Question

Find the perimeter of an isosceles right-angle triangle having an area of 500 $m^2$

Answer 1

REF.Image.

area of $△ABC=\frac{1}{2}\times BC\times AB$

area of $△ABC=\frac{1}{2}{x}^2$ [isosceles ]

$\frac{x^2}{2}=500$ [given]

$x^2=1000$

$x=10m$ ...(1)

finding AC using Pythagoras theorem

$A{C}^2=A{B}^2+B{C}^2$

$A{C}^2=x^2+x^2$

$2x^2=A{C}^2$

$AC=x\sqrt{2}$

from (1)

$AC=10\sqrt{2}m$

perimeter = AB + BC + AC

$=10+10+10\sqrt{2}$

$=20+10\sqrt{2}$

$=10(2+\sqrt{2})m$