Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

Find the surface area of a frustum of cone given below: (Use = $$\pi$$ 3.14)

A
954.6856 m$$^2$$

B
654.6856 m$$^2$$

C
854.6856 m$$^2$$

D
754.6856 m$$^2$$

Solution

Surface area of a frustum of cone = $$\pi$$(r + R)s + $$\pi$$r$$^2$$ + $$\pi$$R$$^2$$
D = 5 m, d = 0.1 m, s = 20 m
Diameter = radius/2
Radius, R = 10 m, r = 0.2 m
$$= \pi (0.2 + 10)20 +\pi 0.2^2 + \pi 10^2$$
$$= 3.14(204 + 0.04 + 100)$$
$$= 3.14(304.04)$$
$$= 954.6856 \,m^2$$
Question 2
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The total surface area of frustum cone is 360 $$ft^2$$. The radius of a top cone is 6 and the bottom cone is 4 ft. What is the curved surface area of a frustum cone? (Use $$\pi$$ = 3.14).

A
116.72 ft

B
126.72 ft

C
106.72 ft

D
146.72 ft

Solution

Total surface area = $$\pi (R+r)s + \pi R^2 + \pi r^2$$
Curved surface area = $$\pi$$(R + r)s
Therefore, 360 = Curved surface area $$+ \pi (6^2 + 4^2)$$
270 = Curved surface area $$+ 3.14 (36 + 16)$$
$$270 - 163.28 = $$Curved surface area
Curved surface area$$ = 106.72 ft$$
Question 3
1 / -0

Find the surface area of a frustum of a given cone: (Use $$\pi$$ $$= 3.14$$)

A
$$1,398.26$$ in$$^2$$

B
$$1,498.26$$ in$$^2$$

C
$$1,298.26 $$ in$$^2$$

D
$$1,598.26$$ in$$^2$$

Solution

Surface area of a frustum of cone $$= \pi(r + R)s +\pi r^2 + \pi R^2$$
$$R = 12 \ \text{in}, r = 5 \ \text{in}, s = 20 \ \text{in}$$
$$= \pi(5 + 12)20 +\pi (5)^2 +\pi 12^2$$
$$= 340 \times 3.14 + 25 \times 3.14 + 144 3.14$$
$$= 1,067.6 + 78.5 + 452.16$$
$$= 1,598.26\, \text{in}^2$$
Question 4
1 / -0

Find the surface area of a frustum of cone given below:

A
87$$\pi$$ mm$$^2$$

B
86$$\pi$$ mm$$^2$$

C
88$$\pi$$ mm$$^2$$

D
85$$\pi$$ mm$$^2$$

Solution

Surface area of a frustum of cone $$=\pi(r + R)s + \pi r^2 + \pi R^2$$
R = 3 mm, r = 2 mm, s = 15 mm
$$= \pi(2 + 3)15 +\pi 2^2 + \pi 3^2$$
$$= 75\pi + 4\pi + 9\pi$$
$$= 88\pi \,mm^2$$
Question 5
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The total surface area of frustum cone is 450 $$m^2$$. The diameter of a top cone is 2 and the bottom cone is 5 m. What is the curved surface area of a frustum cone? (Use $$\pi$$ = 3).

A
410.25 $$m^2$$

B
428.25 $$m^2$$

C
420.32 $$cm^2$$

D
430.02 $$m^3$$

Solution

Total surface area = $$\pi (R+r)s + \pi R^2 + \pi r^2$$
Curved surface area = $$\pi$$(R + r)s
Diameter = radius $$\times 2$$
Radius, R = 2.5 m, r = 1 m
Therefore, 450 = Curved surface area $$+ \pi (2.5^2 + 1^2)$$
450 = Curved surface area $$+ 3 \times (6.25 + 1)$$
$$450 - 21.75 = $$Curved surface area
Curved surface area = 428.25 $$m^2$$
Question 6
1 / -0

Find the surface area of a frustum of cone given below:

A
34.25$$\pi m^2$$

B
33.25$$\pi m^2$$

C
32.25$$\pi m^2$$

D
31.25$$\pi m^2$$

Solution

Surface area of a frustum of cone $$= \pi(r + R)s + \pi r^2 + \pi R^2$$
$$R = 2m, r = 0.5m, s = 12m$$
$$= \pi(0.5 + 2)12 +\pi 0.5^2 + \pi 2^2$$
$$= 30\pi + 0.25\pi + 4\pi$$
$$= 34.25\pi m^2$$
Question 7
1 / -0

Find the surface area of a frustum of cone given below:

A
122.25$$\pi$$ cm$$^2$$

B
112.25$$\pi$$ cm$$^2$$

C
132.25$$\pi$$ cm$$^2$$

D
142.25$$\pi$$ cm$$^2$$

Solution

Surface area of a frustum of cone $$= \pi(r + R)s + \pi r^2 + \pi R^2$$
R = 2.5 cm, r = 1 cm, s = 30 cm
$$= \pi (1 + 2.5)30 +\pi 1^2 + \pi 2.5^2$$
$$= 105\pi + \pi + 6.25\pi$$
$$= 112.25= \pi (1 + 2.5)30 +\pi 1^2 + \pi 2.5^2$$
$$= 105\pi + \pi + 6.25\pi$$
$$= 112.25\pi \,cm^2$$
Question 8
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The frustum cone curved surface area is 600 $$mm^3$$. The upper and lower circle area of a cone is 25 and 35 mm. Find the total surface area.

A
660 $$mm^3$$

B
650 $$mm^3$$

C
560 $$mm^3$$

D
660 $$mm$$

Solution

Total surface area = $$\pi (R+r)s + A_1 + A_2 $$
Curved surface area = $$\pi (R + r)s \Rightarrow 600 mm^2$$
$$A_1 = 25 mm, A_2 = 35 mm$$
Therefore, Total surface area = 600 + 25 + 35
TSA = 660 $$mm^3$$
Question 9
1 / -0

The surface area of a frustum cone is 2,400 m$$^2$$. The larger and smaller radius of the cone is 12 and 4 m. find its slant height. (Use $$\pi$$ = 3.14).

A
37.77 m

B
38.77 m

C
39.77 m

D
51.77 m

Solution

Surface area of a frustum of cone = $$\pi$$(r + R)s + $$\pi$$r$$^2$$ + $$\pi$$R$$^2$$
R = 12m, r = 4m, SA = 2,400 m2
2,400 = 3.14 $$\times$$ (4 +12)s +$$\pi$$4$$^2$$ + $$\pi$$12$$^2$$
2,400 = 3.14 $$\times$$ [16s +16 + 144]
2,400 = 3.14 $$\times$$ [16s +160]
2,400/3.14 = 16s + 160
764.33 160 = 16s
604.33/16 = s
S = 37.77 m
Therefore, the slant height is 37.77 m
Question 10
1 / -0

The frustum cone curved surface area is 389.13 $$cm^3$$. The upper and lower circle area of a cone is 2.5 and 6.5 cm. Find the total surface area.

A
368.18 $$cm^3$$

B
378.18 $$cm^3$$

C
388.18 $$cm^3$$

D
398.18 $$cm^3$$

Solution

Total surface area = $$\pi (R+r)s + A_1 + A_2 $$
Curved surface area = $$\pi (R + r)s \Rightarrow 389.13 cm^2$$
$$A_1 = 2.5 cm, A_2 = 6.5 cm$$
Therefore, Total surface area$$ = 389.13 + 2.5 +6.5$$
TSA = 398.18 $$cm^3$$