Question

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

Answer 1

Answer:

10 cm

Step-by-step explanation:

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 =   1 2    × AB × L

By Heron's formula, area of ΔABC =  

√ S ( S − a ) ( S −b ) ( S − c ) √ S ( S − a ) (S− b ) ( S − c )

where, S =   AB + BC + CA 2    =   13 + 5 + 12 2    = 15 cm

and a,b, and c are the length of three sides of the triangle. So, area of ΔABC =  √ 15 ( 15 − 13 ) ( 15 − 5 ) ( 15 − 12 ) √ 15 ( 15 − 13 ) ( 15 − 5 ) ( 15 − 12 )  = 30 cm

Area of ΔABC = 30 cm =   1 2    × AB × L =   1 2    × 12 × L

Therefore, L =   2 × 30 12    = 5 cm

Length of CD = 2L = 2 × 5 cm = 10 cm

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