GMAtprep question
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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
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
Ankur Tyagi
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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.
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.
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.
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.