Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 43

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Weekly Quiz Competition

Question 1
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The surface area of the frustum cone is $$297 cm^{2}$$ whose top circle radius is $$2$$ cm and bottom circle radius is $$5$$ cm. Find its slant height. (Use $$\pi = 3$$)

A
$$10$$ m

B
$$8$$ cm

C
$$9$$ cm

D
$$10$$ cm

Solution

Surface area of frustum cone $$=\pi (r + R)s + \pi r^{2}+\pi R^{2}$$
$$297 =\pi (2 + 5)s + \pi 2^{2}+\pi 5^{2}$$
$$297 = (3 \times 7)s + 3 * 4 + 3 \times 25$$
$$297 = 21s + 12 + 75$$
$$297 = 21s + 87$$
$$297 - 87 = 21s$$
$$210\div21 = s$$

Slant height, $$s = 10$$ cm
Question 2
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The surface area of the frustum cone is $$450 m^{2}$$ whose top circle radius is $$3$$ m and bottom circle radius is $$2$$ m. Find its slant height. (Use $$\pi$$ = 3)

A
$$27.4$$ m

B
$$26.4$$ m

C
$$25.4$$ m

D
$$24.4$$ m

Solution

Surface area of frustum cone = $$\pi (r + R)s + \pi r^{2}+\pi R^{2}$$
$$450 =\pi (2 + 3)s + \pi 2^{2}+\pi 3^{2}$$
$$450 = 3 \times 5s + 3 \times 4 + 3 \times 9$$
$$450 = 15s + 12 + 27$$
$$450 = 15s + 39$$
$$450 - 39 = 15s$$
$$411\div15 = s$$
Slant height, $$s = 27.4$$ m
Question 3
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A rocket has a cylindrical part which is converted into conical part at the front. The cylindrical part of the rocket radius is 12 cm and its height is 20 cm. The conical part of the rocket slant height is 4 cm and its radius is 7 cm. Find the surface area of rocket ( $$\pi$$ = 3)

A
1428 $$cm^2$$

B
1524 $$cm^2$$

C
1380 $$cm^2$$

D
1200 $$cm^2$$

Solution

Surface area of rocket = CSA of cylindrical part + CSA of conical part
CSA of cylindrical part = $$2 \pi r h$$
CSA of conical part = $$\pi r l$$
Surface area of rocket = $$2 \times 3\times 12 \times 20 + 3 \times 7 \times 4$$
= $$1,440 + 84$$
= 1524 $$cm^2$$
Question 4
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Find the surface area of a frustum of cone, whose larger and smaller radius is $$12$$ cm and $$8$$ cm. The height of the cone is $$3$$ cm. (Use $$\pi = 3$$)

A
$$914 cm^{2}$$

B
$$924 cm^{2}$$

C
$$934 cm^{2}$$

D
$$944 cm^{2}$$

Solution

Surface area of frustum cone

$$=\pi (r + R)\sqrt{(R - r)^{2}+ h^{2}} + \pi r^{2}+\pi R^{2}$$

$$= \pi (8 + 12)\sqrt{(12 - 8)^{2}+ 3^{2}} + \pi *8^{2}+\pi 12^{2}$$

$$= 3 \times20\sqrt{16 + 9}+ 3\times64 + 3 \times 144$$

$$= 60 \times 5 + 192 + 432$$

$$= 924 cm^{2}$$
Question 5
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A tent of 20 mm in radius and 14 mm in height has to be made. If the cost of canvas is Rs. 14 per $$mm^2$$, total cost of canvas used to make the tent is Rs. ____________. ( Use $$\pi$$ = 22/7)

A
Rs. 20,400.50

B
Rs. 21,480.76

C
Rs. 23,490.56

D
Rs. 30,000.26

Solution

Radius = 20 mm
Slant height = $$\sqrt{r^2 + h^2} = \sqrt{20^2 + 14^2}$$
= $$\sqrt{400 + 196} = 24.41 mm$$
Curved surface area = $$\displaystyle \pi r l \Rightarrow \frac{22}{7} \times 20 \times 24.41$$
= $$1,534.34 mm^2$$
$$\therefore $$ Cost of canvas = $$1,534.34 \times 14 = Rs. 21,480.76$$
Question 6
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What is the surface area of a frustum of cone, whose larger and smaller radius is $$6$$ in. and $$2$$ in. The height of the cone is $$3$$ in. (Use $$\pi = 3.14$$)

A
$$231.2 in^{2}$$

B
$$241.2 in^{2}$$

C
$$221.2 in^{2}$$

D
$$251.2 in^{2}$$

Solution

Surface area of frustum cone = $$\pi (r + R)\sqrt{(R - r)^{2}+ h^{2}} + \pi r^{2}+\pi R^{2}$$

$$=\pi (6 + 2)\sqrt{(6 - 2)^{2}+ 3^{2}} + \pi *6^{2}+\pi *2^{2}$$

$$=3.14 \times8\sqrt{16 + 9}+ 3.14\times36 + 3.14 \times 4$$

$$=25.12 \times 5 + 113.04 + 12.56$$

$$= 251.2 in^{2}$$
Question 7
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Find the surface area of a frustum of cone, whose larger and smaller radius is $$8$$ m and $$4$$ m. The height of the cone is $$3$$ cm. (Use $$\pi= 3$$)

A
$$420m^{2}$$

B
$$421 m^{2}$$

C
$$430 m^{2}$$

D
$$440 m^{2}$$

Solution

TSA of frustum = $$\pi (R+r)s+\pi \left( { R }^{ 2 }+{ r }^{ 2 } \right) $$

= $$\pi \left( R+r \right) \sqrt { { \left( R-r \right) }^{ 2 }+{ h }^{ 2 } } +\pi \left( { R }^{ 2 }+{ r }^{ 2 } \right) $$

= $$3\times\left( 8+4 \right) \sqrt { { 4 }^{ 2 }+{ 3 }^{ 2 } } +3 \left( { 8 }^{ 2 }+{ 4 }^{ 2 } \right) $$

= $$3 \times 12\times 5+3 \times 80$$

= $$420$$
Question 8
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A bucket is in the form of a frustum of a cone. The curved surface area of the bucket is $$270 \pi \space\ cm^2$$. The top and bottom radius of the bucket is $$3$$cm and $$6$$cm. What is the slant height?

A
$$10$$ cm

B
$$20$$ cm

C
$$30$$ cm

D
$$40$$ cm

Solution

Curved surface area $$=\pi (R + r)s$$
$$270 \pi = \pi(6 + 3) \times s$$
$$\cfrac{270}{9} = s$$
$$s = 30$$ cm
Question 9
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A bucket in the shape of a frustum with the top and bottom circle area is $$120m^2$$ and $$280m^2$$. The height of the bucket is $$15$$ m. Find its volume.

A
$$1916m^3$$

B
$$2916 m^3$$

C
$$3916m^3$$

D
$$4916 m^3$$

Solution

Volume of frustum cone = $$\dfrac{h}{3}[A_{1}+A_{2}+\sqrt{A_{1}A_{2}}]$$

$$=\dfrac{15}{3}[120+280+\sqrt{120\times 280}]$$

$$= 5[400 + 183.30]$$

$$= 2916 m^3$$
Question 10
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A truncated drum is in the shape of a frustum of a cone of height $$10$$ cm. The radius of its two circular ends are $$4$$ cm and $$7$$ cm. Find the volume of the drum.

A
$$673.4 cm^3$$

B
$$573.4 cm^3$$

C
$$973.4 cm^3$$

D
$$873.4 cm^3$$

Solution

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

$$=\dfrac{1}{3}\pi \times 10(7^2+4^2+28)$$

$$=\dfrac{1}{3}\pi \times 930$$

$$=973.4 cm^3$$

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Re-Attempt Weekly Quiz Competition

Surface Areas and Volumes Test - 43

Result

Surface Areas and Volumes Test - 43

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SHARING IS CARING

Question 1

$$10$$ m

$$8$$ cm

$$9$$ cm

$$10$$ cm

Solution

Question 2

$$27.4$$ m

$$26.4$$ m

$$25.4$$ m

$$24.4$$ m

Solution

Question 3

1428 $$cm^2$$

1524 $$cm^2$$

1380 $$cm^2$$

1200 $$cm^2$$

Solution

Question 4

$$914 cm^{2}$$

$$924 cm^{2}$$

$$934 cm^{2}$$

$$944 cm^{2}$$

Solution

Question 5

Rs. 20,400.50

Rs. 21,480.76

Rs. 23,490.56

Rs. 30,000.26

Solution

Question 6

$$231.2 in^{2}$$

$$241.2 in^{2}$$

$$221.2 in^{2}$$

$$251.2 in^{2}$$

Solution

Question 7

$$420m^{2}$$

$$421 m^{2}$$

$$430 m^{2}$$

$$440 m^{2}$$

Solution

Question 8

$$10$$ cm

$$20$$ cm

$$30$$ cm

$$40$$ cm

Solution

Question 9

$$1916m^3$$

$$2916 m^3$$

$$3916m^3$$

$$4916 m^3$$

Solution

Question 10

$$673.4 cm^3$$

$$573.4 cm^3$$

$$973.4 cm^3$$

$$873.4 cm^3$$

Solution

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