AD is a diameter of a circle and AB is a chord. If AD = 26 cm, AB = 10 cm, find the perpendicular distance of AB from the centre of the circle.


Answer:

12 cm

Step by Step Explanation:
  1. Let us draw the figure as follows:

    We are told that the diameter AD = 26 cm and the length of the chord AB = 10 cm.
    We are asked to find the perpendicular distance of AB from the center O.
    This is the length of the line segment OC.
  2. We can see that △OAC is a right-angled triangle.
    We also know that OA is the radius, i.e half of the diameter AD.
    Therefore, AO = 13 cm.
  3. We know that the perpendicular to the chord from the center i.e. OC, divides the chord into two equal parts.
    Therefore, AC = CB =  
    AB
    2
      =  
    10
    2
      = 5 cm
  4. Since, △OAC is a right angle triangle, OC2 + AC2 = OA2
    ⇒ OC2 + 52 = 132
    ⇒ OC2 = 169 - 25
    ⇒ OC2 = 144
    ⇒ OC = 12 cm
  5. Therefore, perpendicular distance of AB from the centre of the circle is 12 cm.

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