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

(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”

P

_{0}P_{2}= a; P_{1}P_{2}= b;
P

_{1}P_{2}= h => length of the chord = 2h
Radius
of bigger circle = r

_{o}= 20
Radius
of smaller circle = r

_{1}= 15
Applying
Pythagoras’ to the right triangle P

_{0}P_{2}P_{4},
a

^{2}+h^{2}= r_{0}^{2}= 20^{2}= 400 ---(1)
Applying
Pythagoras’ to the right triangle P

_{1}P_{2}P_{4},
b

^{2}+h^{2}= r_{1}^{2}= 15^{2}= 225 ---(2)
(1)-(2)
gives,

a

^{2}-b^{2}= 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),

16

^{2}+h^{2}= 400 => h^{2}= 400-256 = 144 => h =12
Hence length
of chord = 2h = 24

**Answer (2)**
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