The lengths of two adjacent sides of a cyclic quadrilateral are 2 units and 5 units and the angle between them is 60 o . If the area of the quadrilateral is 4 3 sq. units, then the perimeter of the quadrilateral is

Step-by-step solution

cos ⁡ 60 o = 4 + 25 - c 2 2.2 . 5 ⇒ 10 = 29 - c 2 ⇒ c 2 = 19 ⇒ c = 19 Now, cos 120 ° = a 2 + b 2 − c 2 2 a b ⇒ - 1 2 = a 2 + b 2 - 19 2 a b ⇒ a 2 + b 2 - 19 = - a b ⇒ a 2 + b 2 + a b = 19 Area of quadrilateral = 1 2 × 2 × 5 sin ⁡ 60° + 1 2 a b sin ⁡ 120° = 4 3 ⇒ 5 3 2 + a b 3 4 = 4 3 ⇒ a b 4 = 4 - 5 2 = 3 2 ⇒ a b = 6 ⇒ a 2 + b 2 = 13 ⇒ a = 2 , b = 3 Perimeter = 2 + 5 + 2 + 3 = 12