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A right triangle with side 3 cm and 4 cm is revolved around its hypotenuse . Find the volume of double cone thus genrated .

\sf \left(use \pi =3.14 \right)

Answers

Answered by itzbangtanarmy7
11

Answer:

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 . ⇒ OB = 12 / 5 = 2.4 cm.

Answered by MysticalRainbow
34

Answer

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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