Question

In the rectangle $WXYZ$, $XY+YZ=17cm$ and $XZ+YW=26cm$, calculate the length and breadth of the rectangle.

Answer 1

$XZ+YW=26cm$
$d_1+d_2=26cm$
$2d=26cm$ $[\because d_1=d_2]$
$\therefore d=13cm$
$\therefore XZ=YW=13cm$
Let length $XY$ be $xcm$ $\therefore$ breadth $=XW=(17-x)cm$
In $△WXY$, ${WX}^2+{XY}^2={WY}^2$ [$\because$ pythagoras theorem]
${(17-x)}^2+x^2={13}^2$
$(289-34x+x^2)+x^2=169$
$2x^2-34x+120=0$

On dividing above equation by $2$ we get,
$x^2-17x+60=0$
$x^2-12x-5x+60=0$
$x(x-12)-5(x-12)=0$
$(x-12)(x-5)=0$
$\therefore x-12=0$ or $x-5=0$
$\therefore x=12$ or $x=5$
$\therefore$ Length $=12cm$, breadth $=5cm$