Question
In the given figure, MN is the common tangent to both the inner circles with M and N lying on the circumference of the biggest circle. MN=16. The centers of the smaller circles lie on the diameter of the biggest circle. Find the area of the shaded region
Answer: Option D
:
D
e. 32π
If the radii of the three circles be a, b and c decreasing order of size,then
2b+2c=2a; ⇒ a= b+c.
ΔAMC and ΔANC are congruent. ΔABM and ΔABN are similar; so ΔBMC and ΔBNC are congruent. Thus MB=BN=8.
Thus, AB.BC = MB.BN=64.
AB.BC= 2c.2b=64. (since triangle AMB=triangle MCB)
The required area is π ( (b+c)2 - b2 - c2); = 2π bc = 2π.16=32π.
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:
D
e. 32π
If the radii of the three circles be a, b and c decreasing order of size,then
2b+2c=2a; ⇒ a= b+c.
ΔAMC and ΔANC are congruent. ΔABM and ΔABN are similar; so ΔBMC and ΔBNC are congruent. Thus MB=BN=8.
Thus, AB.BC = MB.BN=64.
AB.BC= 2c.2b=64. (since triangle AMB=triangle MCB)
The required area is π ( (b+c)2 - b2 - c2); = 2π bc = 2π.16=32π.
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