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Question

(A) Let the two sides forming the right angle be $x \, cm$ and $(x + 7) \, cm$.By the Pythagorean theorem,the hypotenuse is $\sqrt{x^2 + (x + 7)^2} =

Vedclass · Accepted Answer

$5 \, cm, 12 \, cm, 13 \, cm$

Vedclass · Answer

(A) Let the two sides forming the right angle be $x \, cm$ and $(x + 7) \, cm$.By the Pythagorean theorem,the hypotenuse is $\sqrt{x^2 + (x + 7)^2} = \sqrt{x^2 + x^2 + 14x + 49} = \sqrt{2x^2 + 14x + 49}$.The perimeter is given as $30 \, cm$,so $x + (x + 7) + \sqrt{2x^2 + 14x + 49} = 30$.$2x + 7 + \sqrt{2x^2 + 14x + 49} = 30$.$\sqrt{2x^2 + 14x + 49} = 23 - 2x$.Squaring both sides: $2x^2 + 14x + 49 = (23 - 2x)^2$.$2x^2 + 14x + 49 = 529 - 92x + 4x^2$.$2x^2 - 106x + 480 = 0$.$x^2 - 53x + 240 = 0$.$(x - 5)(x - 48) = 0$.Since $x$ cannot be $48$ (as the perimeter is $30$),$x = 5$.The sides are $5 \, cm$,$5 + 7 = 12 \, cm$,and the hypotenuse is $\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \, cm$.

The difference between the lengths of the sides of a right triangle which form the right angle is $7 \, cm$. If its perimeter is $30 \, cm$,find the lengths of the sides.

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