In rectangle ABCD, diagonal BD = 26 cm and cotangent of angle ABD = 1.5. Find the area and the perimeter of the rectangle ABCD.

Given:

cot ∠ ABD = 1.5 = $\dfrac{15}{10} = \dfrac{3}{2}$

cot ∠ ABD = $\dfrac{Base}{Perpendicular} = \dfrac{3}{2}$

∴ If length of AB = 3x cm, length of AD = 2x cm.

In Δ ABD,

⇒ BD² = AB² + AD² (∵ BD is hypotenuse)

⇒ BD² = (3x)² + (2x)²

⇒ BD² = 9x² + 4x²

⇒ BD² = 13x²

⇒ BD = $\sqrt{13\text{x}^2}$

⇒ BD = x $\sqrt{13}$

It is given that BD = 26 cm

So, x $\sqrt{13}$ = 26

⇒ x = $\dfrac{26}{\sqrt{13}}$

⇒ x = $\dfrac{26 \times \sqrt{13}}{\sqrt{13} \times \sqrt{13}}$

⇒ x = $\dfrac{26 \sqrt{13}}{13}$

⇒ x = $2 \sqrt{13}$

AB = DC = 3x = 3 x $2 \sqrt{13}$ = $6 \sqrt{13}$ cm and AD = BC = 2 x $2 \sqrt{13}$ = $4 \sqrt{13}$ cm

Perimeter of rectangle ABCD = sum of all sides of rectangle

= AB + BC + CD + DA

= $6 \sqrt{13} + 4 \sqrt{13} + 6 \sqrt{13} + 4 \sqrt{13}$ cm

= $20 \sqrt{13}$ cm

Area of rectangle = length x breadth

= AB x BC

= $6 \sqrt{13} \times 4 \sqrt{13}$ cm²

= 24 $\times$ 13 cm²

= 312 cm²

Hence, area of rectangle = 312 cm² and perimeter = $20 \sqrt{13}$ cm.