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.


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Answer:

10 cm

Step by Step Explanation:
  1. 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 =  
    1
    2
      × AB × L
  2. By Heron's formula, area of ΔABC = √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) = 30 cm
  3. 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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