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# Geometry 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

2. 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.

3. 24π

4. Abhishek Tandon

24π

5. 24 π

6. 7. 24pie

8. Gaurav Raj

24π

(6÷2r)=(r÷8)
So r=✓24
A= 24π

9. Could you please tell me how is CFB a right angle triangle?

1. Hello Vernica !

triangle CFD not triangle CFB .

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

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