The hypotenuse and the semi-perimeter of a ri | vedclass

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Question

(A) Let the two sides of the right triangle be $a$ and $b$,and the hypotenuse be $c = 20\, cm$.The semi-perimeter $s = \frac{a + b + c}{2} = 24\, cm$.

Vedclass · Accepted Answer

$16\, cm, 12\, cm$

Vedclass · Answer

(A) Let the two sides of the right triangle be $a$ and $b$,and the hypotenuse be $c = 20\, cm$.The semi-perimeter $s = \frac{a + b + c}{2} = 24\, cm$.Therefore,$a + b + 20 = 2 	imes 24 = 48$,which gives $a + b = 28\, cm$ $...(1)$.By the Pythagorean theorem,$a^2 + b^2 = c^2 = 20^2 = 400$.We know that $(a + b)^2 = a^2 + b^2 + 2ab$.Substituting the values,$28^2 = 400 + 2ab$.$784 = 400 + 2ab$,so $2ab = 384$,which means $ab = 192$.Now,$(a - b)^2 = (a + b)^2 - 4ab = 28^2 - 4(192) = 784 - 768 = 16$.Thus,$a - b = \sqrt{16} = 4\, cm$ $...(2)$.Adding $(1)$ and $(2)$,$2a = 32$,so $a = 16\, cm$.Subtracting $(2)$ from $(1)$,$2b = 24$,so $b = 12\, cm$.Thus,the other two sides are $16\, cm$ and $12\, cm$.

The hypotenuse and the semi-perimeter of a right triangle are $20\, cm$ and $24\, cm$,respectively. The other two sides of the triangle are

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