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Source: — Problem Solving |
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by kishore » Sat Jun 21, 2008 2:40 pm
Since radius from center O = 1

diameter of the circle = 2

Base = AC = diamter = 2

Height = BC = 1


Area of Traingle = 1/2 * Base * Height
= 1/2 * 2 * 1 = 1
= 1
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by niraj_a » Sat Jun 21, 2008 3:48 pm
How is height BC? Its not perpendicular to the base....
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by antyagi » Sat Jun 21, 2008 5:23 pm
Using property a circle, a triangle defined with one segment as a diameter is a right-angled triangle it's pretty simple.
Diameter(Hyp.) is 2 and one base is 1.
Another side is root(3)

Hence area = 1/2*b*h = 1/2*1*root(3) = root(3)/2
Thanks,
Ankur Tyagi
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by niraj_a » Sun Jun 22, 2008 2:24 am
Ankur,

They have never said that that this was a right triangle anywhere. GMAT diagrams cannot be assumed to be to scale and even if this one was assumed to be one, it doesn't look like a right triangle by any means.

Please explain.
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by vcb » Sun Jun 22, 2008 4:17 am
Though its not mentioned explicitly, the triangle is a right angled triangle. If a triangle, inscribed in a circle, has one of the sides as the diameter of the circle, the triangle becomes a right triangle, with the right angle opposite to the diameter. Now, instead of dropping a line from B to the diameter, consider BC as the base(1) and AB as the height (root 3), and use the formula, 1/2 * B * H. Otherwise, u could also use the formula:
Area=SQRT(s(s-a)(s-b)(s-c)),
where s=(a+b+c)/2 or perimeter/2. (havent calculated, but pretty sure it should be what antyagi mentioned.
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by niraj_a » Sun Jun 22, 2008 7:22 am
gotcha now...thank you so much

i should have been an engineer.....lol
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