﻿ Similar Solids Worksheet | Problems & Solutions

# Similar Solids Worksheet

Similar Solids Worksheet
• Page 1
1.
Two similar cylinders have heights 3 cm and 2 cm. What is the similarity ratio?.
 a. 3 : 2 b. 4 : 3 c. 3 : 4 d. 2 : 3

#### Solution:

Heights of the two similar cylinders are a = 3 cm and b = 2 cm.
[Given.]

The ratio of corresponding dimensions of two similar solids is the similarity ratio.
[Definition.]

Similarity ratio = Height of the first cylinderHeight of the second cylinder

a : b = 3 : 2
[Substitute 3 for a and 2 for b.]

2.
Two similar cones have heights 8 cm and 4 cm. What is the ratio of their surface areas?
 a. 2 : 1 b. 6 : 1 c. 8 : 1 d. 4 : 1

#### Solution:

Heights of the two similar cylinders are a = 8 cm and b = 4 cm.
[Given.]

The ratio of corresponding dimensions of two similar solids is the similarity ratio.
[Definition.]

Similarity ratio = Height of the first coneHeight of the second cone

= ab = 8 / 4 = 2 : 1
[Substitute 8 for a and 4 for b.]

The ratio of their corresponding surface area is a2 : b2.
[Theorem.]

= (ab)² = 4 : 1
[Substitute and simplify.]

3.
The radii of two spheres are 3 cm and 2 cm. Find the ratio of their volumes approximately.
 a. 9 : 4 b. 27 : 8 c. 3 : 2 d. 2 : 3

#### Solution:

The radii of the two spheres are a = 3 cm and b = 2 cm.
[Given.]

[Formula.]

= ab = 3 / 2
[Substitute 3 for a and 2 for b.]

The ratio of Volumes of the solids is a3 : b3 = 27 : 8 .
[Formula.]

4.
The ratio of areas of two similar solids is 9 : 1. What is the similarity ratio of the solids?
 a. 3 : 1 b. 9 : 1 c. 1 : 3 d. 1 : 9

#### Solution:

a²b² = 91
[The ratio of the areas is a2 : b2.]

ab = 31
[Find the square root of each side.]

So, the similarity ratio of the two solids is a : b = 3 : 1

5.
The ratio of volumes of two similar solids is 27 : 8. What is the similarity ratio of the solids?
 a. 3 : 2 b. 9 : 4 c. 4 : 9 d. 2 : 3

#### Solution:

a3b3 = 278
[The ratios of the volumes is a3 : b3.]

ab = 32
[Find the cube root of each side.]

So, the similarity ratio of the two solids is a : b = 3 : 2 .

6.
The ratio of the base areas of two similar cylinders is 1 : 4. What is the ratio of the volumes of the solids?
 a. 1 : 2 b. 8 : 1 c. 1 : 4 d. 1 : 8

#### Solution:

a2b2 = 1 / 4
[The ratios of the base areas is a2 : b2.]

ab = 1 / 2
[Find the square root on each side.]

So, a3b3 = (ab)3 = (12)3 = 1 / 8
[Substitute 1 / 2 for ab.]

The ratio of volumes of the two solids is a3 : b3 = 1 : 8.

7.
The ratio of volume of two similar cones is 1 : 512. Find the ratio of the base areas of the two solids.
 a. 64 : 1 b. 1 : 64 c. 1 : 8 d. 8 : 1

#### Solution:

a3b3 = 1 / 512
[The ratios of the volume is a3 : b3.]

ab = 1 / 8
[Find the cube root of each side.]

So, a2b2=(ab)2=(18)2 = 164.
[Substitute (1 / 8) for ab.]

The ratio of base areas of the two solids is a2 : b2 = 1 : 64.

8.
The ratio of the volumes of two similar solids is 8 : 27. What is the volume of the smallest one?
 a. 3 cubic units b. 2 cubic units c. 8 cubic units d. cannot be determined

#### Solution:

The ratio of the volumes is 8 : 27.
[Given.]

a3b3=827
[The ratio of the volume is a3 : b3.]

Without knowing the volume of the larger solid b3, the volume of the smaller solid cannot be determined.

9.
The heights of two similar solids are 4 cm and $x$ cm. The similarity ratio of these two similar solids is 2 : 3. Find the value of $x$.
 a. 4 b. 7 c. 6 d. 8

#### Solution:

ab=23
[Similarity ratio is a : b.]

4x =23
[Substitute 4for a and x for b.]

x = 6
[Cross product property.]

10.
The surface area of two similar spheres are 144$\pi$ cm2 and 16$\pi$ cm2. The volume of the smaller solid is $\frac{32}{3}$$\pi$ cm3. What is the volume of the larger sphere?
 a. 144 $\pi$ cm3 b. 288 $\pi$ cm3 c. 288 cm3 d. 144 cm3

#### Solution:

a2b2=144π16π
[The ratio of the surface areas is a2 : b2.]

a2b2 = 9
[Divide numerator and denominator by the common factor 16π.]

ab = 3
[Find the square root of each side.]

Let v1, v2 be the volumes of the two similar spheres.

v1v2=33 v1 = v2(27)
[The ratios of the volumes is a3 : b3.]

v1 = 32 / 3π(27)
[Substitute 32 / 3π for v2.]

v1 = 288 π
[Cross product property.]

The volume of the larger sphere is 288 π cm3.