1. The length of the common chord of two circles of radii 15 cm and 20 cm, whose centres are 25 cm apart, is
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By: anil on 05 May 2019 02.30 am
The radii of both the circles and the line joining the centers of the two circles form a right angled triangle. So, the length of the common chord is twice the length of the altitude dropped from the vertex to the hypotenuse.
Let the altitude be h and let it divide the hypotenuse in two parts of length x and 25-x
So, $$h^2 + x^2 = 15^2$$ and $$h^2 + (25-x)^2 = 20^2$$
=> $$225 - x^2 = 400 - x^2 + 50x - 625$$
=> 50x = 450 => x = 9 and h = 12
So, the length of the common chord is 24 cm.
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Let the altitude be h and let it divide the hypotenuse in two parts of length x and 25-x
So, $$h^2 + x^2 = 15^2$$ and $$h^2 + (25-x)^2 = 20^2$$
=> $$225 - x^2 = 400 - x^2 + 50x - 625$$
=> 50x = 450 => x = 9 and h = 12
So, the length of the common chord is 24 cm.