AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre, prove that 4q² = p² + 3r².

Solution:

Given, AB and AC are two chords of a circle of radius r

AB = 2AC

The distance of AB and AC from the centre are p and q.

We have to prove that 4q² = p² + 3r²

Let AC = a

AB = 2a

A perpendicular is drawn to the chords AC and AB from centre O at M and N

AM = MC = a/2

AN = NB = a

In triangle OAM,

By Pythagorean theorem,

AO² = AM² + MO²

AO² = (a/2)² + q² ------------------------ (1)

In triangle OAN,

By Pythagorean theorem,

AO² = AN² + NO²

AO² = a² + p² ----------------------- (2)

In triangle OAN,

By Pythagorean theorem,

r² = a² + p² --------------------------- (3)

From (1) and (2),

(a/2)² + q² = a² + p²

a²/4 + q² = a² + p²

a² + 4q² = 4a² + 4p²

4q² = 4a² - a² + 4p²

4q² = 3a² + 4p²

4q² = p² + 3a² + 3p²

4q² = p² + 3(a² + p²)

From (3),

Therefore, 4q² = p² + 3r²

✦ Try This: In a circle of radius 45 cm, an arc subtends an angle of 60° at the centre. Find length of the arc.

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 10

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NCERT Exemplar Class 9 Maths Exercise 10.4 Problem 12

AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre, prove that 4q² = p² + 3r².

Summary:

AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre, it is proven that 4q² = p² + 3r²

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☛ Related Questions:

• Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD
• In Fig. 10.20,O is the centre of the circle, ∠BCO = 30°. Find x and y
• In Fig. 10.21, O is the centre of the circle, BD = OD and CD ⊥ AB. Find ∠CAB