In the given figure, ABCD is a trapezium with DC//AB, ∠AOB = 126° and ∠DCQ = ∠CDP = 52°.
Find the values of x and y.

Question

In the given figure, ABCD is a trapezium with DC//AB, ∠AOB = 126° and ∠DCQ = ∠CDP = 52°. Find the values of x and y.

KnowledgeBoat · Accepted Answer

Given: ABCD is a trapezium with DC//AB, ∠AOB = 126° and ∠DCQ = ∠CDP = 52°. To find: The values of x and y. Construction: Extend P and Q such that they meet at R. In Δ RDC, ⇒ ∠RDC = ∠RCD = 52° (since DCQ = CDP = 52°) ...................(1) Since the base angles are equal, Δ RDC is isosceles, which means: ⇒ DR = CR. In quadrilateral ABCD, since AB ∥ DC, we use the property of corresponding angles: ⇒ ∠RAB = ∠RDC = 52° ( Corresponding angles) ....................(2) ⇒ ∠RBA = ∠RCD = 52° ( Corresponding angles) ....................(3) Since ∠RAB = ∠RBA, we conclude that: ⇒ AR = RB From the figure, we also note: AD = AR - DR = RB - CR = BC Therefore, we can say that ABCD is an isosceles trapezium. Since, OA = OB, we conclude that: ⇒ ∠OAB = ∠OBA Using the angle sum property in Δ AOB: ⇒ ∠OAB + ∠OBA + ∠AOB = 180° ⇒ ∠OAB + ∠OAB + 126° = 180° ⇒ 2∠OAB + 126° = 180° ⇒ 2∠OAB = 180° - 126° ⇒ 2∠OAB = 54° ⇒ ∠OAB = = 27° Now, ∠x = ∠DAB - ∠OAB = 52° - 27° = 25° Using angle sum property in Δ ABC: ∠ABC + ∠BAC + ∠ACB = 180° ⇒ 52° + 27° + y = 180° ⇒ 79° + y = 180° ⇒ y = 180° - 79° = 101° Hence, the value of x = 25° and y = 101°.

Mathematics

In the given figure, ABCD is a trapezium with DC//AB, ∠AOB = 126° and ∠DCQ = ∠CDP = 52°.
Find the values of x and y.

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Mathematics

In the given figure, ABCD is a trapezium with DC//AB, ∠AOB = 126° and ∠DCQ = ∠CDP = 52°.Find the values of x and y.

Rectilinear Figures

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