Saturday, 29 October 2011

Geometry-11 (CAT-2002)

There is a common chord of two circles with radius 15 and 20.The distance between the two centres is 25.The length of the chord is:
(1)48   (2)24   (3)36   (4)28

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Solution:

“For intersecting circles, the line joining the centres is perpendicular to and bisects the common chord”



P0P2 = a;  P1P2 = b;
P1P2 = h => length of the chord = 2h
Radius of bigger circle = ro = 20
Radius of smaller circle = r1 = 15
Applying Pythagoras’ to the right triangle P0P2P4,
a2+h2 = r02 = 202 = 400   ---(1)
Applying Pythagoras’ to the right triangle P1P2P4,
b2+h2 = r12 = 152 = 225   ---(2)
(1)-(2) gives,
a2-b2 = 175        ---(3)
a+b = distance between the centres = 25   ----(4)
(3)/(4) gives:
a-b = 7                ---(5)
Solving (4) and (5), a = 16; b = 9;
Substituting ‘a’ value in (1),
162+h2 = 400 => h2 = 400-256 = 144 => h =12
Hence length of chord = 2h = 24
Answer (2)





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