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Geometry question
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mendiratta
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mendiratta
- Junior | Next Rank: 30 Posts
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I think it is sqrt(3)/2.
Angle subtended by diameter is always 90 degree.
So it is right angled triangle with 2 sides as 1 & 2(twice the radius).
There the third side is sqrt(3). ( 1,2,sqrt(3) ) kind of triangle.
therefore
area = (base*height)/2
area = ( sqrt(3) * 1 )/2
area = sqrt(3)/2.
Angle subtended by diameter is always 90 degree.
So it is right angled triangle with 2 sides as 1 & 2(twice the radius).
There the third side is sqrt(3). ( 1,2,sqrt(3) ) kind of triangle.
therefore
area = (base*height)/2
area = ( sqrt(3) * 1 )/2
area = sqrt(3)/2.
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Cybermusings
- Legendary Member
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Remember the theorem for your exam...An angle in a semicircle is a right-angle...
Here angle ABC is a right angle...
AC = Diameter = 2
BC = 1
So AB = sqr rt. 3
Now Area = 1/2 * sqr rt 3 * 1 = sqr rt. 3/ 2
Hence Choice C
Here angle ABC is a right angle...
AC = Diameter = 2
BC = 1
So AB = sqr rt. 3
Now Area = 1/2 * sqr rt 3 * 1 = sqr rt. 3/ 2
Hence Choice C
