# Volume of Pyramids and Cones Worksheet - Page 2

Volume of Pyramids and Cones Worksheet
• Page 2
11.
If the side of each cube in the shown pyramid is 4 cm, then find the total volume occupied by the cubes.

 a. 7676 cm3 b. 7670 cm3 c. 7680 cm3 d. 7665 cm3

#### Solution:

Assume the pyramid as the layers of cubes.

Number of cubes in the first layer = 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36

Number of cubes in the second layer = 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28

Number of cubes in the third layer = 6 + 5 + 4 + 3 + 2 + 1 = 21

Number of cubes in the fourth layer = 5 + 4 + 3 + 2 + 1 = 15

Number of cubes in the fifth layer = 4 + 3 + 2 + 1 = 10

Number of cubes in the sixth layer = 3 + 2 + 1 = 6

Number of cubes in the seventh layer = 2 + 1 = 3

Number of cubes in the eighth layer = 1

So, total number of cubes in the pyramid = 36 + 28 + 21 + 15 + 10 + 6 + 3 + 1 = 120

Volume of each cube = 64 cm3
[Side of each cube = 4 cm.]

The volume occupied by the cubes of the pyramid = Total number of cubes × volume of each cube = 120 × 64 = 7680 cm 3

12.
Find the volume of a cone and round the answer to the nearest whole unit.

 a. 2366.3 in3 b. 2366 in3 c. 2365 in3 d. 2377 in3

#### Solution:

Volume of a cone, V = 13 π r2 h
[Formulae.]

= 13 × 3.141 × 10.5 × 10.5 × 20.5
[ r =10.5; h = 20.5 and substituting the values.]

= 2366.3 in.3 = 2366 in.3
[ Simplify and round the answer to the nearest whole unit.]

The volume of the cone to the nearest whole unit is 2366 in.3.

13.
Find the volume of a cone, if the diameter of the cone is 12.4 cm. and the height of the cone is 30 cm.
 a. 1206 $\mathrm{cm}$3 b. 1207.4 $\mathrm{cm}$3 c. 1207 $\mathrm{cm}$3 d. 1208 $\mathrm{cm}$3

#### Solution:

Volume of a cone, V = 13 π r2 h
[Formulae.]

r = 6.2 cm.; h = 30 cm.
[ Radius is half of diameter.]

V = 13 × 3.141 × 6.2 × 6.2 × 30
[Substituting the values.]

= 1207.4 cm.3
[Simplify.]

Therefore, the volume of the cone is 1207.4 cm.3.

14.
Find the volume of the pyramid.

 a. 52.26 cm3 b. 52 cm3 c. 53 cm3 d. 51 cm3

#### Solution:

Volume of the pyramid, V = 13 B h
[Formulae.]

V = 13 × 49 × 3.2
[Here, B = s × s (i.e., area of a square) and substituting the values.]

= 52.26 cm.3
[Simplify.]

So, the volume of the pyramid is 52.26 cm.3.

15.
Find the volume of the figure given.

 a. 314 π in3 b. 110 π in3 c. 3.14 π in3 d. 100 π in3

#### Solution:

Volume of a cone, V= 13 π r2 h
[Formula.]

Here, r = 5 in.; h = 6 + 6 = 12 in.

V = 13 × π × 5 × 5 × 12
[Substituting the values.]

= 100 π in.3
[Simplify.]

The volume of a given figure is 100 π in.3.

16.
Base radius of a cone is 5 in. and its height is 15 in. If the height and the radius of the cone are doubled, then find the volume of the new cone.[Use $\pi$ = 3.]
 a. 1500 in.3 b. 9000 in.3 c. 900 in.3 d. 3000 in.3

#### Solution:

Base radius of the new cone = 2 × radius of the initial cone = 2 × 5 = 10 in.
[Since base radius of new cone is double the radius of initial cone.]

Height of the new cone = 2 × height of the initial cone = 2 × 15 = 30 in.
[Since height of new cone is double the height of initial cone.]

Volume of cone = 1 / 3πr2h
[Formula.]

= 3000
Volume of new cone = 1 / 3× 3 × 102 × 30
[Substitute the values in the formula and simplify.]

Volume of the new cone is 3000 in.3.

17.
The circumference of the base of a cone of height 15 in. is 44 in. Find the volume of the cone. (Use π = $\frac{22}{7}$)
 a. 770 in.3 b. 766 in.3 c. 784 in.3 d. 774 in.3

#### Solution:

Circumference of the base of a cone = 2πr
[Formula.]

r = 44 in.
[Since circumference of a cone is 44 in.]

2 × 227 × r = 44

r = 7 in.
[Simplify.]

Volume of the cone = 1 / 3πr2h
[Formula.]

= 13 × 227 × 72 × 15
[Substitute the values.]

= 770
[Simplify.]

Volume of the cone = 770 in.3.

18.
A bucket is in the shape of a frustum of a cone with a height($x$) of 12 cm, diameter of the top portion($a$) 48 cm and diameter of bottom portion($b$) 16 cm. Find the capacity of the bucket. [Take $\pi$ = 3.]

 a. 29952 cm3 b. 2688 cm3 c. 9984 cm3 d. 2304 cm3

#### Solution:

The bucket can be considered as a cone with the upper part removed.
[Analysis.]

ΔFBA represents the full cone, ΔEBD represents the part of the cone removed.
[From the figure.]

ΔOAB ~ ΔCDB.
[From the figure.]

OA / CD= OB / BC
[From step 3.]

482162=12+yy
[Substitute.]

y = 16×1248-16 6
[Simplify.]

The volume of the bucket = Volume of the cone with base diameter 48 cm - Volume of the cone with base diameter 16 cm
[Formula.]

Height of cone with diameter 48 cm = 6 + 12 18 cm

Height of cone with diameter 16 cm = 6 cm

Volume of the cone with diameter 48 cm = 1 / 3× 3 × 4824 × 18 10368 cm3
[Volume of the cone = 1 / 3π r2 h.]

Volume of the cone with diameter 16 cm = 1 / 3× 3 × 4824 × 6 = 384 cm3
[Formula.]

Volume of the bucket = 10368 - 384 9984 cm3
[Simplify.]

Capacity of the bucket = 9984 cm3

19.
The circumference of the base of a cone of height 18 in. is 44 in. Find the volume of the cone.
 a. 939.18 in.3 b. 924.38 in.3 c. 919.58 in.3 d. 929.18 in.3

#### Solution:

Circumference of the base of a cone = 2πr
[Formula.]

r = 44 in.
[Since circumference of a cone is 44 in.]

2 × π × r = 44

r = 7.0029 in.
[Simplify.]

Volume of the cone = 1 / 3πr2h
[Formula.]

= 924.38
[Simplify.]

Volume of the cone = 924.38 in.3.

20.
Which of the following statements is true?
 a. The volume of a pyramid is $\frac{1}{3}$ times the product of base area and height. b. The volume of a prism is $\frac{1}{3}$ times the sum of base area and height. c. The volume of a pyramid is $\frac{1}{2}$ times the product of base area and height. d. The volume of a prism is $\frac{1}{2}$ times the product of base area and its height.

#### Solution:

The volume of a pyramid is 1 / 3 times the product of base area and height.

So, the statement, 'The volume of a pyramid is 1 / 3 times the product of base area and height' is true.