PQRS is a cyclic quadrilateral. Find \(x\) + \(y\)
50
60
15
0
Correct answer is D
∠PQR + ∠PSR = 180o (opp. angles of cyclic quad. are supplementary)
⇒ 5\(x\) - \(y\) + 10 + (-2\(x\) + 3\(y\) + 145) = 180
⇒ 5\(x\) - \(y\) + 10 - 2\(x\) + 3\(y\) + 145 = 180
⇒ 3\(x\) + 2\(y\) + 155 = 180
⇒ 3\(x\) + 2\(y\) = 180 - 155
⇒ 3\(x\) + 2\(y\) = 25 ----- (i)
∠QPS + ∠QRS = 180o (opp. angles of cyclic quad. are supplementary)
⇒ -4\(x\) - 7\(y\) + 150 + (2\(x\) + 8\(y\) + 105) = 180
⇒ -4\(x\) - 7\(y\) + 75 + 2\(x\) + 8\(y\) + 180 = 180
⇒ -2\(x\) + \(y\) + 255 = 180
⇒ -2\(x\) + y = 180 - 255
⇒ -2\(x\) + \(y\) = -75 ------- (ii)
⇒ \(y\) = -75 + 2\(x\) -------- (iii)
Substitute (-75 + 2\(x\)) for \(y\) in equation (i)
⇒ 3\(x\) + 2(-75 + 2\(x\)) = 25
⇒ 3\(x\) - 150 + 4\(x\) = 25
⇒ 7\(x\) = 25 + 150
⇒ 7\(x\) = 175
⇒ \(x = \frac{175}{7} = 25\)
From equation (iii)
⇒ \(y\) = -75 + 2(25) = -75 + 50
⇒ \(y\) = -25
∴ \(x\) + \(y\) = 25 + (-25) = 0
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