Space Figures and Cross Sections

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Space Figures and Cross Sections GEOMETRY LESSON 11-1 (For help, go to Lesson 1-3.) For each exercise, make a copy of the cube at the right. Shade the plane that contains the indicated points. 1. A, B, and C 2. A, B, and G

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Space Figures and Cross Sections

GEOMETRY LESSON 11-1

(For help, go to Lesson 1-3.)

For each exercise, make a copy of the cube

at the right. Shade the plane that contains

the indicated points.

1. A, B, and C 2. A, B, and G

3. A, C, and G 4. A, D, and G

5. F, D, and G 6. B, D, and G

7. the midpoints of AD CD, EH, and GH

Check Skills You’ll Need

11-1

Space Figures and Cross Sections

GEOMETRY LESSON 11-1

1.2.

3. 4.

5. 6.

7.

Solutions

11-1

AF, BG, CH, DI, EJ, AB, BC, CD,

DE, EA, FG, GH, HI, IJ, and JF.

Space Figures and Cross Sections

GEOMETRY LESSON 11-1

How many vertices, edges, and faces of the

polyhedron are there? List them.

There are 10 vertices:

A, B, C, D, E, F, G, H, I, and J.

There are 15 edges:

There are 7 faces:

pentagons: ABCDE and FGHIJ, and

quadrilaterals: ABGF, BCHG, CDIH, DEJI, and EAFJ

Quick Check

11-1

Space Figures and Cross Sections

GEOMETRY LESSON 11-1

Use Euler’s Formula to find the number of edges on a solid with 6 faces and 8 vertices.

F+V= E+ 2 Euler’s Formula

6 + 8 = E+ 2 Substitute the number of faces and vertices.

12 = ESimplify.

A solid with 6 faces and 8 vertices has 12 edges.

Quick Check

11-1

Draw a net.

Space Figures and Cross Sections

GEOMETRY LESSON 11-1

Use the pentagonal prism from Example 1 to verify

Euler’s Formula. Then draw a net for the figure and verify

Euler’s Formula for the two-dimensional figure.

Use the faces F = 7, vertices V = 10, and edges E = 15.

F+V= E+ 2 Euler’s Formula

7 + 10 = 15 + 2 Substitute the number of faces and vertices.

Count the regions: F = 7

Count the vertices: V = 18

Count the segments: E = 24

F + V = E + 1 Euler’s Formula in two dimensions

7 + 18 = 24 + 1 Substitute.

Quick Check

11-1

Space Figures and Cross Sections

GEOMETRY LESSON 11-1

Describe this cross section.

The plane is parallel to the triangular base of the figure, so the cross section is also a triangle.

Quick Check

11-1

Space Figures and Cross Sections

GEOMETRY LESSON 11-1

Draw and describe a cross section formed by a vertical plane intersecting the top and bottom faces of a cube.

If the vertical plane is parallel to opposite faces, the cross section is a square.

Sample: If the vertical plane is not parallel to opposite faces, the cross section is a rectangle.

Quick Check

11-1

Space Figures and Cross Sections

GEOMETRY LESSON 11-1

1. Draw a net for the figure.

Sample:

Use Euler’s Formula to solve.

2. A polyhedron with 12 vertices and 30 edges has how many faces?

20

• 3. A polyhedron with 2 octagonal faces and 8 rectangular faces has how many vertices?
• 4. Describe the cross section.
• 5. Draw and describe a cross section formed by a vertical plane cutting the left and back faces of a cube.

16

Circle

Check students’ drawings; rectangle.

11-1

Surface Areas of Prisms and Cylinders

GEOMETRY LESSON 11-2

(For help, go to Lessons 1-9 and 10-3.)

Find the area of each net.

1.2.3.

Check Skills You’ll Need

11-2

Solutions

1. Each square has an area of (4)(4) = 16 cm2. The total area is 6(16) = 96 cm2.

2. The area of each circle is r 2= (2)2 = 4 cm2. The area of the rectangle is bh = (4 )(8) = 32 cm2. The total area of the two circles and the rectangle is 2(4 ) + 32 = 8 + 32 = 40 , 125.7 cm2.

3. An altitude from any vertex of a triangle measures 3 3. The

area of each triangle is bh = (6)(3 3) = 9 3 m2. The total area of the four triangles is 4(9 3) = 36 3 or about 62 m2.

1

2

1

2

Surface Areas of Prisms and Cylinders

GEOMETRY LESSON 11-2

11-2

Draw a net for the cube.

Surface Area = sum of areas of lateral faces + area of bases

= (121 + 121 + 121 + 121) + (121 + 121)

Surface Areas of Prisms and Cylinders

GEOMETRY LESSON 11-2

Quick Check

Use a net to find the surface area of the cube.

Find the area of one face. 112= 121

The area of each face is 121 in.2.

= 6 • 121

= 726

Because there are six identical faces, the surface area is 726 in.2.

11-2

S.A. = L.A. + 2B to find the surface area of the prism. The area B of the base is ap, where a is the apothem and p is the perimeter.

1

2

Draw the base.

Use 30°-60°-90° triangles to find the apothem.

Surface Areas of Prisms and Cylinders

GEOMETRY LESSON 11-2

Find the surface area of a 10-cm high right prism with triangular bases having 18-cm edges. Round to the nearest whole number.

The triangle has sides of length 18 cm, so p = 3 • 18 cm, or 54 cm.

11-2

9 = 3 alonger leg  3  shorter leg

9 3

3

3

3

a==

= 33Rationalize the denominator.

3354 =

81 3

1

2

1

2

B=ap =

The area of each base of the prism is 81 3 cm2.

S.A. =L.A. + 2BUse the formula for surface area.

= 540 + 162 3

2(81 3 ) Substitute

=ph + 2B

820.59223Use a calculator.

= (54)(10) +

9

3

Surface Areas of Prisms and Cylinders

GEOMETRY LESSON 11-2

Quick Check

(continued)

Rounded to the nearest whole number, the surface area is 821 cm2.

11-2

2( r2)

=2rh + Substitute the formula for lateral area of a cylinder and area of a circle.

= 2 (6)(9) + 2 (62) Substitute 6 for r and 9 for h.

= 108 + Simplify.

= 180

The surface area of the cylinder is 180 ft2.

72

Surface Areas of Prisms and Cylinders

GEOMETRY LESSON 11-2

The radius of the base of a cylinder is 6 ft, and its height is 9 ft. Find its surface area in terms of .

S.A. =L.A. + 2BUse the formula for surface area of a cylinder.

Quick Check

11-2

Cornmeal Container Barley Container

S.A. =L.A. + 2B

Use the formula for surface area of a cylinder.

S.A. =L.A. + 2B

Substitute the formulas for lateral area of a cylinder and area of a circle.

=2rh + 2 r 2

=2rh + 2 r 2

d

2

Find the surface area of each container. Remember that r= .

Surface Areas of Prisms and Cylinders

GEOMETRY LESSON 11-2

A company sells cornmeal and barley in cylindrical containers. The diameter of the base of the 6-in. high cornmeal container is 4 in. The diameter of the base of the 4-in. high barley container is 6 in. Which container has the greater surface area?

11-2

Substitute the formulas for lateral area of a cylinder and area of a circle.

=rh + r 2

=rh + r 2

2

2

2

2

Substitute

for r and h.

= (2)(6) + (22)

= (3)(4) + (32)

2

2

2

2

Simplify.

= 24 +

= 24 +

8

= 32

= 42

Because 42 in.2 32 in.2, the barley container has the greater surface area.

18

Surface Areas of Prisms and Cylinders

GEOMETRY LESSON 11-2

(continued)

Cornmeal Container Barley Container

S.A. =L.A. + 2B

Use the formula for

surface area of a cylinder.

S.A. =L.A. + 2B

Quick Check

11-2

Use the prism below for Exercises 1 and 2.

1. Use a net to find the surface area.

2. Use a formula to find the surface area.

3. The height of a prism is 5 cm. Its rectangular bases have 3-cm and

9-cm sides. Find its surface area.

4. The radius of the base of a cylinder is 16 in., and its height is 4 in. Find

its surface area in terms of .

5. A contractor paints all but the bases of a 28-ft high cylindrical water

tank. The diameter of the base is 22 ft. How many square feet are

painted? Round to the nearest hundred.

S.A. = 216 ft2

640 in.2

Surface Areas of Prisms and Cylinders

GEOMETRY LESSON 11-2

S.A. = L.A. + 2B = 168 + 48 = 216; 216 ft2

174 cm2

1900 ft2

11-2

Surface Areas of Pyramids and Cones

GEOMETRY LESSON 11-3

(For help, go to Lesson 8-1.)

Find the length of the hypotenuse in simplest radical form.

1.2.3.

Check Skills You’ll Need

11-3

Surface Areas of Pyramids and Cones

GEOMETRY LESSON 11-3

Solutions

1. Use the Pythagorean Theorem: a2 + b2 = c2 (8)2 + (13)2 = x2

64 + 169 = x2x2 = 233 x = 233 in.

2. Use the Pythagorean Theorem: a2 + b2 = c2 (7)2 + (9)2 = x2

49 + 81 = x2x2 = 130 x = 130 m

3. Use the Pythagorean Theorem: a2 + b2 = c2 (12)2 + (13)2 = x2

144 + 169 = x2x2 = 313 x = 313 cm

11-3

You are given = 12 ft and you found that p= 30, so you can find the lateral area.

1

2

L.A. = p Use the formula for lateral area of a pyramid.

1

2

= (30)(12)Substitute.

Surface Areas of Pyramids and Cones

GEOMETRY LESSON 11-3

Find the surface area of a square pyramid with base edges 7.5 ft and slant height 12 ft.

The perimeter p of the square base is 4 X 7.5 ft, or 30 ft.

= 180Simplify.

11-3

Find the area of the square base.

S.A. = L.A. BUse the formula for surface area of a pyramid.

= 180  56.25Substitute.

Surface Areas of Pyramids and Cones

GEOMETRY LESSON 11-3

(continued)

Because the base is a square with side length 7.5 ft,

Bs2 7.52 56.25.

= 236.25Simplify.

The surface area of the square pyramid is 236.25 ft 2.

Quick Check

11-3

1

2

Use the formula L.A. =p to find the lateral area of the pyramid.

The altitude of the pyramid, apothem of the base, and altitude of a lateral face form a right triangle, so you can use the Pythagorean Theorem to find the slant height .

= 202 + (4 3)2 = 400 + 48 = 448

Surface Areas of Pyramids and Cones

GEOMETRY LESSON 11-3

Find the lateral area of the hexagonal pyramid below. Round your answer to the nearest whole number.

11-3

1

2

L.A. =pUse the formula for lateral area.

1

2

= (48)( 448) Substitute.

507.98425 Use a calculator.

Surface Areas of Pyramids and Cones

GEOMETRY LESSON 11-3

(continued)

The lateral area of the hexagonal pyramid is about 508 m2.

Quick Check

11-3

= r+r 2Substitute the formulas for L.A. and B.

= (5)(13) + (5)2Substitute.

= 65 + 25Simplify.

= 90

The surface area of the cone is 90 in.2.

Surface Areas of Pyramids and Cones

GEOMETRY LESSON 11-3

Find the surface area of the cone in terms of .

S.A. = L.A.+ BUse the formula for surface area of a cone.

Quick Check

11-3

The altitude of the cone, radius of the base, and slant height form a right triangle. Use the Pythagorean Theorem to find the slant height .

= 0.52  1.52 = 0.25  2.25 = 2.5

Surface Areas of Pyramids and Cones

GEOMETRY LESSON 11-3

Leandre uses paper cones to cover her plants in the early spring. The diameter of each cone is 1 ft, and its height is 1.5 ft. How much paper is in the cone? Round your answer to the nearest tenth.

The cone’s diameter is 1 ft, so its radius r is 0.5 ft.

11-3

L.A. =r Use the formula for lateral area of a cone.

= (0.5) 2.5 Substitute 0.5 for r and 2.5 for .

2.4836471 Use a calculator.

Surface Areas of Pyramids and Cones

GEOMETRY LESSON 11-3

(continued)

Find the lateral area.

The lateral area of the cone is about 2.5 ft.2

Quick Check

11-3

1. Find the slant height of a square pyramid with base edges 12 cm and altitude 8 cm.

2. Find the lateral area of the regular

square pyramid to the right.

3. Find the surface area of the pyramid to the right

whose base is a regular hexagon. Round to

the nearest whole number.

4. Find the surface area of a cone with radius 8 cm and slant height 17 cm in terms of .

5. The roof of a building is shaped like a cone with diameter 40 ft and height 20 ft. Find the area of the roof. Round to the nearest whole number.

200 cm2

Surface Areas of Pyramids and Cones

GEOMETRY LESSON 11-3

10 cm

56 in.2

1517 ft2

1777 ft2

11-3

Volumes of Prisms and Cylinders

GEOMETRY LESSON 11-4

(For help, go to Lessons 1-9 and 10-1.)

Find the area of each figure. Round to the nearest tenth if necessary.

1. square - side length 7 cm

3. circle - radius 10 mm

5. rectangle - 14 in. by 11 in.

7. equilateral triangle - side length 8 in.

2. circle - diameter 15 in.

4. rectangle - 3 ft by 1 ft

6. triangle - base 11 cm, height 5 cm

Check Skills You’ll Need

11-4

1.A = s2 = 72 = 49 cm2

2.A = r2 = (7.5)2 = 56.25 176.7 in.2

3.A = r2 = (10)2 = 100 314.2 mm2

4.A = w = (3)(1) = 3 ft2

5.A = w = (14)(11) = 154 in.2

6.A = bh = (11)(5) = 27.5 cm2

7. A = bh = (8)(4 3) = 16 3 27.7 in.2

1

2

1

2

1

2

1

2

Volumes of Prisms and Cylinders

GEOMETRY LESSON 11-4

Solutions

11-4

The area of the base B=w= 3  5 = 15.

Volumes of Prisms and Cylinders

GEOMETRY LESSON 11-4

Find the volume of the prism below.

V= Bh Use the formula for volume.

= 15 • 5Substitute 15 for B and 5 for h.

= 75Simplify.

The volume of the rectangular prism is 75 in.3.

Quick Check

11-4

292 – 202 = 841  400 = 441  21

Volumes of Prisms and Cylinders

GEOMETRY LESSON 11-4

Find the volume of the prism below.

The prism is a right triangular prism with triangular bases.

The base of the triangular prism is a right triangle where one leg is the base and the other leg is the altitude.

11-4

1

2

1

2

The area B of the base is bh= (20)(21) = 210. Use the area of the base to find the volume of the prism.

Volumes of Prisms and Cylinders

GEOMETRY LESSON 11-4

(continued)

V= Bh Use the formula for the volume of a prism.

= 210 •40Substitute.

= 8400Simplify.

The volume of the triangular prism is 8400 m3.

Quick Check

11-4

The formula for the volume of a cylinder is V= r 2h. The diagram shows h and d, but you must find r.

1

2

r= d = 8

V= r 2h Use the formula for the volume of a cylinder.

= • 82•9Substitute.

= 576Simplify.

The volume of the cylinder is 576 ft3.

Volumes of Prisms and Cylinders

GEOMETRY LESSON 11-4

Find the volume of the cylinder below. Leave your answer in terms of .

Quick Check

11-4

You can use three rectangular prisms to find the volume.

Volumes of Prisms and Cylinders

GEOMETRY LESSON 11-4

Find the volume of the composite space figure.

Each prism’s volume can be found using the formula V = Bh.

11-4

Volumes of Prisms and Cylinders

GEOMETRY LESSON 11-4

(continued)

Volume of prism I = Bh = (14 • 4) • 25 = 1400

Volume of prism II = Bh = (6 • 4) • 25 = 600

Volume of prism III = Bh = (6 • 4) • 25 = 600

Sum of the volumes = 1400 + 600 + 600 = 2600

The volume of the composite space figure is 2600 cm3.

Quick Check

11-4

Volumes of Prisms and Cylinders

GEOMETRY LESSON 11-4

Find the volume of each figure to the nearest whole number.

1. 2. 3.

4.5.

62 m3

1800 ft3

45 in.3

63 m3

1800 mm3

11-4

Volumes of Pyramids and Cones

GEOMETRY LESSON 11-5

(For help, go to Lesson 8-1.)

Use the Pythagorean Theorem to find the value of the variable.

1. 2. 3.

Check Skills You’ll Need

11-5

Volumes of Pyramids and Cones

GEOMETRY LESSON 11-5

Solutions

1. a2 + b2 = c2h2 + (9)2 = (15)2 h2 + 81 = 225h2 = 225 – 81 = 144 h = 144 = 12 cm

2. One leg is half the length of the side of the base, 6 in. a2 + b2 = c2h2 + (6)2 = (10)2h2 + 36 = 100 h = 64 = 8 in.

3. a2 + b2 = c2 (2)2 + (1.5)2 = 2 4 + 2.25 = 2 2 = 6.252 = 6.25 = 2.5 m

11-5

1

3

1

3

V=BhUse the formula for volume of a pyramid.

= (225)(22) Substitute 225 for B and 22 for h.

Volumes of Pyramids and Cones

GEOMETRY LESSON 11-5

Find the volume of a square pyramid with base edges 15 cm and height 22 cm.

Because the base is a square, B= 15 • 15 = 225.

= 1650 Simplify.

The volume of the square pyramid is 1650 cm3.

Quick Check

11-5

Volumes of Pyramids and Cones

GEOMETRY LESSON 11-5

Find the volume of a square pyramid with base edges 16 m and slant height 17 m.

The altitude of a right square pyramid intersects the base at the center of the square.

11-5

Because each side of the square base is 16 m, the leg of the right triangle along the base is 8 m, as shown below.

172= 82h2Use the Pythagorean Theorem.

289 = 64 h2Simplify.

Volumes of Pyramids and Cones

GEOMETRY LESSON 11-5

(continued)

Step 1: Find the height of the pyramid.

225 =h2Subtract 64 from each side.

h= 15 Find the square root of each side.

11-5

1

3

1

3

V=BhUse the formula for the volume of a pyramid.

= (16  16)15Substitute.

Volumes of Pyramids and Cones

GEOMETRY LESSON 11-5

(continued)

Step 2: Find the volume of the pyramid.

= 1280 Simplify.

The volume of the square pyramid is 1280 m3.

Quick Check

11-5

Find the volume of the cone below in terms of .

1

2

r= d = 3 in.

1

3

V=r 2hUse the formula for volume of a cone.

1

3

= (32)(11) Substitute 3 for r and 11 for h.

= 33 Simplify.

The volume of the cone is 33 in.3.

Volumes of Pyramids and Cones

GEOMETRY LESSON 11-5

Quick Check

11-5

1

3

d

2

V= r 2h Use the formula for the volume of a cone.

r= = 2

1

3

V= (22)(7)Substitute 2 for r and 7 for h.

29.321531Use a calculator.

Volumes of Pyramids and Cones

GEOMETRY LESSON 11-5

An ice cream cone is 7 cm tall and 4 cm in diameter. About how much ice cream can fit entirely inside the cone? Find the volume to the nearest whole number.

About 29 cm3 of ice cream can fit entirely inside the cone.

Quick Check

11-5

1470 mm3

3 m3

Volumes of Pyramids and Cones

GEOMETRY LESSON 11-5

Find the volume of each figure. When appropriate, leave your answer in

terms of .

1.2.

3. square pyramid with base edges 24 in. long and slant height 15 in.

4. cone with diameter 3 m and height 4 m

5.

60 ft3

1728 in.3

150 ft3

11-5

Surface Areas and Volumes of Spheres

GEOMETRY LESSON 11-6

(For help, go to Lesson 1-9.)

Find the area and circumference of a circle with the given radius. Round your answers to the nearest tenth.

1. 6 in. 2. 5 cm 3. 2.5 ft

4. 1.2 m 5. 15 yd 6. 12 mm

Check Skills You’ll Need

11-6

2 r

2 r

2 r

2 r

2 r

2 r

Surface Areas and Volumes of Spheres

GEOMETRY LESSON 11-6

Solutions

1. A = r 2 = (6)2 = 36 113.1 in.2, C = = 2 (6) = 12 37.7 in.

2. A = r 2 = (5)2 = 25 78.5 cm2, C = = 2 (5) = 10 31.4 cm

3. A = r 2 = (2.5)2 = 6.25 19.6 ft2, C = = 2 (2.5) = 5 15.7 ft

4. A = r 2 = (1.2)2 = 1.44 4.5 m2, C = = 2 (1.2) = 2.4 7.5 m

5. A = r 2 = (15)2 = 225 706.9 yd2, C = = 2 (15) = 30 94.2 yd

6. A = r 2 = (12)2 = 144 452.4 mm2, C = = 2 (12) = 24 75.4 mm

11-6

18

2

= 4 • 92Substitute r== 9.

= 324 Simplify.

The surface area of the sphere is 324 ft2.

Surface Areas and Volumes of Spheres

GEOMETRY LESSON 11-6

Find the surface area of the sphere below. Leave your answer in terms of .

Quick Check

11-6

C=2 r Use the formula for circumference.

13=2 r Substitute 13 for C.

13

2

13

2

= r Solve for r.

S.A. = 4r 2Use the formula for the surface area of a sphere.

13

2

= 4• ( )2Substitute for r.

169

=Simplify.

53.794371Use a calculator.

Surface Areas and Volumes of Spheres

GEOMETRY LESSON 11-6

Quick Check

The circumference of a rubber ball is 13 cm. Calculate its surface area to the nearest whole number.

Step 2: Use the radius to find the surface area.

To the nearest whole number, the surface area of the rubber ball is 54 cm2.

11-6

4

3

V= r 3Use the formula for the volume of a sphere.

30

2

4

3

= • 153Substitute r = = 15.

= 4500Simplify.

The volume of the sphere is 4500 cm3.

Surface Areas and Volumes of Spheres

GEOMETRY LESSON 11-6

Find the volume of the sphere. Leave your answer in terms of .

Quick Check

11-6

4

3

V= r 3Use the formula for the volume of a sphere.

4

3

1= r 3Substitute.

3

3

4

3

4

= r 3Solve for r 3.

= r Find the cube root of each side.

r0.62035049Use a calculator.

Surface Areas and Volumes of Spheres

GEOMETRY LESSON 11-6

The volume of a sphere is 1 in.3. Find its surface area to the nearest tenth.

Step 1: Use the volume to find the radius r.

11-6

S.A. = 4r 2Use the formula for the surface area of a sphere.

4.8359744Use a calculator.

4(0.62035049)2Substitute.

Surface Areas and Volumes of Spheres

GEOMETRY LESSON 11-6

(continued)

Step 2: Use the radius to find the surface area.

To the nearest tenth, the surface area of the sphere is 4.8 in.2.

Quick Check

11-6

For Exercises 1–3, find the surface area. Leave your answer in terms of .

1. a sphere whose diameter is 13 cm

2. a ball whose circumference is 19 mm

3. a sphere whose volume is 288 in.3

For Exercises 4 and 5, find the volume. Round your answer to the nearest

whole number.

4. sphere with radius 5 m

5. sphere with diameter 7 yd

169 cm2

361 mm2

144 in.2

Surface Areas and Volumes of Spheres

GEOMETRY LESSON 11-6

524 m3

180 yd3

11-6

Areas and Volumes of Similar Solids

GEOMETRY LESSON 11-7

(For help, go to Lessons 8-2 and 11-4.)

Are the figures similar? Explain. Include the similarity ratio.

1. two squares, one with 3-in. sides and the other with 1-in. sides

2. two isosceles right triangles, one with a 3-cm hypotenuse and the other a 1-cm leg

Find the volume of each space figure.

3. cube with a 3-in. edge

4. 3-m by 5-m by 9-m rectangular prism

5. cylinder with radius 4 cm and height 8 cm

Check Skills You’ll Need

11-7

3

2

3 2

2

Areas and Volumes of Similar Solids

GEOMETRY LESSON 11-7

Solutions

1. Yes, four pairs of angles are congruent and the corresponding sides are proportional with a similarity ratio of .

2. Yes, three pairs of angles are congruent and the corresponding sides are proportional with a similarity ratio of .

3. To find the volume, use the formula V = s3: V = (3)3 = (3)(3)(3) = 27 in.3

4. To find the volume, use the formula V = bhl: V = (3)(5)(9) = 135 m3.

3

1

=

5. The volume of the cylinder is V = Bh, where B is the area of the base. The base is a circle whose area is 16 .So, V = 16 (8) = 128  402.1 cm3.

11-7

8

26

8

26

3

9

The ratio of the radii is , and the ratio of the height is .

/

3

9

The cones are not similar because = .

Areas and Volumes of Similar Solids

GEOMETRY LESSON 11-7

Are the two solids similar? If so, give the similarity ratio.

Both solid figures have the same shape. Check that the ratios of the corresponding dimensions are equal.

Quick Check

11-7

98

2

= The ratio of the surface areas is a2 : b2.

49

1

a2

b2

a2

b2

= Simplify.

a

b

7

1

= Take the square root of each side.

Areas and Volumes of Similar Solids

GEOMETRY LESSON 11-7

Find the similarity ratio of two similar cylinders with surface areas of 98 ft2 and 2 ft2.

Use the ratio of the surface areas to find the similarity ratio.

The similarity ratio is 7 : 1.

Quick Check

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48

162

= The ratio of the volumes is a3 : b3.

8

27

= Simplify.

a

b

a3

b3

a3

b3

2

3

= Take the cube root of each side.

Areas and Volumes of Similar Solids

GEOMETRY LESSON 11-7

Two similar square pyramids have volumes of 48 cm3 and 162 cm3. The surface area of the larger pyramid is 135 cm2. Find the surface area of the smaller pyramid.

Step 1: Use the ratio of the volumes to find the similarity ratio.

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22

32

= The ratio of the surface areas is a2 : b2.

S1

S2

S1

S2

4

9

= Simplify.

S1

135

4

9

= Substitute 135 for S2, the surface area of the larger pyramid.

Quick Check

4

9

S1= • 135 Solve for S1.

Areas and Volumes of Similar Solids

GEOMETRY LESSON 11-7

(continued)

Step 2: Use the similarity ratio to find the surface area S1 of the smaller pyramid.

S1= 60 Simplify.

The surface area of the smaller pyramid is 60 cm2.

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Areas and Volumes of Similar Solids

GEOMETRY LESSON 11-7

A box of detergent shaped like a rectangular prism is 6 in. high and holds 3.25 lb of detergent. How much detergent would a similar box that is 8 in. tall hold? Round your answer to the nearest tenth.

The ratio of the heights of the boxes is 6 : 8, so the similarity ratio is 6 : 8, or 3 : 4 in simplest terms.

Because the weights are proportional to the volumes, the ratio of the weights equals the cube of their similarity ratio: 33 : 43, or 27 : 64.

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27

64

3.25x

= Let x  the weight of the larger box of detergent.

x 7.7037037 Divide each side by 27.

Areas and Volumes of Similar Solids

GEOMETRY LESSON 11-7

(continued)

27x= 64  3.25 Cross-Product Property.

27x= 208 Simplify.

The box that is 8 in. tall would hold about 7.7 lb of detergent.

Quick Check

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For Exercises 1 and 2, are the two solids similar? If so, give the similarity

ratio of the smaller figure to the larger figure.

1.2.

3. Find the similarity ratio of two spheres with volumes of 20 m3 and 160 m3.

4. The volumes of two similar solids are 54 ft3 and 250 ft3. The surface area of the smaller solid is 45 ft2. Find the surface area of the larger solid.

5. A solid chocolate rabbit is 6 in. high and weighs 0.25 lb. A similar chocolate rabbit is 12 in. high. How much does it weigh?

Areas and Volumes of Similar Solids

GEOMETRY LESSON 11-7

yes; 2 : 3

no

1 : 2

125 ft2

2 lb

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