Question

PQRS is a cyclic quadrilateral,
P Р
Side PQ = side QR, ZPSR =110°
Find : i) ZPOR
SK
1100
ii) m (arc PQR)
iii) m (arc QR)
R​

Answer 1

Answer:

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° ii. ∠PSR= 1/2 m (arcPQR) [Inscribed angle theorem] 110°= 1/2 m (arcPQR) ∴ m(arc PQR) = 220° iii. 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 = 100°/2 = 55° ∴ ∠PRQ = ∠QPR = 55°….. (i) But, ∠QPR = 1/2 nm(arc QR) [Inscribed angle theorem] ∴ 55° = 1/2 m(arc QR) ∴ m(arc QR) = 110° iv. ∠PRQ = ∠QPR =55° [From (i)] ∴ m ∠PRQ = 55° Read more on Sarthaks.com - https://www.sarthaks.com/851560/in-the-adjoining-figure-quarilateral-pqrs-is-cyclic-side-pq-side-rq-psr-110-find?show=851569#a851569

Answer 2

Answer:

Answer will be 24

Step-by-step explanation:

The Zporq is 90