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SSC CGL Model Paper 18

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Question : 53 of 100

Marks: +1, -0

ABC is a right angled triangle in which ∠A = 90°, AB =5 cm and AC = 12 cm. What is the approximate volume (in cm

^{3}

) of the double cone formed by rotating the triangle about its hypotenuse?

145
290
435
580

Solution:

In right angle ΔABC, by Pythagoras theorem,

(B C )^2 = (A C )^2 + (A B )^2

⇒

(B C )^2=(12)^2+(5)^2

⇒

(B C )^2 = 144 + 25 = 169

∴ BC = 13 cm

Area of

\Delta ABC = \frac{1}{2} \times AB \times AC

⇒

\frac{1}{2} \times AO \times BC = \frac{1}{2} \times 5 \times 12

⇒

A O \times 13 = 60

∴

AO = \frac{60}{13}

Volume of double cone = Volume of first cone + volume of second cone

= \frac{1}{3} \pi r^2 (CO) + \frac{1}{3} \pi r^2 (OB)

= \frac{1}{3} \pi r^2 [CO + OB] = \frac{1}{3} \pi r^2 BC

= \frac{1}{3} \times \frac{22}{7} \times \left(\frac{60}{13}\right)^2 \times 13

= 290.10 \approx 290 \text{ cm}^3

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