Quant Question Of The Day: 25
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.
D. None of the above
Join C – F
Let the radius of the circle be r.
ΔCFD and ΔEOD are similar .
DE/OD = CD/FD
6/r = 2r/8
r² = 24
Area = πr² = 24π
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
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.
24 π is answer
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.