### Two circles of radii 13 cm and 5 cm intersect at two points and the distance between their centres is 12 cm. Find the length of the common chord.

**Answer:**

10 cm

**Step by Step Explanation:**

- Take a look at the following image:

Here, A is the center of the first circle and B the center of the second circle. The common chord is CD.

Join AD and BD. Now, consider ∠ABC and ∠ABD. We have,

AC = AD (Radius of the circle with centre A)

BC = BD (Radius of the circle with centre B)

AB = AB (common)

Hence, ∠ABC ≅ ∠ABD by SSS.

So, ∠AOC = ∠AOD by CPCT. Also, AO = OB and CO = DO by CPCT.

Also, ∠AOC + ∠AOD = 180° (angles on a straight line)

⇒ ∠AOC + ∠AOC = 180°

⇒ 2 × ∠AOC = 180°

⇒ ∠AOC = 90°

Now, we know that the line AB bisects CD, and is perpendicular to it.

Also, the perpendicular from C to AB is half the length of CD. Let us call this length as L.

Area of ΔABC =

× AB × L1 2 - By Heron's formula, area of ΔABC = ^@ \sqrt{ S(S-a)(S-b)(S-c) } ^@,

where, S =

=AB + BC + CA 2

= 15 cm13 + 5 + 12 2

and*a,b,*and*c*are the length of three sides of the triangle. So, area of ΔABC = ^@ \sqrt{ 15(15-13)(15-5)(15-12) } ^@ = 30 cm - Area of ΔABC = 30 cm =

× AB × L =1 2

× 12 × L1 2

Therefore, L =

= 5 cm2 × 30 12

Length of CD = 2L = 2 × 5 cm =**10 cm**