Question

In the figure given, $AandB$ are the centres 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

Given that: The two circles are congurent with radius, $r=17units$

$A$ and $B$ are centers of the circles such that, $AB=30units$

Let point of intersection of $AB$ and $CD$ be $O$.

We know that the line joining the centers of two intersecting circles is the perpendicular bisector of the common chord.

Hence, $CO=OD$

Also $AO=OB=15cm$ as the circles are congruent

In $△AOC$

${AC}^2={AO}^2+{CO}^2$ ... (pythagoras theorem)

${17}^2={15}^2+{CO}^2$

${CO}^2={17}^2-{15}^2$

${CO}^2=(17-15)(17+15)$

${CO}^2=2\times 32=64$

$CO=8$

$CD=CO+OD=2\times CO=2\times 8=16\text{units}$

Answer: $DC=16\text{units}$