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Area of triangular region

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by ferpape » Thu Sep 03, 2009 4:30 pm
In the figure attached, the radius of the circle with center O is 1 and BC=1. What is the area of triangular region ABC?

a) square root of 2 /2
b) square root of 3 /2
c) 1
d) square root of 2
e) square root of 3

Help!
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by tom4lax » Thu Sep 03, 2009 6:31 pm
IMO ans is B.

Since line is the diameter, triangle is 90deg. given two lines of hyp. 2 and short leg 1, we can assume its 30 60 90 triangle, so third leg is root3. 1/2 * root3 * 1 = answer of B
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by Gladiator » Thu Sep 03, 2009 6:35 pm
Any angle formed in a semi-circle is a right angle triangle.

So angle(BCA)=90

so BC =1 and AB=2..using hyp theorem = CA = sqrt(4-1)=sqrt(3)

So area = 1/2 * 1 * sqrt(3)

B is answer
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by ferpape » Fri Sep 04, 2009 8:01 am
Im not sure how you calculated the area of CA??

I'll appriciate your comments.
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by chetanojha » Fri Sep 04, 2009 1:03 pm
ferpape wrote:Im not sure how you calculated the area of CA??

I'll appriciate your comments.
Check image.
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by [email protected] » Wed Jun 08, 2011 1:43 am
Answer is B. Explanation:

AC^2 - BC^2 = AB^2
(2)^2 - (1)^2 = 4 - 1 = 3 = square root of 3

Therefore, area of the triangle = 1/2 X Base X Height
= 1/2 X 1 X square root of 3
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