Question

The two adjacent sides of a cyclic quadrilateral are 2 and 5 and the angle between them is 60^o. If the third side is 3, the remaining fourth side is

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (A)

2

Answer 2

Let $a = P Q = 2$ , $b = Q R = 5$, $R S = c$, $S P = d$

Given, Area of quadrilateral PQRS = $4 \sqrt { 3 }$

⇒ Area of ( $\triangle P Q R + \triangle P R S )$ = $\frac { 1 } { 2 } \left[ 2 \times 5 \sin 60 ^ { \circ } + c \times d \sin 120 ^ { \circ } \right]$

⇒ $4 \sqrt { 3 } = \frac { 5 \sqrt { 3 } } { 2 } + \frac { \sqrt { 3 } } { 4 } c d$ ⇒ $c d = 6$ ........(i)

Now by cosine formula $R S ^ { 2 } + S P ^ { 2 } - 2 R S \cdot S P \cdot \cos 120 ^ { \circ } = P R ^ { 2 }$ $= P Q ^ { 2 } + Q R ^ { 2 } - 2 P Q \cdot Q R \cos 60 ^ { \circ }$

⇒ $c ^ { 2 } + d ^ { 2 } - 2 c d \left( - \frac { 1 } { 2 } \right) = 4 + 25 - 2 ( 2 ) ( 5 ) \left( \frac { 1 } { 2 } \right)$

⇒$c ^ { 2 } + d ^ { 2 } + c d = 19$⇒ $c ^ { 2 } + d ^ { 2 } = 19 - 6 = 13$ ……..(ii)

Solving (i) and (ii) we get $c = 2$and $d = 3$.