## Quant Question Of The Day: 25

# 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

### Solve our previous ‘questions of the day’

1. Quant Question Of The Day: 24

2. Quant Question Of The Day: 23

3. Quant Question Of The Day: 22

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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

6 cant be the radius as DE>OD, radius has to be less than 6

24pi is the answer

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.

24π

24π

24 π

24 π is answer

24pie

24π

(6÷2r)=(r÷8)

So r=✓24

A= 24π

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

Hello Vernica !

triangle CFD not triangle CFB .

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