AB and AC are two chords of a circle such that AB = 2AC. If distances of AB and AC from the centre are 3 cm and 6 cm respectively, find the area of circle. (Assume π =3)


  1. Home
  2. Math
  3. All Questions
Answer:

135 cm2

Step by Step Explanation:
  1. Take a look at the representative image below:


    We are told that AB = 2AC.
    Also, if the perpendicular from O to AC meets the chord at Q, then OQ = 6 cm.
    Similarly, OP = 3 cm.
  2. As the perpendicular from the centre on the chord bisects the chord, OQ bisects AC, and OP bisects AB.
    From the earlier relation AB = 2AC.
    Therefore, BP = 2CQ
    Let us assume CQ = x.
    Then, BP = 2x
  3. Now consider ΔOQC,
    OC = r, the radius of the circle, and OQ = 6 cm
    As, the distance of a chord from the centre is always the perpendicular distance. ΔOQC is a right-angled triangle.
    By using pythagoras theorum, OQ2 + CQ2 = r2
    62 + x2 = r2
    or, 36 + x2 = r2 ------(1)
  4. Similarly, ΔOPB is a right-angled triangle,
    OP2 + BP2 = r2
    32 + (2x)2 = r2
    or, 9 + 4x2 = r2 ------(2)
  5. Subtracting equation (1) from equation (2), we get:
    (9 - 36) + (4x2 - x2) = 0
    or, 3x2 = 27
    or, x2 =  
    27
    3
     
  6. On substituting x2 =  
    27
    3
      in equation (1), we get:
    36 +  
    27
    3
      = r2
    or, r2 =  
    3 × 36 + 27
    3
      =  
    135
    3
     
  7. Therefore, area of the circle = πr2 = 3 ×  
    135
    3
      = 135 cm2

You can reuse this answer
Creative Commons License