Fewpal
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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.

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Solution:

Given that in a right angled triangle, whose sides are 3 cm and 4 cm(other than hypotenuse).

The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse .


Hypotenuse AC = Ö( 32 + 42) = 5 cm

Area of ΔABC = (1/2) x AB xAC
 
⇒ (1/2) x AC x OB = (1/2)x 4 x3
 
⇒ (1/2) x 5 x OB = 6
 
⇒ OB = 12 / 5 = 2.4 cm.

Volume of double cone = Volume of cone 1 + Volume of cone 2

= (1/3) πr2h1 + (1/3)πr2h2

= (1/3) πr2(h1 + h2)
 
= (1/3) πr2(OA +OC)
 
= (1/3) x 3.14x(2.4)2 x(5)

= 30.14 cm3

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

= πrl1 + πrl2  [where l1 and l2 are the slant heights of the cone ABD and BCD respectively]

thus the volume of the double cone is 30.14 cubic cm and surface area of double cone is 52.8 sq cm

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