Question

In the figure given above A and B are the centers of the two congruent circles with radius 17 units If AB = 30 units the length of the common chord DC is

• A. 25 units
• B. 18 units
• C. 10 units
• D. 16 units

Answer 1

The correct option is D 16 units

Let AB and CD intersect at M.
Now, In $△ACD$ and $△BCD$
$AC=BC$
$AD=BD$
$CD=CD$
Hence, $△ACD\cong △BCD$ (SSS rule)
Hence, $\angle ACD=\angle BCD$ (By CPCT)
Now, In $△ACM$ and $△BCM$
$\angle ACM=\angle BCM$ (Proved above)
$CM=CM$ (Common)
$AC=BC$ (Given)
Thus, $△ACM\cong △BCM$ (SAS rule)
Hence, $\angle AMC=\angle BMC$
Since, $\angle AMC+\angle BMC=180$ (Linear pair)
Thus, $\angle AMC=\angle BMC={90}^{\circ }$
and, $AM=BM=\frac{30}{2}=15$ (by CPCT)
Now, In $△BMC$
$B{C}^2=B{M}^2+M{C}^2$
${17}^2={15}^2+M{C}^2$
$M{C}^2=64$
$MC=8$ cm
Similarly, MD = 8 cm
hence, $CD=8+8=16$ cm