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by beat_gmat_09 » Wed Nov 10, 2010 7:03 am
[email protected] wrote:hi everybody got a question see attachment.

I thought asnwer would be 1. because radius is 1, so diameter is 2.
b.h/2=area

so 2.1/2=1?
As the diameter of the circle is one of the sides of triangle , triangle is right angled.
D = Hypotenuse = (2*radius) = 2
One of the sides is = 1 i.e. BC =1.
Apply pythagoras therm 2^2 = 1^2 + x^2
THerefore x = (3)^1/2 i.e. sqrt (3).
Base is BC = 1, height is sqrt (3).
Area = BC*sqrt(3)/2 = (sqrt (3)) / 2
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by Geva@EconomistGMAT » Wed Nov 10, 2010 7:36 am
beat_gmat_09 gave the correct approach. I'll just nip in and clarify what I think is your own misconception - the height to the diameter is NOT the radius, and is actually shorter than the radius.
A height is a line that originates from a vertex and is perpendicular to the base. If you draw a line form vertex B down to the diameter so that the line is perpendicular to the diamter, the line won't hit the center O (since B is not directly "above" the center, but is more to the right), and thus will not be the radius - a radius is, by definition, a line that connects the center O with a point on the circumference.

to find the area of the triangle, use the two legs of the right angle, which are perpendicular to each other, and can thus act as the base/height of each other.
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by MAAJ » Wed Nov 10, 2010 7:57 am
Other responses to the same question:

https://www.beatthegmat.com/circles-and- ... 31103.html

https://www.beatthegmat.com/gmatprep-que ... 12694.html

You can get these links by writting part of the question stem in the search tool. Hope it helps.

Note: you should read the response from "outreach" quoting:
fleckre wrote:
How do you know triangle ABC is a right triangle?

this is becasue there is a std theorem

"states that if A is any point of the circle with diameter BC (except B or C themselves) �-�ABC is a right triangle with A the right angle. The converse states that if a right triangle is inscribed in a circle then the hypotenuse will be a diameter of the circle"
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