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  • 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 4q square equal to p square plus 3r square.

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  

Answers (1)

AB and AC are two chords of a circle of radius r such that AB = 2AC.

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

To Prove: 4q2 = p2 + 3r2

Proof:

Let AB = 2x then AC = x                   (Given, AB = 2AC)

Draw ON perpendicular to AB and OM perpendicular to AC

AM = MC = x/2                                  (perpendicular from the centre bisects the chord)

AN = NB = x                                       (perpendicular from the centre bisects the chord)

In DOAM, applying Pythagoras theorem

AO2 = AM2 + MO2

AO2 = (x/2)2 + q2                                 ….(i)    

In\triangleOAN, applying Pythagoras theorem

AO2 = (AN)2 + (NO)2

AO2 = (x)2 + p2                                      ….(ii)

From equation (i) and (ii)

\\\left ( \frac{x}{2} \right )^{2}+\left ( q \right )^{2}=x^{2}+\left ( p \right )^{2}\\ \frac{x^{2}}{4}+q^{2}=x^{2}+p^{2} 

x2 + 4q2 = 4x2 + 4p2                [Multiply both sides by 4]

4q2 = 3x2 + 4p2

4q2 = p2 + 3 (x2 + p2)

4q2 = p2 + 3r2                                           (In right angle \triangleOAN, r2 = x2 + p2)

Hence Proved.

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