In the adjoining figure, ABCD is a rectangle. Its diagonal AC = 15 cm and ∠ACD = α. If cot α = $\dfrac{3}{2}$, find the perimeter and the area of the rectangle.

Given,

$\text{cot α} = \dfrac{3}{2}$ ………..(1)

$\text{cot α} = \dfrac{\text{Base}}{\text{Perpendicular}} = \dfrac{CD}{AD}$ ……..(2)

From (1) and (2) we get,

$\dfrac{CD}{AD} = \dfrac{3}{2}$

Let CD = 3x and AD = 2x.

In right angle triangle ADC,

$\Rightarrow AC^2 = AD^2 + CD^2 \\[1em] \Rightarrow 15^2 = (2x)^2 + (3x)^2 \\[1em] \Rightarrow 225 = 4x^2 + 9x^2 \\[1em] \Rightarrow 13x^2 = 225 \\[1em] \Rightarrow x^2 = \dfrac{225}{13} \\[1em] \Rightarrow x = \sqrt{\dfrac{225}{13}} \\[1em] \Rightarrow x = \dfrac{15}{\sqrt{13}}.$

Perimeter of rectangle (P) = 2(CD + AD)

Area of rectangle (A) = CD × AD

Substituting values we get :

$P = 2 \times (3x + 2x) \\[1em] = 2 \times 5x \\[1em] = 10x \\[1em] = 10 \times \dfrac{15}{\sqrt{13}} \text{ cm}. \\[1em] = \dfrac{150}{\sqrt{13}} \text{ cm}. \\[1em] A = 3x \times 2x \\[1em] = 6x^2 \\[1em] = 6 \times \Big(\dfrac{15}{\sqrt{13}}\Big)^2 \\[1em] = 6 \times \dfrac{225}{13} \\[1em] = \dfrac{1350}{13} \\[1em] = 103\dfrac{11}{13} \text{ cm}^2.$

Hence, perimeter of rectangle = $\dfrac{150}{\sqrt{13}}$ cm and area = $103\dfrac{11}{13}$ cm².