An isosceles right triangle has an area of 8 | vedclass

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(B) Let the two equal sides of the isosceles right triangle be $a \, cm$. The area of the triangle is given by $\frac{1}{2} 	imes 	ext{base} 	imes

Vedclass · Accepted Answer

$\sqrt{32} \, cm$

Vedclass · Answer

(B) Let the two equal sides of the isosceles right triangle be $a \, cm$. The area of the triangle is given by $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$. Since it is an isosceles right triangle,base $= a$ and height $= a$. Therefore,$\frac{1}{2} 	imes a 	imes a = 8$. $a^{2} = 16$,which gives $a = 4 \, cm$. By the Pythagorean theorem,the hypotenuse $h$ is given by $h = \sqrt{a^{2} + a^{2}} = \sqrt{4^{2} + 4^{2}} = \sqrt{16 + 16} = \sqrt{32} \, cm$.

An isosceles right triangle has an area of $8 \, cm^{2}$. The length of its hypotenuse is

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