If the area of an isosceles right triangle is $8cm^2$, what is the perimeter of the triangle?
A$8+\sqrt{2}\text{cm}^2$
B$8+4{\sqrt{2}}\text{cm}^2$
C$4+8{\sqrt{2}}{}\text{cm}^2$
D$12\sqrt{2}\text{cm}^2$
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B$8+4{\sqrt{2}}\text{cm}^2$
Let each of the two equal sides of an isosceles right triangle be $a\ cm$.
Then, third side $=\text{a}\sqrt{2}\text{cm}$
Area of $\triangle=\frac12\times\text{a}\times\text{a}$
$\Rightarrow8=\frac{\text{a}^2}{2}$
$\Rightarrow a^2 = 16$
$\Rightarrow a = 4cm$
$\Rightarrow$ Perimeter
$\Rightarrow\text{a}+\text{a}+\text{a}\sqrt{2}=4+4+4\sqrt{2}\text{cm}$
Hence, correct option is $(b)$.
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