Question

A right∆ ,whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about it's hypotenuse. Find the volume and surface area of the double cone formed( choose value of π as found appropriate)
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Answer 1

Here it is....................... :D

Answer 2

The right ∠d Δ has sides 3 cm and 4 cm and a hypotenuse .

By Pythagoras theorem :

h² = 3² + 4²

⇒ h² = 9 + 16

⇒ h² = 25

⇒ h = 5

The triangle is revolved about hypotenuse .

Let Δ be ABC .

AC = 5

Let the other Δ be ADC and the lines AC , BD meet at O .

Area of Δ ABC = 1/2 × AB × BC = 1/2 × AC × BO

⇒ AB × BC = AC × BO

⇒ 3 × 4 = 5 × BO

⇒ BO = 12/5

In Δ ABC :

AO² = AB² - OB²

⇒ AO² = 3² - ( 12/5 )²

⇒ AO² = 9 - 144/25

⇒ AO² = ( 225 - 144 )/25

⇒ AO² = 81/25

⇒ AO = 9/5

CO² = 4² - 12²/5²

⇒ CO² = 16 - 144/25

⇒ CO² = 256/25

⇒ CO = 16/5

Volume of double cone :

Volume = 1/3 π r² h + 1/3 π r² h

= 1/3 π ( 12²/5² × 9/5 + ( 12²/5² ) × 16/5 )

= 1/3 π × 144/25 ( 9/5 + 16/5 )

= 1/3 π × 144/25 × 25/5

= 48/5 π

= 48/5 × 3.14

= 30.14 cm³

Surface area :

C.S.A = π r l + π r l

⇒ π ( 12/5 × 3 + 12/5 × 4 )

⇒ π 12/5 × 7

⇒ 3.14 × 84/5

⇒ 3.14 × 16.8

⇒ 52.752 cm²