Multiple choice

A right-angled triangle whose sides are 15 cm and 20 cm, is made to revolve about its hypotenuse. Find the volume and the surface area of the double cone so formed. [Take $\pi \simeq 3.14]$

  1. 3878 $cm^3$,1315.8$cm^2$
  2. 3777 $cm^3$,1312.8 $cm^2$
  3. 3788 $cm^3$ ,1310.8 $cm^2$
  4. 3768 $cm^3 $,1318.8 $cm^2$
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D Correct answer
Explanation

Sides 15, 20 form a hypotenuse of 25. The double cone has a radius equal to the altitude to the hypotenuse: r = (15 * 20) / 25 = 12. The two cones have heights h1 and h2 such that h1 + h2 = 25. Volume = (1/3) * pi * r^2 * (h1 + h2) = (1/3) * 3.14 * 144 * 25 = 3768. Surface area = pi * r * (l1 + l2) = 3.14 * 12 * (15 + 20) = 3.14 * 12 * 35 = 1318.8.

AI explanation

The hypotenuse of the right triangle is sqrt(15^2 + 20^2) = sqrt(625) = 25 cm, and the altitude to the hypotenuse is (15*20)/25 = 12 cm. This altitude forms the common radius of the double cone, while the hypotenuse segments of 9 cm and 16 cm form the heights. The volume is (1/3)pi*12^2(9+16), which equals 1200*pi and approximates to 3768 cm^3. The total surface area is pi*12*(15+20), which equals 420*pi and approximates to 1318.8 cm^2.

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