Question

The length of a rectangle is three times of its width. If the length of the diagonal is $8\sqrt{10}$ m, then the perimeter of the rectangle is

(a) $15\sqrt{10\mathrm{m}}$
(b) $16\sqrt{10\mathrm{m}}$
(c) $24\sqrt{10\mathrm{m}}$
(d) 64 m

Answer 1

(d) 64 m
Let us consider a rectangle ABCD.
Also, let us assume that the width of the rectangle, i.e., BC be x m.
It is given that the length is three times width of the rectangle.
Therefore, length of the rectangle, i.e., AB = 3x m


Now, AC is the diagonal of rectangle.
In right angled triangle ABC.
AC² = AB² + BC² {Using Pythagoras theorem}
${\left(8\sqrt{10}\right)}^2={\left(3x\right)}^2+{\left(x\right)}^2$
640 = 9x² + x²
640 = 10x²
x² = $\frac{640}{10}$ = 64
x = $\sqrt{64}$ = 8 m

Thus, breadth of the rectangle = x = 8 m
Similarly, length of the rectangle = 3x = 3 $\times$ 8 = 24 m
Perimeter of the rectangle = 2 (Length + Breadth)
= 2 (24 + 8)
= 2 $\times$ 32 = 64 m