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Question

# If the area of an isosceles right-angled triangle is 72 cm2, then its perimeter is _________.

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Solution

## Given: Area of an isosceles right-angled triangle is 72 cm2 $\mathrm{Area}\mathrm{of}\mathrm{an}\mathrm{isosceles}\mathrm{right}\mathrm{triangle}=\frac{1}{2}×\mathrm{Base}×\mathrm{Height}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}×\mathrm{Base}×\mathrm{Base}\left(\because \mathrm{Base}=\mathrm{Height}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}×{\left(\mathrm{Base}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒72=\frac{1}{2}×{\left(\mathrm{Base}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒144={\left(\mathrm{Base}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{Base}=12\mathrm{cm}=\mathrm{Perpendicular}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{In}\mathrm{a}\mathrm{right}-\mathrm{angled}\mathrm{triangle},\mathrm{using}\mathrm{pythagoras}\mathrm{theorem}\phantom{\rule{0ex}{0ex}}{\left(\mathrm{Hypotenuse}\right)}^{2}={\left(\mathrm{Base}\right)}^{2}+{\left(\mathrm{Perpendicular}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(12\right)}^{2}+{\left(12\right)}^{2}\phantom{\rule{0ex}{0ex}}=144+144\phantom{\rule{0ex}{0ex}}=288\phantom{\rule{0ex}{0ex}}⇒\mathrm{Hypotenuse}=\sqrt{288}\mathrm{cm}\phantom{\rule{0ex}{0ex}}⇒\mathrm{Hypotenuse}=12\sqrt{2}\mathrm{cm}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\phantom{\rule{0ex}{0ex}}\mathrm{Perimeter}=12+12+12\sqrt{2}=12\left(2+\sqrt{2}\right)\mathrm{cm}$ Hence, its perimeter is $\overline{)12\left(2+\sqrt{2}\right)\mathrm{cm}}.$

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