Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 32

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Weekly Quiz Competition

Question 1
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The height of the frustum cone is 60 cm. The area of the top and bottom circle of the cone is 10 cm$$^2$$ and 10 cm$$^2$$ respectively. Find its volume.

A
500 cm$$^3$$

B
600 cm$$^3$$

C
700 cm$$^3$$

D
800 cm$$^3$$

Solution

Volume of the frustum cone is V = $$\displaystyle{\frac{h}{3}}[\pi R^2 + \pi r^2 + \sqrt{(\pi R^2)(\pi r^2)}$$]
Area of the top cone = 10 cm$$^2$$
Area of the bottom cone = 10 cm$$^2$$
Height = 60 cm
Volume = $$\displaystyle{\frac{60}{3}}[10 + 10 + \sqrt{10*10}$$ ]
$$= 20[20 + \sqrt{100}]$$
$$= 20[20 + 10]$$
$$= 20 \times 30$$
$$= 600 \,cm^3$$
Question 2
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The volume of frustum of a cone is 300 m$$^3$$. The area of the bottom circle of the cone is 90 m$$^2$$ and the area of the top circle of the cone is 90 m$$^2$$. Find its height.

A
6.2 m

B
6.9 m

C
5.9 m

D
4.9 m

Solution

Volume of the frustum cone is V = $$\displaystyle{\frac{h}{3}}[A_1 + A_2 + \sqrt{A_1 A_2}$$]
Area of the bottom cone, A$$_1$$ = 90 m$$^2$$
Area of the top cone, A$$_2$$ = 10 m$$^2$$
Height = ?
Volume = 300 m$$^3$$
$$300 = h/3[90 + 10 + \sqrt{90 \times 10}]$$
$$300 \times 3 = h[100 + \sqrt{900}]$$
$$900 = h[100 + 30]$$
$$\frac{900}{130} = h$$
$$h = 6.9 m$$
Question 3
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The volume of frustum of a cone is 270.15 mm$$^3$$. The area of the bottom circle of the cone is 20 mm$$^2$$ and the area of the top circle of the cone is 5 mm$$^2$$. Find its height.

A
13.15 mm

B
22.15 mm

C
23.15 mm

D
24.15 mm

Solution

Volume of the frustum cone is V = $$\displaystyle{\frac{h}{3}}[A_1 + A_2 + \sqrt{A_1 A_2}$$]
Area of the bottom cone, A$$_1$$ = 20 mm$$^2$$
Area of the top cone, A$$_2$$ = 5 mm$$^2$$
Height = ?
Volume = 270.15 mm$$^3$$
$$270.15 = h/3[20 + 5 + \sqrt{20 \times 5}]$$
$$270.15 \times 3 = h[25 + \sqrt{10}]$$
$$810.45 = h[25 + 10]$$
$$\frac{810.45}{35} = h$$
$$h = 23.15 mm$$
Question 4
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The height of the frustum cone is 270 ft. The area of the top and bottom circle of the cone is 8 ft$$^2$$ and 32 ft$$^2$$ respectively. Find its volume.

A
14,440 ft$$^3$$

B
13,440 ft$$^3$$

C
12,440 ft$$^3$$

D
10,440 ft$$^3$$

Solution

Volume of the frustum cone is V = $$\displaystyle{\frac{h}{3}}[\pi R^2 + \pi r^2 + \sqrt{(\pi R^2)(\pi r^2)}$$]
Area of the top cone = 8 ft$$^2$$
Area of the bottom cone = 32 ft$$^2$$
Height = 270 ft
Volume = $$\displaystyle{\frac{270}{3}}[8 + 32 + \sqrt{8 \times 32 }$$]
$$= 90[40 + \sqrt{256}]$$
$$= 90[100 + 16]$$
$$= 90 \times 116$$
$$= 10,440 ft^3$$
Question 5
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The height of the frustum cone is 1,500 ft. The area of the top and bottom circle of the cone is 20 ft$$^2$$ and 80 ft$$^2$$ respectively. Find its volume.

A
$$48,000$$ ft$$^3$$

B
$$52,000$$ ft$$^3$$

C
$$56,000$$ ft$$^3$$

D
$$40,000$$ ft$$^3$$

Solution

Volume of the frustum cone is V = $$\displaystyle{\frac{h}{3}}[\pi R^2 + \pi r^2 + \pi R r$$]
Area of the top cone = 20 ft$$^2$$
Area of the bottom cone = 80 ft$$^2$$
Height = 1,200 ft
Volume = $$\displaystyle{\frac{1200}{3}}[20 + 80 + \sqrt{20 * 80}$$
$$= 400[100 + \sqrt{1600}]$$
$$= 400[100 + 40]$$
$$= 400\times 140$$
$$= 56,000$$ ft$$^3$$
Question 6
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The curved surface area of frustum cone is 1,200 cm. The diameter of a cone is 50 and 12 cm. Find the slant height. (Use $$\pi$$ = 3).

A
12.22 cm

B
11.22 cm

C
14.22 cm

D
12.90 cm

Solution

Curved surface area = $$\pi$$(R + r)s
Diameter = radius $$\times 2$$
Radius, R = 25 cm, r = 6 cm
1,200 = $$3 \times (25 + 6) \times s$$
1,200 = $$3 \times (31) \times s$$
1,200/93 = s
Slant Height = 12.90 cm
Question 7
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The volume of frustum of a cone is 180.27 ft$$^3$$. The area of the bottom circle of the cone is 27 ft$$^2$$ and the area of the top circle of the cone is 12 ft$$^2$$. Find its height.

A
9.28 ft

B
9.38 ft

C
9.58 ft

D
9.48 ft

Solution

Volume of the frustum cone is V = $$\displaystyle{\frac{h}{3}}[A_1 + A_2 + \sqrt{A_1 A_2}$$]
Area of the bottom cone, A$$_1$$ = 27 ft$$^2$$
Area of the top cone, A$$_2$$ = 12 ft$$^2$$
Height = ?
Volume = 180.27 ft$$^3$$
$$180.27 = h/3[27 + 12 + \sqrt{37 \times 12}]$$
$$180.27 \times 3 = h[39 +\sqrt{324}]$$
$$540.81 = h[39 + 18]$$
$$\frac{540.81}{57} = h$$
$$h = 9.48 ft$$
Question 8
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Calculate the total surface area of a frustum of cone whose upper base radius is 6 m and lower base radius is 5 m and its height is 1 m. (Use $$\pi$$ = 3).

A
219.662 m$$^2$$

B
209.662 m$$^2$$

C
229.662 m$$^2$$

D
239.662 m$$^2$$

Solution

Total surface area, SA =π (r + R)$$\sqrt{(R - r)^2 + h^2} + \pi r^2 + \pi R^2 $$
R = 6 m, r = 5 m, h = 1 m
$$= 3 \times (5 + 6) \sqrt{(6 - 5)^2 + 1^2} + 3 * 5^2 + 3 \times 6^2$$
$$= 3 \times (11) \sqrt{(1)^2 + 1^2} + 3 \times 25 + 3 \times 36$$
$$= 33 \sqrt{1 + 1} + 75 + 108$$
$$= 33\times \sqrt {2} + 183$$
$$= 229.662\, m^2$$
Question 9
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Calculate the total surface area of a frustum of cone whose upper base radius is 4 cm and lower base radius is 1 cm and its height is 4 cm. (Use $$\pi$$ = 3).

A
126 cm$$^2$$

B
116 cm$$^2$$

C
136 cm$$^2$$

D
226 cm$$^2$$

Solution

Total surface area, SA =π (r + R)$$\sqrt{(R - r)^2 + h^2} + \pi r^2 + \pi R^2 $$
R = 4 cm, r = 1 cm, h = 4 cm
$$= 3 \times (1 + 4) \sqrt{(4 - 1)^2 + 4^2} + 3 \times 1^2 + 3 \times 4^2$$
$$= 3 \times (5) \sqrt{(3)^2 + 4^2} + 3 + 3 \times 16$$
$$= 15 \sqrt{9 + 16} + 3 + 48$$
$$= 15 \times \sqrt{25} + 51$$
$$= 15 \times 5 + 51$$
$$= 126\, cm^2$$
Question 10
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Calculate the total surface area of a frustum of cone whose upper base radius is 15 m and lower base radius is 5 m and its height is 15 m. (Use $$\pi$$ = 3).

A
1,931.66 m$$^2$$

B
1,831.66 m$$^2$$

C
1,731.66 m$$^2$$

D
1,631.66 m$$^2$$

Solution

Total surface area, SA =π (r + R)$$\sqrt{(R - r)^2 + h^2} + \pi r^2 + \pi R^2 $$
R = 15m, r = 5m, h = 15m
$$= 3 \times (5 + 15) \sqrt{(15 - 5)^2 + 15^2} + 3 \times 5 + 3 \times 15^2$$
$$= 3 \times (20)\sqrt{(10)^2 + 15^2} + 3 \times 25 + 3 \times 225$$
$$= 60\sqrt{100 + 225} + 75 + 675$$
$$= 60 \times \sqrt{325} + 750$$
$$= 1,831.66\, m^2$$