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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

Step by Step Solutions explained for all the questions in the BIM Geometry Answer Key Chapter 11 Circumference, Area, and Volume are easy to learn and understand as they are prepared by professional subject experts. After referring to the lesson-wise Big Ideas Math Geometry Chapter 11 Circumference, Area, and Volume Solution Key every high school student can become a pro in the concepts and gain better subject knowledge. So, click on the respective links and start preparing each topic of the BIM Geometry Answer Key Chapter 11 Circumference, Area, and Volume efficiently.

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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