In the figure, AB and CD  are diameters of the circle with center O, and AB⊥CD and chord DF intersect AB at E and DE = 6, EF = 2. Find the area of the circle.

A. 25π
B. 24π
C. 36π
D. None of the above


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  1. Mayank Khanna

    D- 36π
    OD is the radius and ODE is right triangle with angle DOE as 90. So using sin90= perpendicular/ hypotenuse. We have 6 as perpendicular which is also radius.
    Area of circle- 36π

    1. 6 cant be the radius as DE>OD, radius has to be less than 6
      24pi is the answer

  2. 24π is the answer.
    Joining CE and CF,∆COE~=∆DOE
    This implies CE=6cm.Using phthagorean theorem in ∆CFB,CF=√32.Then applying Pythagoras theorem in ∆CFD,CD=√96.Then find the area.

    1. Hello Vernica !

      triangle CFD not triangle CFB .

      Triangle inscribed in a semicircle is always a right-angled triangle.

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