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by Deepthi Subbu » Mon Sep 19, 2011 3:09 am
What is the area of shaded region in the figure?OB= 6 and AOB = 60 degree
a. 10pie+27(sqrt3)
b. 10pie+27/4(sqrt3)
c.30pie+27(sqrt3)
d.30pie+9(sqrt3)
e.36pie+27(sqrt3)

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by cans » Mon Sep 19, 2011 3:33 am
AOp = 60. therefore area = 30pi + area of triangle.
perpendicular: cos 30 = height/6 or height = 3root(3)
sin 30 = (base/2)/6 or base = 6
thus area = (1/2)6*3(root(3)) = 9(sqrt3)
IMO D
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by shankar.ashwin » Mon Sep 19, 2011 3:53 am
Cans method is perfect, but if you are not familiar with trigonometry,

In triangle AOB, 2 sides are equal (radius) and other angle is 60, which makes it an equilateral triangle.

So area of eq. triangle is [(sqrt3)/4]*6*6 = 9(sqrt3).

The other sector is 5/6th the total area, so 30pie.

Hence D

Deepthi Subbu wrote:What is the area of shaded region in the figure?OB= 6 and AOB = 60 degree
a. 10pie+27(sqrt3)
b. 10pie+27/4(sqrt3)
c.30pie+27(sqrt3)
d.30pie+9(sqrt3)
e.36pie+27(sqrt3)

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