Question

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

• A. 15
• B. 12
• C. none of the above
• D. 24

Answer 1

Answer: (D)

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