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Question

AD is a diameter of a circle. Two more circles pass through A and intersect AD in B and C respectively, such that AB and AC are diameters of these circle and AD>AC>AB. If the circumference of the middle circle is average of the circumference of the other two, Then given AB=4 units and CD=2 units. Then the area, in sq. units of the largest circle is

A
128π
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B
64π
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C
48π
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D
16π
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Solution

The correct option is D 16π

Given- AD is the diameter of a circle with C3 as circumference. Two other circles of circumferences C1 & C2 pass through A such that that they intersect AD at B & C respectively. AD>AC>AB.AB=4units,CD=2units.C2 is the average of C1&C3.

To find out- ar.C3=?

Solution- Let BC=x.AD=(4+x+2)units=(6+x)units.SoC3=π(6+x)units........(i).

Similarly, AC=(4+x)unitsSoC2=π(4+x)units........(ii). Also AB=4units,

C1=4πunits.........(iii)

Now, by the given condition, C2=C1+C32π(4+x)=4π+π(6+x)2 (from i, ii \& iii)

x=2units. AD=(6+2)units=8units.i.erad.C3=(8÷2)units=4units.

So ar.C3=π42 sq.units=16π sq.units

Ans- Option D.


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