Question 135

The length of the common chord of two circles of radii 30 cm and 40 cm whose centres are 50 cm apart, is (in cm)

Solution

OA = 30 cm , O'A = 40 cm , OO' = 50 cm

Let OC = x => CO' = 50-x

In $$\triangle$$OAC,

=> $$AC^2 = OA^2 - OC^2$$

=> $$AC^2 = 30^2 - x^2$$

Similarly, In $$\triangle$$O'AC => $$AC^2 = 40^2 - (50-x)^2$$

=> $$30^2 - x^2 = 40^2 - (50-x)^2$$

=> $$900 - x^2 = 1600 - 2500 + 100x - x^2$$

=> $$100x = 1800$$ => $$x = 18$$

$$\therefore$$ AC = $$\sqrt{30^2 - 18^2}$$

= $$\sqrt{48*12}$$ = 24

=> AE = 2*24 = 48 cm

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