A right triangle whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its - Maths -

Question

A right triangle whose sides are 3 cm and 4 cm (other than
hypotenuse) is made to revolve about its hypotenuse Find the volume
and surface area of the double cone so formed (Choose value of p as
found appropriate

Ajanta Trivedi · Accepted Answer

in the right angled triangle ABC, (right angled at B),

AB = 4 cm and BC = 3cm

AC = $\sqrt{3^{2} + 4^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5 c m$

area of ΔABC = 1/2 AB*BC = 1/2 AC*BO

AB*BC = AC*BO

4*3 = 5*BO

BO = 12/5 = 2.4 cm

the volume of the required double cone = volume of the cone ABD + volume of the cone BCD

$= \frac{1}{3} π r^{2} h_{1} + \frac{1}{3} π r^{2} h_{2}$ [where h1 and h2 are the heights of the cone ABD and BCD respectively]

$= \frac{1}{3} * π r^{2} (h_{1} + h_{2}) = \frac{1}{3} * \frac{22}{7} * {2.4}^{2} (5) [∵ r = B O = 2.4 c m] = 30.17 c m^{3}$

surface area of double cone = surface area of cone ABD + surface area of cone BCD

$= π r l_{1} + π r l_{2}$ [where $l_{1} a n d l_{2}$ are the slant heights of the cone ABD and BCD respectively]

$= π r (l_{1} + l_{2}) = \frac{22}{7} * 2.4 * (4 + 3) = \frac{22}{7} * 2.4 * 7 = 52.8 c m^{2}$

thus the volume of the double cone is 30.17 cubic cm and surface area of double cone is 52.8 sq cm

hope this helps you.

cheers!!

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A right triangle whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of p as found appropriate.