Question

The difference between the length of the sides of a right triangle which form the right angle is $3cm$. If its perimeter is $36cm$. find the lengths of the sides .

Answer 1

REF.Image

given

$x-y=3$

$x=3+y$

$x+y+z=36$

$3+y+y+z=36$

$2y=33-z$

$z=33-2y$

By Pythagorean theorem

$x^2+y^2=z^2$

$(3+y{)}^2+y^2=(33-2y{)}^2$

$9+y^2+6y+y^2=1089+4y^2-132y$

$2y^2+6y+9=4y^2-132y+1089$

$138y=2y^2+1080$

$y^2+540-69y=0$

$y^2-69y+540=0$

$y^2-60y-9y+540=0$

$y(y-60)-9(y-60)=0$

$(y-9)(y-60)=0$

$y=9,60$

$60\to$ rejected as $p=36cm$

$\therefore$ $y=9cm$

$x=12cm$

$z=15cm$