends to paint is 2×8+2×18+4 = 16+36+4 = 56 ft 2

Question 9. A triangular pyramid has lateral faces with bases of 6 meters and heights of 9 meters. The area of the base of the pyramid is 15.6 square meters. What is the surface area of the pyramid?

Answer: 96.6 m 2

Explanation: The area of one face is 1/2× 6× 9 = 3×9 = 27 m 2 The surface area of the triangular pyramid is 15.6+3×27 = 15.6+ 81 = 96.6 m 2

Solving Surface Area Problems Lesson 11.4 Answer Key Question 10. What is the surface area of a storage box that measures 15 centimeters by 12 centimeters by 10 centimeters? _______ cm 2

Answer: 900 cm 2

Explanation: The area of the two faces is 15×12= 180 cm 2 The area of another two faces is 15×10= 150 cm 2 The area of the other two faces is 10×12= 120 cm 2 So the surface area of the storage box is 2×180+2×150+2×120 cm 2 = 360+300+240 = 900 cm 2

Question 11. A small refrigerator is a cube with a side length of 16 inches. Use the formula S = 6s 2 to find the surface area of the cube. _______ in. 2

Answer: 1,536 in. 2

Explanation: Area = s 2 = 6×(16) 2 = 6× 256 = 1,536 in. 2

Question 1. A prism is filled with 38 cubes with a side length of \(\frac{1}{2}\) unit. What is the volume of the prism in cubic units? _______ \(\frac{□}{□}\) cubic units

Answer: 4.75 cubic units

Explanation: The volume of the cube is S 3 The volume of a cube with S= (1/2) 3 = 1/2×1/2×1/2 = 1/8 = 0.125 cubic units As there are 38 cubes so 38×0.125= 4.75 cubic units.

Question 2. A prism is filled with 58 cubes with a side length of \(\frac{1}{2}\) unit. What is the volume of the prism in cubic units? _______ \(\frac{□}{□}\) cubic units

Answer: 7.25 cubic units.

Explanation: The volume of the cube is S 3 The volume of a cube with S= (1/2) 3 = 1/2×1/2×1/2 = 1/8 = 0.125 cubic units As there are 58 cubes so 58×0.125= 7.25 cubic units.

Find the volume of the rectangular prism.

Answer: 33 cubic units.

Explanation: The volume of the rectangular prism is= Width×Height×Length = 5 1/2 ×3×2 = 11/2 ×3×2 = 33 cubic units.

Answer: 91 1/8 cubic units.

Explanation: The volume of the rectangular prism is= Width×Height×Length = 4 1/2 ×4 1/2×4 1/2 = 9/2 ×9/2×9/2 = 729/8 = 91 1/8 cubic units.

Question 5. Theodore wants to put three flowering plants in his window box. The window box is shaped like a rectangular prism that is 30.5 in. long, 6 in. wide, and 6 in. deep. The three plants need a total of 1,200 in. 3 of potting soil to grow well. Is the box large enough? Explain.

Answer: No, the box is not large enough as the three plants need a total of 1,200 in. 3 and here volume is 1,098 in. 3

Explanation: Volume= Width×Height×Length = 30.5×6×6 = 1,098 in. 3

Question 6. Explain how use the formula V = l × w × h to verify that a cube with a side length of \(\frac{1}{2}\) unit has a volume of \(\frac{1}{8}\) of a cubic unit.

Answer: 1/8 cubic units

Explanation: As length, width and height is 1/2′ so Volume = Width×Height×Length = 1/2 × 1/2 × 1/2 = 1/8 cubic units

Question 7. Karyn is using a set of building blocks shaped like rectangular prisms to make a model. The three types of blocks she has are shown at right. What is the volume of an A block? (Do not include the pegs on top.) \(\frac{□}{□}\) cubic units

Answer: 1/2 cubic units

Explanation: Volume = Width×Height×Length = 1× 1/2 ×1 = 1/2 cubic units

Volume and Surface Area Answer Key Question 8. How many A blocks would you need to take up the same amount of space as a C block? _______ A blocks

Answer: No of blocks required to take up the same amount of space as a C block is 4 A blocks.

Explanation: Volume = Width×Height×Length = 1×2×1 = 2 cubic unit No of blocks required to take up the same amount of space as a C block is 1/2 ÷2 = 2×2 = 4 A blocks

Question 9. Karyn puts a B block, two C blocks, and three A blocks together. What is the total volume of these blocks? _______ \(\frac{□}{□}\) cubic units

Answer: 6 1/2 cubic units

Explanation: The volume of A block is Volume = Width×Height×Length = 1×1 ×1/2 = 1/2 cubic units. As Karyn puts three A blocks together, so 3× 1/2= 3/2 cubic units. The volume of the B block is Volume = Width×Height×Length = 1×1 × 1 = 1 cubic unit. As Karyn puts only one B, so 1 cubic unit. The volume of the C block is Volume = Width×Height×Length = 2×1×1 = 2 cubic units. As Karyn puts two C blocks together, so 2× 2= 4 cubic units. So, the total volume of these blocks is 3/2 + 1+ 4 = 3/2+5 = 13/2 = 6 1/2 cubic units

Question 10. Karyn uses the blocks to make a prism that is 2 units long, 3 units wide, and 1 \(\frac{1}{2}\) units high. The prism is made of two C blocks, two B blocks, and some A blocks. What is the total volume of A blocks used? _______ cubic units

Answer: 3 cubic units.

Explanation: Volume = Width×Height×Length = 2×3×1 1/2 = 2×3× 3/2 = 9 cubic units. The total volume of A block used is 9-(2×2)-(2×1) = 9- 4- 2 = 9-6 = 3 cubic units.

Question 11. Verify the Reasoning of Others Jo says that you can use V = l × w × h or V = h × w × l to find the volume of a rectangular prism. Does Jo’s statement make sense? Explain.

Answer: Yes

Explanation: Yes, Jo’s statement makes sense because by the commutative property, we can change the order of the variables of length, width, and height and both will produce the same result.

Question 12. A box measures 5 units by 3 units by 2 \(\frac{1}{2}\) units. For numbers 12a–12b, select True or False for the statement. 12a. The greatest number of cubes with a side length of \(\frac{1}{2}\) unit that can be packed inside the box is 300. 12b. The volume of the box is 37 \(\frac{1}{2}\) cubic units. 12a. __________ 12b. __________

Answer: 12a True. 12b True.

Explanation: The volume of the cube is S 3 The volume of a cube with S= (1/2) 3 = 1/2×1/2×1/2 = 1/8 cubic units As there are 300 cubes so 300× 1/8= 75/2 = 37 1/2 cubic units.

Answer: 6 3/4 cubic units

Explanation: Volume = Width×Height×Length = 3× 1 1/2× 1 1/2 = 3× 3/2 × 3/2 = 27/4 = 6 3/4 cubic units

Answer: 22 1/2 cubic units

Explanation: Volume = Width×Height×Length = 5×1× 4 1/2 = 5× 9/2 = 45/2 = 22 1/2 cubic units

Answer: 16 1/2 cubic units.

Explanation: Volume = Width×Height×Length = 5 1/2× 1 1/2× 2 = 11/2×3/2×2 = 33/2 = 16 1/2 cubic units.

Answer: 28 1/8 cubic units.

Explanation: Volume = Width×Height×Length = 2 1/2× 2 1/2 × 4 1/2 = 5/2 × 5/2 × 9/2 = 225/8 = 28 1/8 cubic units.

Question 5. Miguel is pouring liquid into a container that is 4 \(\frac{1}{2}\) inches long by 3 \(\frac{1}{2}\) inches wide by 2 inches high. How many cubic inches of liquid will fit in the container? _______ \(\frac{□}{□}\) in. 3

Answer: 31 1/2 cubic units

Explanation: Volume = Width×Height×Length = 4 1/2 × 3 1/2 ×2 = 9/2 × 7/2 × 2 = 63/2 = 31 1/2 cubic units

Go Math Grade 6 Chapter 11 Answer Key Pdf Question 6. A shipping crate is shaped like a rectangular prism. It is 5 \(\frac{1}{2}\) feet long by 3 feet wide by 3 feet high. What is the volume of the crate? _______ \(\frac{□}{□}\) ft 3

Answer: 49 1/2 ft 3

Explanation: Volume = Width×Height×Length = 5 1/2 × 3 × 3 = 11/2 ×9 = 99/2 = 49 1/2 ft 3

Question 7. How many cubes with a side length of \(\frac{1}{4}\) unit would it take to make a unit cube? Explain how you determined your answer.

Answer: There will be 4×4×4= 64 cubes and 1/4 unit in the unit cube.

Explanation: As the unit cube has a 1 unit length, 1 unit wide, and 1 unit height So length 4 cubes = 4× 1/4= 1 unit width 4 cubes = 4× 1/4= 1 unit height 4 cubes = 4× 1/4= 1 unit So there will be 4×4×4= 64 cubes and 1/4 unit in the unit cube.

Question 1. A rectangular prism is 4 units by 2 \(\frac{1}{2}\) units by 1 \(\frac{1}{2}\) units. How many cubes with a side length of \(\frac{1}{2}\) unit will completely fill the prism?

Answer: 120 cubes

Explanation: No of cubes with a side length of 1/2 unit is Length 8 cubes= 8× 1/2= 4 units Width 5 cubes= 5× 1/2= 5/2= 2 1/2 units Height 3 cubes= 3× 1/2= 3/2= 1 1/2 units So there are 8×5×3= 120 cubes in the prism.

Question 2. A rectangular prism is filled with 196 cubes with \(\frac{1}{2}\)-unit side lengths. What is the volume of the prism in cubic units? _______ \(\frac{□}{□}\) cubic units

Answer: 24 1/2 cubic units.

Explanation: As it takes 8 cubes with a side length of 1/2 to form a unit cube, so the volume of the prism in the cubic units is 196÷8= 24 1/2 cubic units.

Question 3. A parallelogram-shaped piece of stained glass has a base measuring 2 \(\frac{1}{2}\) inches and a height of 1 \(\frac{1}{4}\) inches. What is the area of the piece of stained glass? _______ \(\frac{□}{□}\) in. 2

Answer: 3 1/8 in. 2

Explanation: Area of a parallelogram = base×height = 2 1/2 × 1 1/4 = 5/2 × 5/4 = 25/8 = 3 1/8 in. 2

Question 4. A flag for the sports club is a rectangle measuring 20 inches by 32 inches. Within the rectangle is a yellow square with a side length of 6 inches. What is the area of the flag that is not part of the yellow square? _______ in. 2

Answer: 604 in. 2

Explanation: Area of a flag= Length×width = 20×32 = 640 in. 2 Area of the yellow square= S 2 = 6 = 36 in. 2 So the area of the flag that is not a part of the yellow square is 640-36= 604 in. 2

Answer: 80 square units

Explanation: Area of two faces is 12 squares Area of other two faces is 16 squares Area of another two faces is 12 squares So the surface area is 2×12+2×16+2×12 = 24+32+24 = 80 square units

Answer: 161 cm 2

Explanation: The area of the base is 7×7= 49 cm 2 And the area of one face is 1/2 × 7× 8 = 7×4 = 28 cm 2 The surface area of the square pyramid is 49+4×28 = 49+112 = 161 cm 2

Find the volume.

Answer: 3,937 1/2 in. 3

Explanation: Volume= Length× wide× height = 10 1/2 ×15 × 25 = 11/2 × 15 × 25 = 4,125/2 = 3,937 1/2 in. 3

Answer: 27/512 in. 3

Explanation: Volume= Length× wide× height =3/8 ×3/8 × 3/8 = 27/512 in. 3

Find the volume of the prism.

Answer: 690 5/8in. 3

Explanation: Volume= Length× wide× height = 8 1/2 × 6 1/2 × 12 1/2 = 17/2 × 13/2× 25/2 = 5525/2 = 690 5/8in. 3

Answer: 125/4096 in. 3

Explanation: Volume= Length× wide× height = 5/16 ×5/16 × 5/16 = 125/4096 in. 3

Answer: 20 yd 3

Explanation: Area= 3 1/3 yd 2 So Area= wide×height 3 1/3= w × 1 1/3 10/3= w× 4/3 w= 10/3 × 3/4 w= 5/2 w= 2.5 yd Volume= Length×width×height = 6× 2.5× 1 1/3 = 6×2.5× 4/3 = 2×2.5×4 = 20 yd 3

Question 6. Wayne’s gym locker is a rectangular prism with a width and height of 14 \(\frac{1}{2}\) inches. The length is 8 inches greater than the width. What is the volume of the locker? _______ \(\frac{□}{□}\) in. 3

Answer: 4,730 5/8 in. 3

Explanation: As length is 8 inches greater than width, so 14 1/2+ 8 = 29/2+8 = 45/2 = 22 1/2 in Then volume= Length×width×height = 22 1/2 × 14 1/2 × 14 1/2 = 45/2× 29/2× 29/2 = 37845/8 = 4,730 5/8 in. 3

Answer: 33 3/4 ft 3

Explanation: The volume of rectangular prism is= Length×width×height = 4 1/2× 2 1/2× 3 = 9/2 × 5/2× 3 = 135/3 = 33 3/4 ft 3

Question 8. Find the width of Tank 2 and the height of Tank 3.

Answer: Width of Tank 2= 8m, Height of the Tank 3= 10 m

Explanation: The volume of Tank 2= 384 m 3 so V= LWH 384=  12×W×4 W= 384/48 W= 8 m So the width of Tank 2= 8m The volume of Tank 3= 2160 m So V= LWH 2160= 18×12×H H= 2160/216 H= 10 m So the height of Tank 3 = 10 m

Grade 6 Mathematics Unit 11 Area and Surface Area Answer Key Question 9. To keep the fish healthy, there should be the correct ratio of water to fish in the tank. One recommended ratio is 9 L of water for every 2 fish. Find the volume of Tank 4. Then use the equivalencies 1 cm 3 = 1 mL and 1,000 mL = 1 L to find how many fish can be safely kept in Tank 4.

Answer: 35 Fishes

Explanation: Volume of Tank 4 = LWH = 72×55×40 = 1,58,400 cm 3 As 1 cm 3 = 1 mL and 1,000 mL = 1 L 1,58,400 cm 3 = 1,58,400 mL and 1,58,400 mL = 158.4 L So the tank can keep safely (158.4÷ 9)×2 = (17.6)× 2 = 35.2 = 35 Fishes

Question 10. Use Reasoning Give another set of dimensions for a tank that would have the same volume as Tank 2. Explain how you found your answer.

Answer: Another set of dimensions for a tank that would have the same volume as Tank 2 is 8m by 8m by 6m. So when we multiply the product will be 384

Answer: 150 5/16 m 3

Explanation: Volume= Length×width×height = 5× 3 1/4× 9 1/4 = 5× 13/4 × 37/4 = 2405/16 = 150 5/16 m 3

Answer: 27 1/2 in. 3

Explanation: Volume= Length×width×height = 5 1/2 × 2 1/2 × 2 = 11/2 × 5/2 × 2 = 55/2 = 27 1/2 in. 3

Answer: 91 1/8 mm 3

Explanation: Volume= Length×width×height = 4 1/2 × 4 1/2 × 4 1/2 = 9/2 × 9/2 × 9/2 = 729/8 = 91 1/8 mm 3

Answer: 112 1/2 ft 3

Explanation: Volume= Length×width×height = 7 1/2 × 2 1/2 × 6 = 15/2 × 5/2 × 6 = 225/2 = 112 1/2 ft 3

Answer: 36 m 3

Explanation: The area of the shaded face is Length × width= 8 m 2 The volume of the prism= Length×width×height = 8 × 4 1/2 = 8 × 9/2 = 4 × 9 = 36 m 3

Answer: 30 3/8 ft 3

Explanation: Volume of the prism= Length×width×height = 2 1/4 × 6 × 2 1/4 = 9/4 × 6 × 9/4 = 243/8 = 30 3/8 ft 3

Question 7. A cereal box is a rectangular prism that is 8 inches long and 2 \(\frac{1}{2}\) inches wide. The volume of the box is 200 in. 3 . What is the height of the box? _______ in.

Answer: H= 10 in

Explanation: As volume = 200 in. 3 . So V= LWH 200= 8 × 2 1/2 × H 200= 8 × 5/2 × H 200= 20 × H H= 10 in

Question 8. A stack of paper is 8 \(\frac{1}{2}\) in. long by 11 in. wide by 4 in. high. What is the volume of the stack of paper? _______ in. 3

Answer: 374 in. 3

Explanation: The volume of the stack of paper = LWH = 8 1/2 × 11 × 4 = 17/2 × 11 × 4 = 374 in. 3

Question 9. Explain how you can find the side length of a rectangular prism if you are given the volume and the two other measurements. Does this process change if one of the measurements includes a fraction?

Answer: We can find the side length of a rectangular prism if you are given the volume and the two other measurements by dividing the value of the volume by the product of the values of width and height of the prism. And the process doesn’t change if one of the measurements include a fraction.

Question 1. A kitchen sink is a rectangular prism with a length of 19 \(\frac{7}{8}\) inches, a width of 14 \(\frac{3}{4}\) inches, and height of 10 inches. Estimate the volume of the sink.

Answer: 3,000 in. 3

Explanation: Length = 19 7/8 as the number was close to 20 and width 14 3/4 which is close to 15 and height is 10 So Volume= LBH = 20 × 15 × 10 = 3,000 in. 3

Chapter 11 Surface Area and Volume Answer Key Question 2. A storage container is a rectangular prism that is 65 centimeters long and 40 centimeters wide. The volume of the container is 62,400 cubic centimeters. What is the height of the container?

Answer: H= 24 cm

Explanation: Volume of container= LBH Volume= 62,400 cubic centimeters 62,400 = 65× 40 × H 62,400 = 2600 × H H= 62,400/ 2600 H= 24 cm

Question 3. Carrie started at the southeast corner of Franklin Park, walked north 240 yards, turned and walked west 80 yards, and then turned and walked diagonally back to where she started. What is the area of the triangle enclosed by the path she walked? _______ yd 2

Answer: 9,600 yd 2

Explanation: Area of triangle= 1/2 bh = 1/2 × 240 × 80 = 240 × 40 = 9,600 yd 2

Question 4. The dimensions of a rectangular garage are 100 times the dimensions of a floor plan of the garage. The area of the floor plan is 8 square inches. What is the area of the garage?

Answer: 80,000 in 2

Explanation: As 1 in 2 = 10,000 in 2 , so area of the floor plan 8 in = 8×10000 = 80,000 in 2

Question 5. Shiloh wants to create a paper-mâché box shaped like a rectangular prism. If the box will be 4 inches by 5 inches by 8 inches, how much paper does she need to cover the box?

Answer: 184 in 2

Explanation: Area of the rectangular prism= 2(wl+hl+hw) = 2(4×5 + 5×8 + 8×4) = 2(20+40+32) = 2(92) = 184 in 2

Question 6. A box is filled with 220 cubes with a side length of \(\frac{1}{2}\) unit. What is the volume of the box in cubic units? _______ \(\frac{□}{□}\) cubic units

Answer: 27.5 cubic units.

Explanation: The volume of a cube side is (1/2) 3 = 1/8 So 220 cubes= 220× 1/8 = 27.5 cubic units.

Question 1. An aquarium tank in the shape of a rectangular prism is 60 cm long, 30 cm wide, and 24 cm high. The top of the tank is open, and the glass used to make the tank is 1 cm thick. How much water can the tank hold? _______ cm 3

Answer: So tank can hold 37,352 cm 3

Explanation: As Volume= LBH Let’s find the inner dimensions of the tank, so 60-2 × 30-2 × 24-1 = 58×28×23 = 37,352 cm 3

Question 2. What if, to provide greater strength, the glass bottom were increased to a thickness of 4 cm? How much less water would the tank hold? _______ cm 3

Answer: 4,872 cm 3

Explanation: As the glass bottom was increased to a thickness of 4 cm, 60-2 × 30-2 × 24-4 = 58×28×20 = 32,480 cm 3 So the tank can hold 37,352- 32,480= 4,872 cm 3

Question 3. An aquarium tank in the shape of a rectangular prism is 40 cm long, 26 cm wide, and 24 cm high. If the top of the tank is open, how much tinting is needed to cover the glass on the tank? Identify the measure you used to solve the problem. _______ cm 3

Answer: 4,208 cm 3  tinting needed to cover the glass on the tank.

Explanation: The lateral area of the two faces is 26×24= 624 cm 2 The lateral area of the other two faces is 40×24= 960 cm 2 And the area of the top and bottom is 40×26= 1040 cm 2 So the surface area of the tank without the top is 2×624 + 2×960 + 1040 = 1,248+1,920+1,040 = 4,208 cm 3

Question 4. The Louvre Museum in Paris, France, has a square pyramid made of glass in its central courtyard. The four triangular faces of the pyramid have bases of 35 meters and heights of 27.8 meters. What is the area of glass used for the four triangular faces of the pyramid?

Answer: 1946 m 2

Explanation: The area of one face is 1/2 × 35 × 27.8= 486.5 m 2 And the area of glass used for the four triangular faces of the pyramid is 4×486.5= 1946 m 2

Question 5. A rectangular prism-shaped block of wood measures 3 m by 1 \(\frac{1}{2}\) m by 1 \(\frac{1}{2}\) m. How much of the block must a carpenter carve away to obtain a prism that measures 2 m by \(\frac{1}{2}\) m by \(\frac{1}{2}\) m? _______ \(\frac{□}{□}\) m 3

Answer: 6 1/4 m 3

Explanation: The volume of the original block= LWH = 3 × 1 1/2 × 1 1/2 = 3× 3/2 × 3/2 = 27/4 = 6 3/4 m 2 And volume of carpenter carve is 2× 1/2 × 1/2 = 1/2 m 2 So, the carpenter must carve 27/4 – 1/2 = 25/2 = 6 1/4 m 3

Question 6. The carpenter (Problem 5) varnished the outside of the smaller piece of wood, all except for the bottom, which measures \(\frac{1}{2}\) m by \(\frac{1}{2}\) m. Varnish costs $2.00 per square meter. What was the cost of varnishing the wood? $ _______

Answer: $8.50

Explanation: The area of two lateral faces are 2×1/2= 1 m 2 The area of the other two lateral faces are 2×1/2= 1 m 2 The area of the top and bottom is 1/2×1/2= 1/4 m 2 And the surface area is 2×1 + 2×1 + 1/4 = 2+2+1/4 = 17/4 = 4.25 m 2 And the cost of vanishing the wood is $2.00× 4.25= $8.50

Question 7. A wax candle is in the shape of a cube with a side length of 2 \(\frac{1}{2}\) in. What volume of wax is needed to make the candle? _______ \(\frac{□}{□}\) in. 3

Explanation: The Volume of wax is needed to make the candle is= LWH = 2 1/2 × 2 1/2 × 2 1/2 = 5/2 × 5/2 × 5/2 = 125/8 = 15 5/8 in. 3

Question 8. Describe A rectangular prism-shaped box measures 6 cm by 5 cm by 4 cm. A cube-shaped box has a side length of 2 cm. How many of the cube-shaped boxes will fit into the rectangular prismshaped box? Describe how you found your answer.

Answer: 12 cube-shaped boxes

Explanation: As 6 small boxes can fit on the base i.e 6 cm by 5 cm, as height is 4cm there can be a second layer of 6 small boxes. So, there will be a total of 12 cube-shaped boxes and will fit into a rectangular prism-shaped box

Question 9. Justin is covering the outside of an open shoe box with colorful paper for a class project. The shoe box is 30 cm long, 20 cm wide, and 13 cm high. How many square centimeters of paper are needed to cover the outside of the open shoe box? Explain your strategy _______ cm 2

Answer: 1,900 cm 2

Explanation: The area of the two lateral faces of the shoebox is 20×13= 260 cm 2 The area of another two lateral faces of the shoebox is 30×13= 390 cm 2 The area of the top and bottom is 30×20= 600 cm 2 So, the surface area of the shoebox without the top is 2×260 + 2× 390 + 600 = 520+780+600 = 1,900 cm 2

Read each problem and solve.

Question 1. The outside of an aquarium tank is 50 cm long, 50 cm wide, and 30 cm high. It is open at the top. The glass used to make the tank is 1 cm thick. How much water can the tank hold? _______ cm 3

Answer: So water tank can hold 66,816 cm 3

Explanation: The volume of inner dimensions of the aquarium is 50-2 × 50-2 × 30-1 = 48×48×29 = 66,816 cm 3 So water tank can hold 66,816 cm 3

Question 2. Arnie keeps his pet snake in an open-topped glass cage. The outside of the cage is 73 cm long, 60 cm wide, and 38 cm high. The glass used to make the cage is 0.5 cm thick. What is the inside volume of the cage? _______ cm 3

Answer: The volume of the cage is 1,59,300 cm 3

Explanation: The volume of inner dimensions is 73-1 × 60-1 × 38-0.5 = 72×59×37.5 = 1,59,300 cm 3 So, the volume of the cage is 1,59,300 cm 3

Question 3. A display number cube measures 20 in. on a side. The sides are numbered 1–6. The odd-numbered sides are covered in blue fabric and the even-numbered sides are covered in red fabric. How much red fabric was used? _______ in. 2

Answer: 1200 in. 2

Explanation: The area of each side of a cube is 20×20= 400 in. 2 , as there are 3 even-numbered sides on the cube. So there will be 3×400= 1200 in. 2

Question 4. The caps on the tops of staircase posts are shaped like square pyramids. The side length of the base of each cap is 4 inches. The height of the face of each cap is 5 inches. What is the surface area of the caps for two posts? _______ in. 2

Answer: 112 in. 2

Explanation: The area of the base is 4×4= 16 in. 2 The area of one face is 1/2×5×4= 10 in. 2 The surface area of one cap is 16+4×10 = 16+40 = 56 in. 2 And the surface area of the caps for two posts is 2×56= 112 in. 2

Question 5. A water irrigation tank is shaped like a cube and has a side length of 2 \(\frac{1}{2}\) feet. How many cubic feet of water are needed to completely fill the tank? _______ \(\frac{□}{□}\) ft 3

Answer: 15 5/8 ft 3

Explanation: Volume= LWH = 2 1/2 × 2 1/2 × 2 1/2 = 5/2 × 5/2 × 5/2 = 125/8 = 15 5/8 ft 3

Question 6. Write and solve a problem for which you use part of the formula for the surface area of a triangular prism.

Answer: In a triangular prism, the triangular end has a base of 5cm and the height is 8 cm. The length of each side is 4 cm and the height of the prism is 10 cm. What is the lateral area of this triangular prism?

Explanation: The area of two triangular faces is 1/2 × 5 × 8 = 5×4 = 20 cm 2 The area of two rectangular faces is 4×10= 40 cm 2 The lateral area is 2×20+2×40 = 40+80 = 120 cm 2

Question 1. Maria wants to know how much wax she will need to fill a candle mold shaped like a rectangular prism. What measure should she find?

Answer: Maria needs to find the volume of the mold.

Question 2. The outside of a closed glass display case measures 22 inches by 15 inches by \(\frac{1}{2}\) inches. The glass is 12 inch thick. How much air is contained in the case? _______ in. 3

Answer: 3381 in. 3

Explanation: The inner dimensions are 22-1× 15-1 × 12- 1/2 = 21 ×14×23/2 = 3381 in. 3

Question 3. A trapezoid with bases that measure 5 centimeters and 7 centimeters has a height of 4.5 centimeters. What is the area of the trapezoid? _______ cm 2

Answer: 27 cm 2

Explanation: Area of trapezoid= 1/2 ×(7+5)×4.5 = 6×4.5 = 27 cm 2

Question 4. Sierra has plotted two vertices of a rectangle at (3, 2) and (8, 2). What is the length of the side of the rectangle? _______ units

Answer: 5 units.

Explanation: The length of the side of the rectangle is 8-3= 5 units.

Answer: 104 m 2

Explanation: The area of the base 4×4= 16 The area of the one face is 1/2 × 4 × 11 = 2×11 = 22 m 2 The surface area of the square pyramid is 16+4×22 = 16+88 = 104 m 2

Question 6. A shipping company has a rule that all packages must be rectangular prisms with a volume of no more than 9 cubic feet. What is the maximum measure for the height of a box that has a width of 1.5 feet and a length of 3 feet? _______ feet

Answer: 2 feet.

Explanation: As given volume = 9 cubic feet So 1.5×3×H < 9 4.5×H < 9 H< 9/4.5 and H<2 So maximum measure for the height of the box is 2 feet.

Question 1. Elaine makes a rectangular pyramid from paper. The base is a _____. The lateral faces are _____. The base is a ___________ . The lateral faces are ___________ .

Answer: The base is a rectangle. The lateral faces are triangles.

Answer: b,c

Explanation: The expressions that show how to find the surface area is 2(20 × 12) + 2(15 × 12) + (20 × 15), (20 × 12) + (20 × 12) + (15 × 12) + (15 × 12) + (20 × 15)

Question 3. A prism is filled with 44 cubes with \(\frac{1}{2}\)-unit side lengths. What is the volume of the prism in cubic units? _______ \(\frac{□}{□}\) cubic unit

Explanation: The volume of a cube with S= (1/2) 3 = 1/2×1/2×1/2 = 1/8 = 0.125 cubic units As there are 44 cubes so 44×0.125=5.5 cubic units.

Question 4. A triangular pyramid has a base with an area of 11.3 square meters, and lateral faces with bases of 5.1 meters and heights of 9 meters. Write an expression that can be used to find the surface area of the triangular pyramid.

Answer: 11.3+ 3 × 1/2+ 5.1×9

Explanation: The expression that can be used to find the surface area of the triangular pyramid is 11.3+ 3 × 1/2+ 5.1×9

Question 5. Jeremy makes a paperweight for his mother in the shape of a square pyramid. The base of the pyramid has a side length of 4 centimeters, and the lateral faces have heights of 5 centimeters. After he finishes, he realizes that the paperweight is too small and decides to make another one. To make the second pyramid, he doubles the length of the base in the first pyramid. For numbers 5a–5c, choose Yes or No to indicate whether the statement is correct. 5a. The surface area of the second pyramid is 144 cm 2 . 5b. The surface area doubled from the first pyramid to the second pyramid. 5c. The lateral area doubled from the first pyramid to the second pyramid. 5a. ___________ 5b. ___________ 5c. ___________

Answer: 5a. True. 5b. False 5c. True.

Explanation: The area of the base is 4×4= 16 cm 2 . The area of one face is 1/2×4×5 = 2×5 = 10 cm 2 . The surface area of the First pyramid is 16+ 4×10 = 16+40 = 56 cm 2 . The area of the base is 8×8= 64 The area of one face is 1/2×8×5 = 4×5 = 20 cm 2 . The surface area od the second pyramid is 64+ 4×20 = 64+80 = 144 cm 2 .

Answer: 369 in 2

Explanation: The area of the base is 9×9= 81 in 2 The area of one face is 1//2 × 16× 9 = 8×9 = 72 in 2 The surface area of a square pyramid is 81+ 4× 72 = 81+ 288 = 369 in 2

Question 7. Dominique has a box of sewing buttons that is in the shape of a rectangular prism. The volume of the box is 2 \(\frac{1}{2}\) in. × 3 \(\frac{1}{2}\) in. × _____ = _____.

Answer: 17.5 in 3

Explanation: The volume of the box is 2 1/2 × 3 1/2 × 2 = 5/2 × 7/2 × 2 = 5/2 × 7 = 35/2 = 17.5 in 3

Question 8. Emily has a decorative box that is shaped like a cube with a height of 5 inches. What is the surface area of the box? _______ in. 2

Explanation: Surface area of the box is 6 a 2 So 6 × 5 2 = 6×5×5 2 = 150 in. 2

Answer: c, d

Explanation: 2 units ×1/2 unit × 2 1/2 units and 2.5 cubic units

Answer: 11a. True. 11b. True. 11c. False. 11d. False.

Explanation: The area of the face A is 4×2= 8 square units The area of the face B is 5×2= 10 square units The area of the face C is 5×4= 20 square units So the surface area is 2×8+2×10+2×20 = 16+20+40 = 76 square units

Question 12. Stella received a package in the shape of a rectangular prism. The box has a length of 2 \(\frac{1}{2}\) feet, a width of 1 \(\frac{1}{2}\) feet, and a height of 4 feet. Part A Stella wants to cover the box with wrapping paper. How much paper will she need? Explain how you found your answer

Answer: 39.5 ft 2

Explanation: The area of two lateral faces is 4 × 2 1/2 = 4 × 5/2 = 2×5 = 10 ft 2 The area of another two lateral faces is 4 × 1 1/2 = 4 × 3/2 = 2×3 = 6 ft 2 The area of the top and bottom is 2 1/2 × 1 1/2 = 5/2 × 3/2 = 15/4 = 3 3/4 ft 2 So Stella need 2×10+ 2×6 + 2 × 15/4 = 20+ 12+15/2 = 20+12+7.5 = 39.5 ft 2

Question 12. Part B Can the box hold 16 cubic feet of packing peanuts? Explain how you know

Answer: The box cannot hold 16 cubic feet of the packing peanuts

Explanation: Volume = LWH = 2 1/2 ×1 1/2 × 4 = 5/2 × 3/2 ×4 = 5×3 = 15 ft 3 So the box cannot hold 16 cubic feet of the packing peanuts.

Question 13. A box measures 6 units by \(\frac{1}{2}\) unit by 2 \(\frac{1}{2}\) units. For numbers 13a–13b, select True or False for the statement. 13a. The greatest number of cubes with a side length of \(\frac{1}{2}\) unit that can be packed inside the box is 60. 13b. The volume of the box is 7 \(\frac{1}{2}\) cubic units. 13a. ___________ 13b. ___________

Answer: 13a. True 13b. True.

Explanation: Length is 12 × 1/2= 6 units Width is 1× 1/2= 1/2 units Height is 5× 1/2= 5/2 units So, the greatest number of cubes with a side length of 1/2 unit that can be packed inside the box is 12×1×5= 60 The volume of the cube is S 3 The volume of a cube with S= (1/2) 3 = 1/2×1/2×1/2 = 1/8 = 0.125 cubic units As there are 60 cubes so 60×0.125= 7.5cubic units.

Question 14. Bella says the lateral area of the square pyramid is 1,224 in. 2 . Do you agree or disagree with Bella? Use numbers and words to support your answer. If you disagree with Bella, find the correct answer.

Answer: 900 in 2

Explanation: Area= 4× 1/2 bh = 4× 1/2 × 18 × 25 = 2× 18 × 25 =  900 in 2 So lateral area is 900 in 2 , so I disagree

Question 15. Lourdes is decorating a toy box for her sister. She will use self-adhesive paper to cover all of the exterior sides except for the bottom of the box. The toy box is 4 feet long, 3 feet wide, and 2 feet high. How many square feet of adhesive paper will Lourdes use to cover the box? _______ ft 2

Answer: 40 ft 2

Explanation: The area of two lateral faces is 4×2= 8 ft 2 The area of another two lateral faces is 3×2= 6 ft 2 The area of the top and bottom is 4×3= 12 ft 2 So Lourdes uses to cover the box is 2×8 + 2×6 + 12 = 16+12+12 = 40 ft 2

Answer: Length of shed 1= 12 ft Width of shed 2= 12 ft Height of shed 3= 6 ft Volume of shed 4= 1200 ft 3

Explanation: Volume= LWH Volume of shed1= 960 ft So 960= L×10×8 960= 80×L L= 960/80 L= 12 ft Volume of shed2= 2160 ft So 2160= 18×W×10 960= 180×W W= 2160/180 W= 12 ft Volume of shed3= 288 ft So 288= 12×4×H 288= 48×H H= 288/48 W= 6 ft Volume of shed2= 10×12×10 So V= 10×12×10 V= 1200 ft 3

Question 17. Tina cut open a cube-shaped microwave box to see the net. How many square faces does this box have? _______ square faces

Answer: The box has 6 square faces.

Answer: a and b can be used to represent Charle’s treasure box.

Question 19. Julianna is lining the inside of a basket with fabric. The basket is in the shape of a rectangular prism that is 29 cm long, 19 cm wide, and 10 cm high. How much fabric is needed to line the inside of the basket if the basket does not have a top? Explain your strategy. _______ cm 2

Answer: 1511 cm 2

Explanation: The surface area= 2(WL+HL+HW) The surface area of the entire basket= 2(19×29)+2(10×29)+2(10×19) = 2(551)+2(290)+2(190) = 1102+580+380 = 2,062 cm 2 The surface area of the top is 29×19= 551 So Julianna needs 2062-551= 1511 cm 2

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