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Question
In the given figure, ▢PQRS is cyclic. side PQ ≅ side RQ. ∠PSR = 110°, Find -
(1) measure of ∠PQR
(2) m(arc PQR)
(3) m(arc QR)
(4) measure of ∠PRQ
Solution
(1) ▢PQRS is a cyclic quadrilateral. ...[Given]
∴ ∠PSR + ∠PQR = 180° ...[Opposite angles of a cyclic quadrilateral are supplementary]
∴ 110° + ∠PQR = 180°
∴ ∠PQR = 180° − 110°
∴ m∠PQR = 70°
(2) ∠PSR = `1/2` m(arc PQR) .....[Inscribed angle theorem]
∴ 110° = `1/2` m(arc PQR)
∴ m(arc PQR) = 220°
(3) In ∆PQR,
Side PQ ≅ side RQ ...[Given]
∴ ∠PRQ ≅ ∠QPR ...[Isosceles triangle theorem]
Let ∠PRQ = ∠QPR = x
Now, ∠PQR + ∠QPR + ∠PRQ = 180° ...[Sum of the measures of angles of a triangle is 180°]
∴ ∠PQR + x + x = 180°
∴ 70° + 2x = 180°
∴ 2x = 180° − 70°
∴ 2x = 110°
∴ x = `(110°)/2`
∴ x = 55°
∴ ∠PRQ = ∠QPR = 55° ......(i)
But, ∠QPR = `1/2` m(arc QR) .....[Inscribed angle theorem]
∴ 55° = `1/2` m(arc QR)
∴ m(arc QR) = 110°
(4) In ∆PQR,
∠PQR + ∠PRQ + ∠QPR = 180° ...[Sum of the measures of angles of a triangle is 180°]
70° + ∠PRQ + 55° = 180°
∠PRQ = 180° - 125°
∠PRQ = 55°.
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Solution:
MRPN is cyclic
The opposite angles of a cyclic square are `square`
∠R + ∠N = `square`
∴ (5x – 13)° + (4x + 4)° = `square`
∴ 9x = 189°
∴ x = `square`
∴ ∠R = (5x – 13)° = `square`
∴ ∠N = (4x + 4)° = `square`
Prove the following theorems:
Opposite angles of a cyclic quadrilateral are supplementary.
In the figure, PQRS is cyclic, side PQ ≅ side RQ, ∠PSR = 110°. Find
(i) measure of ∠PQR
(ii) m(arc PQR)
(iii) m(arc QR)
If two consecutive angles of cyclic quadrilateral are congruent, then prove that one pair of opposite sides is congruent and other is parallel.
