ab and ac are two chords of the circle having radius R if p and q are distance of the chord ab and AC from the centre respectively and if AB is equal to 2 AC prove that 4 Q square is equal to p square + 3R square.
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ab and ac are two chords of the circle having radius R if p and q are distance of the chord ab and AC from the centre respectively and if AB is equal to 2 AC prove that 4 Q square is equal to p square + 3R square.
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Given, AB and AC are two chords of a circle of radius r
Again, AB = 2AC and the distances of AB and AC from the center are p and q.
Let AC = x, then AB = 2x
We know that the perpendicular drawn from the centre to the chord bisect the chord.
So, CL = x/2 and AM = x
Now, in ΔOLC,
r² = q² + (x/2)² (By Pythagoras theorem)
=> r² = q² + x² /4
Again, in ΔOAM,
4q² = 4(r² - x² /4) (using 1)
4r² = 4r² - x²
=> 4q² = 4r² - (r² - p² ) (using 2)
=> 4q² = 4r² - r² + p²
=> 4q² = 3r² + p²
Hope this helps!