Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 30

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Weekly Quiz Competition

Question 1
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The volume of the frustum of a cone is 313 cm$$^3$$. The bottom and top radius are 0.2 cm and 0.1 cm. Find its height. (Use $$\pi$$ = 3.14). Round off your answer to nearest whole number.

A
4,172 cm

B
4,372 cm

C
4,472 cm

D
4,272 cm

Solution

Volume of the frustum cone is V = $$\displaystyle{\frac{\pi h}{3}}(R^2 + Rr + r^2)$$
R = 0.2 cm, r = 0.1 cm, h = ?
Volume = 313 $$cm^3$$
$$313 = (3.14 \times h)/(3)[0.2^2 + 0.2 \times 0.1 + 0.1^2]$$
$$313 \times 3/3.14 = h[0.04 + 0.02 + 0.01]$$
$$299.044 = h[0.07]$$
$$4,272 = h$$
$$h = 4,272\, cm$$
Question 2
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The volume of the frustum of a cone is 690.21 m$$^3$$. The bottom and top diameters are 12 m and 8 m. Find its height. (Use $$\pi$$ = 3). Round off your answer to nearest whole number.

A
9 m

B
10 m

C
5 m

D
8 m

Solution

Volume of the frustum cone is $$V = \displaystyle{\frac{\pi h}{3}}(R^2 + Rr + r^2)$$
Let us consider D is the bottom diameter and d is top diameter
D = 12 m, d = 8 m, h = ?
Volume = 690.21 m$$^3$$
$$Diameter = radius / 2$$
$$R = 6 m$$, r = 4 m
$$690.21 = (3\times h)/(3)[6^2+6\times 4 + 4^2]$$
$$690.21= h[36 + 24 + 16]$$
$$690.21 = h[76]$$
$$h = 690.21 / 76$$
$$h = 9 m$$
Question 3
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Calculate the volume of a frustum cone given below: (Use $$\pi$$
= 3.14)

A
126,556.8 cm$$^3$$

B
136,556.8 cm$$^3$$

C
146,556.8 cm$$^3$$

D
116,556.8 cm$$^3$$

Solution

Volume of the frustum cone is $$V = \displaystyle{\frac{\pi h}{3}}(R^2 + Rr + r^2)$$
D = 15 cm, r = 1 cm, h = 120 cm
Diameter = radius / 2
R = 30 cm
$$V = (3.14 \times 120)/(3)[30^2 + 30\times 1 + 1^2]$$
$$= 3.14 \times 40[900 + 24 + 4]$$
$$= 125.6[928]$$
$$= 116,556.8 \,cm^3$$
Question 4
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The volume of the frustum of a cone is 4,200 mm$$^3$$. The bottom and top radius are R = 20 mm and r = 10 mm. Find its height. (Use $$\pi$$ = 3). Round off your answer to nearest whole number.

A
6 mm

B
7 mm

C
8 mm

D
10 mm

Solution

Volume of the frustum cone is V = $$\displaystyle{\frac{\pi h}{3}}(R^2 + Rr + r^2)$$
R = 20 mm, r = 10 mm, h = ?
Volume = 4,200 mm$$^3$$
$$4,200 = (3 \times h)/(3)[20^2 + 20 \times 10 + 10^2]$$
$$4,200 = h[400 + 200 + 100]$$
$$4,200 = h[700]$$
$$\frac{4,200}{700} = h$$
h = 6 mm
Question 5
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Find the curved surface area of frustum cone radii 2.5 and 1.2 cm and a slant height 12.5 cm.

A
45.25 $$\pi cm^3$$

B
47.25 $$\pi cm^3$$

C
46.25 $$\pi cm^2$$

D
48.25 $$\pi cm^3$$

Solution

Curved surface area = $$\pi$$(R + r)s
= $$\pi (2.5 + 1.2) \times 12.5$$
= $$\pi (3.7) \times 12.5$$
= 46.25 $$\pi cm^2$$
Question 6
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Calculate the volume of a frustum cone given below: (Use = 3.14)

A
8,301.2 m$$^3$$

B
8,201.2 m$$^3$$

C
8,101.2 m$$^3$$

D
8,001.2 m$$^3$$

Solution

Volume of the frustum cone is V = $$\displaystyle{\frac{\pi h}{3}}(R^2 + Rr + r^2)$$
R = 12 m, r = 2 m, h = 45 m
$$V = (3.14 \times 45)/(3)[12^2 + 12 \times 2 + 2^2]$$
$$= 3.14 \times 15[144 + 24 + 4]$$
$$= 47.1[172]$$
$$= 8,101.2\, m^3$$
Question 7
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Find the curved surface area of frustum cone radii $$12$$ and $$6$$ cm and a slant height $$20$$ cm.

A
$$350\pi ~cm^2$$

B
$$360\pi ~cm^2$$

C
$$370\pi ~cm^2$$

D
$$380\pi ~cm^2$$

Solution

Curved surface area of frustum of a cone$$= \pi(R + r)l$$
$$=\pi (12 + 6) \times 20$$
$$=\pi (18) \times 20$$
$$=360\pi ~cm^2$$
Question 8
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The height of the frustum cone is $$18$$ in. The area of the top and bottom circle of the cone is $$4$$ in$$^2$$ and $$16$$ in$$^2$$ respectively. Find its volume.

A
$$168$$ in$$^3$$

B
$$158$$ in$$^3$$

C
$$146$$ in$$^3$$

D
$$148$$ in$$^3$$

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 = 4 in$$^2$$
Area of the bottom cone = 16 in$$^2$$
Height $$= 18$$in
Volume $$=$$ $$\displaystyle{\frac{18}{3}}[16 + 4 + \sqrt{16*4}$$]
$$= 6[20 + \sqrt{64}]$$
$$= 6[20 + 8]$$
$$= 6 \times 28$$
$$=168\, in^3$$
Question 9
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The height of the frustum cone is 12 ft. The area of the top and bottom circle of the cone is 4 ft$$^2$$ and 4 ft$$^2$$ respectively. Find its volume.

A
125 ft$$^3$$

B
126 ft$$^3$$

C
127 ft$$^3$$

D
128 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 = 4 ft$$^2$$
Area of the bottom cone = 4 ft$$^2$$
Height = 12 ft
Volume = $$\displaystyle{\frac{12}{3}}[4 + 4 + \sqrt{4* 4}$$]
= 4[8 + $$\sqrt{16}$$]
= 4[8 + 4]
= 4 [8 + 4]
= 4 x 32
= 128 ft$$^3$$
Question 10
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The volume of the frustum of a cone is 1,200 mm$$^3$$. The bottom and top radius are D = 5 mm and d = 1 mm. Find its height. (Use $$\pi$$ = 3).

A
20 mm

B
30 mm

C
40 mm

D
9.67 mm

Solution

Volume of the frustum cone is $$V = \displaystyle{\frac{\pi h}{3}}(R^2 + Rr + r^2)$$
D = 5 mm, d = 1 mm, h = ?
Diameter = radius/2
Radius, R = 10 mm, r = 2mm
Volume = 1,200 mm$$^3$$
$$1,200 = (3 \times h)/(3)[10^2 + 10 \times 2 + 2^2]$$
$$1,200 = h[100 + 20 + 4]$$
$$1,200 = h[124]$$
$$\frac{1,200}{124} = h$$
$$h = 9.67 \,mm$$