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I like this question, but my answer choice doesn't match any of the choices; I am getting \(area=\frac{64\pi}{3}.\) What is the source of this question?

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Both AB and CD are diameters of circle .
so OA=OB=OC=OD=r radius of the circle.
In circles have a property if two chords bisect each other
we can multiply bisect portion each chord.
Here DF chord bisects AB
so DE*EF=AE*EB.

DE=6, EF=2 so AE*EB=12.

AE= AO+OE
EB= OB-OE . Take OA=x, OE=y.
OA=OB radius of the circle. then x^2-y^2=12----1

Here right angled triangle DOE at O , DE=6, OD=r radius of the circle so x
then DE^2=OD^2+OE^2
x^2+y^2=36------2

sum bothe equation 1 and 2 then 2x^2=48=> x= 2root6 radius of the circle

Area of circle pi r^2=24pi.
So answer C is correct
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Both AB and CD are diameters of circle .
so OA=OB=OC=OD=r radius of the circle.
In circles have a property if two chords bisect each other
we can multiply bisect portion each chord.
Here DF chord bisects AB
so DE*EF=AE*EB.

DE=6, EF=2 so AE*EB=12.

AE= AO+OE
EB= OB-OE . Take OA=x, OE=y.
OA=OB radius of the circle. then x^2-y^2=12----1

Here right angled triangle DOE at O , DE=6, OD=r radius of the circle so x
then DE^2=OD^2+OE^2
x^2+y^2=36------2

sum bothe equation 1 and 2 then 2x^2=48=> x= 2root6 radius of the circle

Area of circle pi r^2=24pi.
So answer C is correct

I have never heard of the intersecting chords theorem before your post!
Without considering the circle, Right triangle CFD with median FO and with side DF = 8, and sides CO and OD congruent can have numerous solutions. Two such solutions are CO=OD=\(\sqrt{24}\) with area of circle equal \(area=24\pi\), and CO=OD=\(8/\sqrt{3}\) with area of circle equal \(area=\frac{64\pi}{3}.\)
I partially neglected the impact the circle has in this question. It is the intersecting chords theorem that ensures an unique solution.
Thanks! I learned something new.
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mmelendez
In the figure, AB and CD are diameters of the circle with center O and AB bisects CD, and chord DF intersects AB at E. if DE=6 and EF=2, Then the area of the circle is?

A 23π
B 47π/2
C 24π
D 49π/2
E 25π

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Theorem of intersecting chords say DE*EF=DO*OC
6*8=DO*DO
48=2*DO^2
DO(radius)=Sq root 24
area =π*(sq.rt 24)^2
=24π
Ans C
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Psiva00734
Both AB and CD are diameters of circle .
so OA=OB=OC=OD=r radius of the circle.
In circles have a property if two chords bisect each other
we can multiply bisect portion each chord.
Here DF chord bisects AB
so DE*EF=AE*EB.

DE=6, EF=2 so AE*EB=12.

AE= AO+OE
EB= OB-OE . Take OA=x, OE=y.
OA=OB radius of the circle. then x^2-y^2=12----1

Here right angled triangle DOE at O , DE=6, OD=r radius of the circle so x
then DE^2=OD^2+OE^2
x^2+y^2=36------2

sum bothe equation 1 and 2 then 2x^2=48=> x= 2root6 radius of the circle

Area of circle pi r^2=24pi.
So answer C is correct


x^2-y^2=12----1 How?
Would you please explain in details?

Bunuel, would you help us?



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