What is the length of the common chord of two circles of radii 15 cm and 20 cm whose centre are 25 cm apart?

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Solution

The correct option is D 24
Let P and Q be the centre of the two circles respectively and AB be the common chord
The $O P ⊥ A B$ and bisects AB, i.e, AO=OB
Also, $O Q ⊥ A B$
GIven AP=15 cm,AQ=20cm and PQ=25 cm
Let OP=x cm.Then OQ=(25-x)cm
In rt. $∠ d △ A O P, A O^{2} = A P^{2} - O P^{2} = 15^{2} - x^{2}$
In rt. $∠ d △ A O Q,$
$A O^{2} = A Q^{2} - O Q^{2} = 20^{2} - (25 - x)^{2}$ ....(ii)
From eq.(i) and(ii)
$15^{2} - x^{2} = 20^{2} - (25 - x)^{2}$
$\Rightarrow 225 - x^{2} = 400 - (625 - 50 x + x^{2} \Rightarrow 225 - x^{2} = 400 - 625 + 50 x - x^{2}$
$\Rightarrow 50 x = 450 \Rightarrow x = 9$
$∴ A O = \sqrt{15^{2} - 9^{2}} = \sqrt{225 - 81} = \sqrt{144} = 12 c m (F r o m (i))$
$\Rightarrow A B = 2 \times 12 c m = 24 c m$

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