Question

) AB and AC are the two chords of a circle whose radius is r . If p and q are the distance of chord AB and CD, from the centre respectively and if AB = 2AC then proove that 4 q 2 = p 2 + 3 r 2 .

Answer 1

Answer:

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

Ab = 2 AC

distance between AB and AC from the centre are P  and q

Let AC = x , then AB = 2x

The perpendicular from the centre  to the chord bisect the chord

CL = x/2 and AM = x

In Δ OLC,

r² = q² + (x/2)²

r² = q² + x²/4

In ΔOAM

4q² = 4( r² -x² /4)

4r² = 4r² -x²

4q² = 4r² - (r²-p²)

4q² = 4r² -r² +p²

4q² = 3r² +p²

Hence proved