Question

The area of a rectangle is 192 cm² and its perimeter is 56 cm. Find the dimensions of rectangle.

Answer 1

Let l cm and b cm be the length and breadth of the rectangle, respectively.
Given:
Area of the rectangle = 192 cm²
Perimeter of the rectangle = 56 cm

Now, area of the rectangle = length × breadth
$\Rightarrow$ 192 = lb
$\Rightarrow l=\frac{192}{b}$ ...(1)
Also, perimeter of the rectangle = 2(length + breadth)
$\Rightarrow$ 56 = 2 (l + b)
$\Rightarrow$ l + b = 28
$$\begin{gathered} \Rightarrow \frac{192}{b}+b=28\left\{\mathrm{From}\left(1\right)\right\} \\ \Rightarrow 192+b^2=28b \\ \Rightarrow b^2-28b+192=0 \\ \Rightarrow b^2-16b-12b+192=0 \\ \Rightarrow b(b-16)-12(b-16)=0 \\ \Rightarrow (b-12)(b-16)=0 \end{gathered}$$

i.e., b = 16 cm or 12 cm

Substituting b = 16 cm in eq. (1), we get:

$l=\frac{192}{16}=12\mathrm{cm}$

Substituting b = 12 cm in eq. (1), we get:

$l=\frac{192}{12}=16\mathrm{cm}$

∴ Length = 12 cm or 16 cm
and breadth = 16 cm or 12 cm