In a right triangle,the length of the hypoten | vedclass

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(D) Let the length of the hypotenuse be $h\,cm$,the base be $b\,cm$,and the altitude be $a\,cm$.Given: $h = b + 2 \implies b = h - 2$.Also,$h = 2a + 1

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(D) Let the length of the hypotenuse be $h\,cm$,the base be $b\,cm$,and the altitude be $a\,cm$.Given: $h = b + 2 \implies b = h - 2$.Also,$h = 2a + 1 \implies a = \frac{h - 1}{2}$.By the Pythagorean theorem,$h^2 = b^2 + a^2$.Substituting the expressions for $b$ and $a$: $h^2 = (h - 2)^2 + \left(\frac{h - 1}{2}\right)^2$.$h^2 = h^2 - 4h + 4 + \frac{h^2 - 2h + 1}{4}$.Multiply by $4$: $4h^2 = 4h^2 - 16h + 16 + h^2 - 2h + 1$.$0 = h^2 - 18h + 17$.Factoring the quadratic equation: $(h - 17)(h - 1) = 0$.Since $h$ must be greater than $2$ (as $b = h - 2 > 0$),we have $h = 17$.Thus,the length of the hypotenuse is $17\,cm$.

In a right triangle,the length of the hypotenuse is $2\,cm$ more than the length of the base and $1\,cm$ more than twice the length of the altitude. Find the length of the hypotenuse. (in $,cm$)

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