Surface Areas and Volumes Test

Result

Surface Areas and Volumes Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

Calculate the volume of a frustum cone given below: (Use $$\pi$$ = 3.14)

A
163,280 cm$$^3$$

B
163,280 m$$^3$$

C
163,280 mm

D
163,280 mm$$^3$$

Solution

Volume of the frustum cone is $$V = \displaystyle{\frac{\pi h}{3}}(R^2 + Rr + r^2)$$
R = 50 mm, r = 5 mm, h = 60 mm
$$V = (3.14 \times 60)/(3)[50^2 + 15 \times 5 + 5^2]$$
$$= 3.14 \times 20[2,500 + 75 + 25]$$
$$= 62.8[2,600]$$
$$= 163,280\, mm^3$$
Question 2
1 / -0

Calculate the volume of a frustum cone given below: (Use $$\pi$$ = 3).

A
6,770 m$$^3$$

B
8,770 m$$^3$$

C
7,770 m$$^3$$

D
9,770 m$$^3$$

Solution

Volume of the frustum cone is $$V = \displaystyle{\frac{\pi h}{3}}(R^2 + Rr + r^2)$$
R = 15 m, r = 2 m, h = 30 m
$$V = (3 \times 30)/(3)[15^2 + 15 \times 2 + 2^2]$$
$$= 30[225 + 30 + 4]$$
$$= 30[259]$$
$$= 7,770 \,m^3$$
Question 3
1 / -0

Calculate the volume of a frustum cone given below: (Use $$\pi$$ = 3.14)

A
419,552 ft$$^3$$

B
439,552 ft$$^3$$

C
449,552 ft$$^3$$

D
429,552 ft$$^3$$

Solution

Volume of the frustum cone is $$V = \displaystyle{\frac{\pi h}{3}}(R^2 + Rr + r^2)$$
R = 36 ft, d = 12 ft, h = 150 ft
Diameter = radius/2r = 24 ft
$$V = (3.14 \times 150)/(3)[36^2 + 36 \times 24 + 24^2]$$
$$= 3.14 \times 50[1,296 + 864 + 576]$$
$$= 157[2,736]$$
$$= 429,552\, ft^3$$
Question 4
1 / -0

The surface area of the frustum cone is given its base radius, R = 12 m and top radius, r = 9 m. The height of the cone is 4 m. Find the slant height of the cone.

A
$$2 \sqrt{7}$$ cm

B
$$\sqrt{5}$$ cm

C
5 cm

D
$$2 \sqrt{5}$$ cm

Solution

Slant height of a cone = $$\sqrt{(R-r)^2 + h^2}$$
= $$\sqrt{(12-9)^2 + 4^2}$$
= $$\sqrt{(3)^2 + 16}$$
= $$\sqrt{9 + 16}$$
= $$\sqrt{25}$$ cm
= 5 cm
Question 5
1 / -0

The surface area of the frustum cone is given its base radius, R = 4 cm and top radius, r = 2 cm. The height of the cone is 4 cm. Find the slant height of the cone.

A
$$\sqrt{5}$$ cm

B
$$3\sqrt{5}$$ cm

C
24 cm

D
$$2\sqrt{5}$$ cm

Solution

Slant height of a cone = $$\sqrt{(R-r)^2 + h^2}$$
= $$\sqrt{(4-2)^2 + 4^2}$$
= $$\sqrt{(2)^2 + 16}$$
= $$\sqrt{20}$$
= $$2\sqrt{5}$$ cm
Question 6
1 / -0

The surface area of the frustum cone is given its base radius, R = 6 cm and top radius, r = 3 cm. The height of the cone is 4 cm. Find the slant height of the cone.

A
$$2 \sqrt{7}$$ cm

B
$$\sqrt{5}$$ cm

C
5 cm

D
$$2 \sqrt{5}$$ cm

Solution

Slant height of a cone = $$\sqrt{(R-r)^2 + h^2}$$
= $$\sqrt{(6-3)^2 + 4^2}$$
= $$\sqrt{(3)^2 + 16}$$
= $$\sqrt{9 + 16}$$
= $$\sqrt{25}$$ cm
= 5 cm
Question 7
1 / -0

Calculate the volume of a frustum cone given below: (Use $$\pi$$ = 3.14)

A
$$113,829.168$$ m$$^3$$

B
$$2873.10$$ m$$^3$$

C
$$123,829.168$$ in$$^3$$

D
$$133,829.168$$ in$$^3$$

Solution

Volume of frustum $$=\dfrac { 1 }{ 3 } \pi \left( { R }^{ 2 }+{ r }^{ 2 }+Rr \right) \times h$$

$$=\dfrac { 1 }{ 3 } \times 3.14\left( { \left( 2.5 \right) }^{ 2 }+{ 2 }^{ 2 }+2.5\times 2 \right) \times 180$$

$$=188.40(6.25+4+5)$$

$$=188.40\times 15.25=2873.10{ m }^{ 3 }$$
Question 8
1 / -0

The volume of the frustum of a cone is 7,740 cm$$^3$$. The bottom and top radius are 14 cm and 7 cm. Find its height. (Use $$\pi$$ = 22/7). Round off your answer to nearest whole number.

A
21 cm

B
22 cm

C
23 cm

D
24 cm

Solution

Volume of the frustum cone is V = $$\displaystyle{\frac{\pi h}{3}}(R^2 + Rr + r^2)$$
$$R = 14 cm, r = 7 cm, h = ?$$
$$Volume = 7,740 cm$$$$^3$$
$$7,740 = (22 \times /' h)/(7 \times 3)[14^2 + 14 \times 7 + 7^2]$$
$$7,740 \times 21= 22 \times h[196 + 98 + 49]$$
$$\frac{162,540}{22} = h[343]$$
$$\frac{7,388.18182}{343} = h$$
$$h = 22 cm$$
Question 9
1 / -0

The surface area of the frustum cone is given its base radius, R = 18 m and top radius, r = 9 m. The height of the cone is 12 m. Find the slant height of the cone.

A
$$2 \sqrt{7}$$ cm

B
$$\sqrt{15}$$ cm

C
15 cm

D
$$2 \sqrt{15}$$ cm

Solution

Slant height of a cone = $$\sqrt{(R-r)^2 + h^2}$$
= $$\sqrt{(18-9)^2 + 12^2}$$
= $$\sqrt{(9)^2 + 144}$$
= $$\sqrt{81 + 144}$$
= $$\sqrt{225}$$ cm
= 15 cm
Question 10
1 / -0

Find the curved surface area of frustum cone radii 3 and 9 cm and a slant height 12 cm.

A
144 $$\pi cm^2$$

B
134 $$\pi cm^3$$

C
124 $$\pi cm^3$$

D
114 $$\pi cm^3$$

Solution

Curved surface area = $$\pi$$(R + r)s
= $$\pi (9 + 3) \times 12$$
= $$\pi (12) \times 12$$
= 144 $$\pi cm^2$$