Question

The length of a rectangle is three times its width and the length of its diagonal is $6\sqrt{10}$ cm. The perimeter of the rectangle is
(a) 48 cm
(b) 36 cm
(c) 24 cm
(d) $24\sqrt{10}$ cm

Answer 1

(a) 48 cm
Let the width and the length of the rectangle be x cm and 3x cm, respectively.

Applying Pythagoras theorem:

$$\begin{gathered} (\mathrm{Diagonal}{)}^2=(\mathrm{Length}{)}^2+(\mathrm{Width}{)}^2 \\ \Rightarrow (6\sqrt{10}{)}^2=(3\mathrm{x}{)}^2+(\mathrm{x}{)}^2 \\ \Rightarrow 360=9{\mathrm{x}}^2+{\mathrm{x}}^2 \\ \Rightarrow 360=10{\mathrm{x}}^2 \\ \Rightarrow {\mathrm{x}}^2=\frac{360}{10} \\ \Rightarrow {\mathrm{x}}^2=36 \\ \Rightarrow \mathrm{x}=\pm 6 \\ \mathrm{Since}\mathrm{the}\mathrm{width}\mathrm{cannot}\mathrm{be}\mathrm{negative},\mathrm{we}\mathrm{will}\mathrm{neglect}-6. \end{gathered}$$

So, width of the rectangle is 6 cm.
Length of the rectangle = $\left(3\times 6\right)=18\mathrm{cm}$
Perimeter of the rectangle = 2(Length + Breadth) = 2(18 + 6) = 48 cm