A right triangle,whose sides are 3, cm and 4, | vedclass

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Question

(C) Let the right triangle be $ABC$ with sides $AB = 3\, cm$,$BC = 4\, cm$,and $\angle B = 90^\circ$.Hypotenuse $AC = \sqrt{3^2 + 4^2} = \sqrt{9 + 16}

Vedclass · Accepted Answer

$52.75$

Vedclass · Answer

(C) Let the right triangle be $ABC$ with sides $AB = 3\, cm$,$BC = 4\, cm$,and $\angle B = 90^\circ$.Hypotenuse $AC = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\, cm$.When the triangle is revolved about the hypotenuse $AC$,a double cone is formed where the radius $r$ of the common base is the altitude $OB$ of the triangle $ABC$ to the hypotenuse $AC$.Area of $	riangle ABC = \frac{1}{2} 	imes AB 	imes BC = \frac{1}{2} 	imes 3 	imes 4 = 6\, cm^2$.Also,Area of $	riangle ABC = \frac{1}{2} 	imes AC 	imes OB = \frac{1}{2} 	imes 5 	imes OB$.So,$\frac{1}{2} 	imes 5 	imes OB = 6 \implies OB = \frac{12}{5} = 2.4\, cm$.The surface area of the double cone is the sum of the curved surface areas of the two cones: $S = \pi r l_1 + \pi r l_2$,where $l_1 = AB = 3\, cm$ and $l_2 = BC = 4\, cm$.$S = \pi r (l_1 + l_2) = 3.14 	imes 2.4 	imes (3 + 4) = 3.14 	imes 2.4 	imes 7 = 52.752\, cm^2 \approx 52.75\, cm^2$.

$A$ right triangle,whose sides are $3\, cm$ and $4\, cm$ (other than the hypotenuse) is made to revolve about its hypotenuse. Find the surface area of the double cone so formed. (Use $\pi = 3.14$) (in $cm^2$)

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