Question

If the area of a rectangle is $48sq.m$ and its diagonal is $10m$, then find its length, breadth, and perimeter.

Answer 1

Let length and breadth of rectangle be ${}^{\prime }{l}^{\prime }$ and ${}^{\prime }{b}^{\prime }$ respectively. Then,

Area of rectangle $=48m^2$

$=l\times b=48m^2\Rightarrow l=\frac{48}{b}⟶\left(1\right)$

and $l^2+b^2={\left(10\right)}^2=100$

from (1), $l=\frac{48}{b}$, we get

${\left(\frac{48}{b}\right)}^2+b^2=100\Rightarrow \frac{2304}{b^2}+b^2=100$

$\Rightarrow b^4-100b^2+2304=0$

after solving, $b=8or6$

If $b=8$, then $l=\frac{48}{8}=6$ &

If $b=6$, then $l=\frac{48}{6}=8$

Perimeter $=2(l+b)$

$=2(6+8)=28m$