Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

A bucket is in the shape of the frustum of a right circular cone, whose radii are $$6$$mm and $$24$$ mm. The curved surface area is $$450$$mm. Find the slant height. (Use $$\pi = 3$$).

A
$$5$$ mm

B
$$4$$ mm

C
$$3$$ mm

D
$$2$$ mm

Solution

Curved surface area $$=\pi (R + r)s$$

$$450 = 3\times (24 + 6) \times s$$

$$\dfrac{450}{90} = s$$

$$s = 5$$ mm
Question 2
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A vessel is in the form of a frustum of a cone. Its radius at top end is $$10$$ m and the bottom end is $$5$$ m. Its volume is $$402 \pi$$ cubic meter. Find its height.

A
$$5.68$$ m

B
$$6.89$$ m

C
$$7.21$$ m

D
$$8.34$$ m

Solution

Volume of a frustum cone $$=\pi \times \cfrac{h}{3}(R^2 + Rr + r^2)$$
$$402 = \pi \times \cfrac{h}{3}(10^2 + 50 + 5^2)$$
$$402 \times 3 = h(100 + 50 + 25)$$
$$1,206 = h (175)$$
$$h = 6.89$$ m
Question 3
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A flower bucket in the shape of a frustum cone with the top and bottom circles of radii $$5$$ in. and $$10$$ in. It's depth is $$15$$ in. Find its volume.

A
$$375\pi \space\ in^3$$

B
$$475\pi \space\ in^3$$

C
$$875\pi \space\ in^3$$

D
$$675\pi \space\ in^3$$

Solution

Volume of the frustum cone = $$\cfrac{1}{3}\pi h(R^2+r^2+Rr)$$
= $$\cfrac{1}{3}\pi \times 15(10^2+5^2+50)$$
= $$\cfrac{1}{3}\pi \times 2625$$
= $$875\pi \space\ in^3$$
Question 4
1 / -0

A drinking glass is in the shape of a frustum of a cone of height $$12$$ cm. The diameters of its two circular ends are $$2$$ cm and $$1$$ cm. Find the capacity of the glass.

A
$$4 \pi \space\ cm^3$$

B
$$3 \pi \space\ cm^3$$

C
$$6 \pi \space\ cm^3$$

D
$$7 \pi \space\ cm^3$$

Solution

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

$$=$$ $$\dfrac{1}{3}\pi \times 12(1^2+0.5^2+0.5)$$

$$=$$ $$\dfrac{1}{3}\pi \times 21$$

$$=$$ $$7 \pi \space\ cm^3$$
Question 5
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The curved surface area of the frustum cone shaped bucket is $$360 \pi \space\ cm^2$$. The top and bottom diameter of the bucket is $$6$$cm and $$12$$ cm. What is the slant height?

A
$$10$$ cm

B
$$20$$ cm

C
$$30$$ cm

D
$$40$$ cm

Solution

Radius $$=\dfrac{\text {diameter}}{2}$$

$$D = 12, d = 6$$
$$R = 6, r = 3$$

Curved surface area $$=\pi (R + r)s$$
$$360\pi = \pi(6 + 3) \times s$$

$$s=\dfrac{360}{9}$$

$$s = 40$$ cm
Question 6
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A flower pot in the shape of a frustum cone with the top and bottom circles of radii $$20$$m and $$40$$m. It's depth is $$15$$in. Find its surface area.

A
$$3100$$ $$m^2$$

B
$$3400\pi $$ $$m^2$$

C
$$3500 $$ $$m^2$$

D
$$3500\pi $$ $$m^2$$

Solution

If top and bottom radius of the frustum is $$R$$ snd $$r$$, then its surface area id given as,

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

$$=\pi (40+20)\sqrt{(40-20)^2+15^2}+\pi 20^2+\pi 40^2$$

$$=\pi (60)\sqrt{400+225}+\pi 400+\pi 1600$$

$$=\pi\times [60\times 25+2000]$$

$$=3500\pi $$ $$m^2$$
Question 7
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Find the volume of the frustum cone with $$r=2$$m, $$R=4$$m and $$h=10$$m. (Use $$\pi = \cfrac{22}{7}$$).

A
$$2640 m$$

B
$$2640 m^2$$

C
$$293m^3$$

D
$$2640 cm^3$$

Solution

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

$$=$$ $$\dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \times 10\left( { 4 }^{ 2 }+{ 2 }^{ 2 }+4\times 2 \right) $$

$$=$$ $$\dfrac { 220 }{ 21 } (16+4+8)$$

$$=$$ $$\dfrac { 220 }{ 21 } \times 28=\dfrac { 880 }{ 3 } =293{ m }^{ 3 }$$
Question 8
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The volume of a frustum cone is $$4,200 cm^3$$, whose base and upper area of a circle is $$10cm^2$$ and $$90cm^2$$. Find the height of the cone.

A
$$67$$

B
$$77$$

C
$$87$$

D
$$97$$

Solution

Volume of a frustum cone = $$\dfrac{1}{3}(A_1+A_2+\sqrt{A_{1}A_{2}})h$$

$$4200= \cfrac{1}{3}(90 + 10 + \sqrt{90\times 10})h$$

$$4200= \cfrac{1}{3}(100 + 30)h$$

$$\cfrac{4200}{43.33}= h$$

$$h \approx 97$$ cm
Question 9
1 / -0

A table lamp is in the shape of the frustum of a right circular cone whose curved area is $$1200 m^2$$ and the area of the base and top is $$2400 m^2$$. Find the total surface area of the frustum cone.

A
$$1600 m^2$$

B
$$2600 m^2$$

C
$$3600 m^2$$

D
$$4600 m^2$$

Solution

Total surface area $$=$$ Curved surface area + Area of the base and top
$$= 1200 + 2400$$
$$= 3600 m^2$$
Question 10
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A loudspeaker diaphragm is in the form of a frustum of a cone. The slant height is $$15$$ cm. The curved surface area is $$1200 cm^2$$. The base diameter is $$3$$ cm. Find the top diameter of the loud speaker.

A
$$25$$ m

B
$$50$$ m

C
$$30$$ m

D
$$10$$ m

Solution

Curved surface area $$=\pi (R + r)l$$
$$1200 = 3\times (1.5 + r) \times 15$$
$$1200 = 67.5 + 45r$$
$$1200 - 67.5 = 45r$$
$$1132.5 = 45 r$$
$$r \approx 25$$ m
Top diameter $$= 25 \times 2 = 50$$ m

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Re-Attempt Weekly Quiz Competition

Surface Areas and Volumes Test - 44

Result

Surface Areas and Volumes Test - 44

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SHARING IS CARING

Question 1

$$5$$ mm

$$4$$ mm

$$3$$ mm

$$2$$ mm

Solution

Question 2

$$5.68$$ m

$$6.89$$ m

$$7.21$$ m

$$8.34$$ m

Solution

Question 3

$$375\pi \space\ in^3$$

$$475\pi \space\ in^3$$

$$875\pi \space\ in^3$$

$$675\pi \space\ in^3$$

Solution

Question 4

$$4 \pi \space\ cm^3$$

$$3 \pi \space\ cm^3$$

$$6 \pi \space\ cm^3$$

$$7 \pi \space\ cm^3$$

Solution

Question 5

$$10$$ cm

$$20$$ cm

$$30$$ cm

$$40$$ cm

Solution

Question 6

$$3100$$ $$m^2$$

$$3400\pi $$ $$m^2$$

$$3500 $$ $$m^2$$

$$3500\pi $$ $$m^2$$

Solution

Question 7

$$2640 m$$

$$2640 m^2$$

$$293m^3$$

$$2640 cm^3$$

Solution

Question 8

$$67$$

$$77$$

$$87$$

$$97$$

Solution

Question 9

$$1600 m^2$$

$$2600 m^2$$

$$3600 m^2$$

$$4600 m^2$$

Solution

Question 10

$$25$$ m

$$50$$ m

$$30$$ m

$$10$$ m

Solution

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