Question

The area of the rectangle is 192 $c{m}^2$ and its perimeter is 56 cm.Find the dimensions of the rectangle.

Answer 1

The area of a rectangle = l $\times$ b = 192 $c{m}^2$

$l=\frac{192}{b}$

The perimeter $=2l+2b=56$

$2\times \left(\frac{192}{b}\right)+2b=56$

Multiplying throughout by b,

$384+2b^2=56b$

Rearranging&dividing by 2

$b^2-28b+192=0$

$b^2-16b-12b+192=0$

$b(b-16)-12(b-16)=0$

$(b-16)(b-12)=0$

so $b-16=0$

i.e b= 16

or
$b-12=0$

i.e b = 12

if l = 12 then b = 16 or vice versa

Since length is greater that breadth, l = 16 and b = 12

Area $=l\times b=16\times 12=192$

Perimeter $=(2\times 16)+(2\times 12)=52+24=76$