NCERT Solutions for Class 9th Maths Chapter 5 In-text Questions — Angles Subtended by an Arc

Book page 107 Updated on2026-09-08

Q1.

Exercise: A circle with centre O is drawn, and A, B, C, D are points on the circle (see Fig. 5.19). Measure the angles subtended by arc AKB and arc CLD at the centre O. If the angle at the centre is less than 180°, it is a minor arc. If the angle at the centre is greater than 180°, it is a major arc. State whether arcs AKB and CLD are minor arcs or major arcs.

Answer

Measuring the two angles in Fig. 5.19 with a protractor:

Arc	Angle swept at O	Less / greater than 180°	Type
arc AKB	about 100°	less than 180°	minor arc
arc CLD	about 205° (reflex)	greater than 180°	major arc

How to read the figure correctly. The angle an arc subtends is the angle swept as the radius turns from one end of the arc to the other along that arc.

For arc AKB you turn from OA to OB passing through OK. K sits between A and B on the short way round, so you sweep the ordinary (non-reflex) angle AOB — about 100°.
For arc CLD you must go from OC to OD passing through OL. L lies on the long way round, so you sweep the reflex angle COD — about 205°. The direct angle COD is about 155°, but that belongs to the other arc from C to D, the one not containing L.

Angle for arc CLD + angle for the other arc from C to D = 360°
205° + 155° = 360° ✓

Why the naming letter matters: "arc CD" is ambiguous — there are two arcs joining C and D. Writing CLD names the middle point L and so fixes which of the two you mean. This is exactly why the book insists on three letters for an arc.

Q2.

Activity: Draw a circle and a chord AB. Fix an arc AKB formed by AB and a point K between A, B on the circle. Measure the angle subtended at the centre by arc AKB. Take three points P, Q, R on the circle outside arc AKB. Measure the angles subtended by arc AKB at points P, Q, R. What do you notice?

Answer

Two things emerge, and together they are Theorem 9.

The three angles ∠APB, ∠AQB and ∠ARB come out equal to one another — it makes no difference where on the circle (outside arc AKB) the point is taken.
Each of them is exactly half the angle that arc AKB subtends at the centre.

Sample reading (draw your own and check):
angle at the centre, ∠AOB = 110°
∠APB = ∠AQB = ∠ARB = 55° = ½ × 110°

Why it must be so: join P to the centre O and extend PO to meet the circle again. That splits ∠APB into two pieces, and each piece sits in an isosceles triangle whose two equal sides are radii. The exterior-angle theorem then doubles each piece at the centre. Adding the two pieces gives ∠AOB = 2∠APB — and the argument never used where P was, only that it lies on the circle outside the arc. That is why every position of P gives the same answer.

Try This: now take a point F inside the circle and a point G outside it, and measure ∠AFB and ∠AGB. You will find ∠AGB < ∠APB < ∠AFB. The constancy is special to points that lie on the circle — see Fig. 5.25.

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