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

**Answer :**

In right ∆ADC

cot α = CD/AD = 3/2

Take CD = 3x then AD = 2x

AC^{2} = CD^{2} + AD^{2}

(15)^{2} = (3x)^{2} + (2x)^{2}

13x^{2} = 225

⇒ x^{2} = 225/13

x = √225/13 = 15/√13

Length of rectangle (l) = 3x = (3 × 15)/ √13 = 45/√13 cm

Breadth of rectangle (b) = 2x = (2×15)/√13 = 30/√13 cm

**(i)** Perimeter of rectangle = 2 (l + b)

= 2 (45/√13 + 30/√13)

So we get

= 2 × 75/√13

= 150/√13 cm

**(ii)** Area of rectangle = l × b

Substituting the values of l and b

= 45/√13 × 30/√13

= 1350/13

= 103 (11/13) cm^{2}