Question

Find the perimeter of an isosceles right triangle having an area of $100c{m}^2$.

• A. $20(\sqrt{2}+1)$
• B. $20(\sqrt{2}-1)$
• C. $2(\sqrt{2}+1)$
• D. $10(\sqrt{2}+1)$

Answer 1

The correct option is A $20(\sqrt{2}+1)$

Area $=\frac{1}{2}\times a\times a=\frac{a^2}{2}=100㎠$

$a=10\sqrt{2}㎝$

Perimeter$=2a+a\sqrt{2}$

$=20\sqrt{2}+10\times {\left(\sqrt{2}\right)}^2$

$=20(\sqrt{2}+1)㎝$