Question

AB and AC are the two chords of a circle whose radius is r . If p and q are the distance of a chord AB and AC from the centre respectively and if AB=2AC then prove that 4q²= p² + 3r²

Answer 1

Answer: Proved

Step-by-step explanation:  Given : AB and AC are two chords of a circle of radius r .

such that AB = 2 AC , p and q are the distance of AB and AC from

 OM = p

ON = q

To prove : 4q² = p² +3r²

Construction : Join OA

Proof : OM and ON are the perpendicular distance of chords AB and AC from O.

∴ AM = 1/2 AB

AN = 1/2 AC

[ ∵ perpendicular from the centre bisects the chord ]

AM = 1/2 AB = 2AC /2 = AC = 2 AN

∴ AM = 2 AN......(1)

In right Δ AMO

AO² = AM² + OM²

r² = AM² + p²

AM² = r² - p²

( 2AN)² = r² - p².. from (1)

4AN ² = r² - p²

AN² = 1/4 (r² - p²)...(2)

In Δ ANO

AO² = AN² + ON²

r² = AN² + q²

AN² = r² - q²...(3)

From equation (2) and (3), we get

1/4 (r²⁻p²) = (r²- q²)

r² ⁻p² = 4 ( r² - q ² )

4q² = 4r² -r² + p²

4q² = 3r² + p²

Hence proved