Question

Find the perimeter of rectangle whose length is $40$ cm and diagonal is $41$ cm.

Answer 1

In right $△ADC$, by Pythagoras theorem,

$A{C}^2=A{D}^2+C{D}^2$
${41}^2={40}^2+C{D}^2$
$C{D}^2={41}^2-{40}^2=(41-40)(41+40)=81$
$CD=9$ cm
Thus, the breadth $CD=9$ cm
Perimeter = $2(l+b)=2\times (40+9)=98$ cm