Angle Subtended by an Arc of a Circle | Circles | Assessment | English | Grade 9

The angle subtended at the center by the diameter of a semicircle is of ____.

• 60⁰

• 360⁰

• 180⁰

• 90⁰

Tags

CCSS.4.MD.C.5B

CCSS.HSG.CO.A.1

Equal ____ of the congruent circles subtend equal angles at the centers.

• Segments

• Chords

• Radii

• Tangents

Answer explanation

Equal chords of the congruent circles subtend equal angles at the centers.

Tags

CCSS.HSG.C.A.2

Find the value of x in the image.

• 60⁰

• 160⁰

• 120⁰

• 40⁰

Answer explanation

Using the theorem that the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle. ∠MOQ = 2∠MNQ ∠MNQ = 1/2∠MOQ = 1/2 x 80 = 40⁰

Tags

CCSS.HSG.C.A.2

In the given figure ∠ABC = 30⁰, then ∠AOC is equal to ____.

• 120⁰

• 90⁰

• 60⁰

• 80⁰

Answer explanation

Using the theorem the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle. ∠AOC = 2∠ABC = 2 x 30⁰ = 60⁰

Tags

CCSS.HSG.C.A.2

Find the value of x in the image.

• 52⁰

• 68⁰

• 48⁰

• 72⁰

Answer explanation

The angle in a semicircle is right angle. ∠PRQ = 90⁰ Sum of interior angles in a triangle is 180⁰ ∠QPR + ∠PRQ + ∠PQR = 180⁰ x + 90 + 38 = 180⁰ x + 128 = 180⁰ x = 52⁰

Tags

CCSS.HSG.C.A.2

Find the value of ∠QOR in the image.

• 110⁰

• 65⁰

• 130⁰

• 85⁰

Answer explanation

OP = OQ = OR (Radius) ∠ORP = ∠OPR = 30⁰ ∠OQP = ∠OPQ = 35⁰ ∠QOR = 2∠QPR ∠QPR = ∠QPO + ∠OPR = 35⁰ + 30⁰ = 65⁰ ∠QOR = 2∠QPR (the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle) = 2 x 65⁰ = 130⁰

Tags

CCSS.HSG.C.A.2

Find the value of x in the image.

• 52⁰

• 65⁰

• 68⁰

• 42⁰

Answer explanation

∠POQ = 2∠PRQ (the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle) ∠POQ = 2(38⁰) = 76⁰ In triangle POQ, ∠OPQ + ∠PQO + ∠QOP = 180⁰ OP = OQ = OR = Radius ∠OPQ = ∠OQP = x x + x + 76 = 180⁰ 2x = 104⁰ x = 52⁰

Tags

CCSS.HSG.C.A.2