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ps - triangle in a circle from GMAT Prep Test

by ccassel » Tue Jun 07, 2011 9:02 am
https://www.beatthegmat.com/area-of-tria ... tml#187497

This question has been discussed but there are differing opinions on what angle in the triangle is 90 degrees and therefore, what the hypot represents. Can anyone clarify? The link is above.

Thanks,

Q. In the figure above, the radius of the circle with the center O is 1 and Bc=1. What is the area of triangular region ABC?

a. root2/2
b. roog3/2
c. 1
d. root2
e. root3

Answer is b
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by jainnikhil02 » Tue Jun 07, 2011 9:10 am
given that radius =1
so Hypo =2 (diameter in the figure)

so the remaining AB =root3

then area of traingle will be = 1/2 length*breadth
= 1/2*1*root3

=root3/2
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by Frankenstein » Tue Jun 07, 2011 9:11 am
Hi,
There should be no conflict in opinions. The angle opposite to the diameter is always 90 degrees.
AC = 2, BC=1.
As AC^2 = AB^2 + BC^2, BC = sqrt(3).
Area of triangle is (1/2).AB.BC = (root3)/2

Hence, B
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by ccassel » Tue Jun 07, 2011 9:16 am
That is what I thought. Some previous posts were confusing the PT for 1:1:root3, illogically comming to the, coincidentally, correct conclusion. However, as stated the correct PT ratio is 1:root3:2 based on the given information.

Thanks.
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by GMATGuruNY » Tue Jun 07, 2011 9:33 am
ccassel wrote:That is what I thought. Some previous posts were confusing the PT for 1:1:root3, illogically comming to the, coincidentally, correct conclusion. However, as stated the correct PT ratio is 1:root3:2 based on the given information.

Thanks.
A central angle is formed by two radii.
An inscribed angle is formed by two chords.
When an inscribed angle intercepts the same arc as a central angle, the inscribed angle = 1/2 * the central angle.

Since a diameter is formed by two radii, a diameter is a central angle. The degree measurement of this central angle is 180.
When an inscribed angle intercepts the endpoints of a diameter, it intercepts the same arc on the circle as does the diameter/central angle.
Thus, the degree measurement of the inscribed angle = (1/2)*180 = 90:

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by cans » Tue Jun 07, 2011 10:07 pm
AC is the diameter. Also diameters subtend 90 degree angle at the circle.
Thus <ABC is 90
And AC is hypotenuse. AC=2, BC=1
AB^2 = 4-1=3
AB=root(3)
Thus area = (1/2)AB*BC = root(3)/2
IMO B
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by kruthika » Tue Jun 07, 2011 11:01 pm
Given that AC is the diameter, and angle subtended in a semicircle is a right angle, therefore <ABC = 90.
Using Pythogoras theorem, we get AB=square root of 3.
Area of a triangle = 1/2 * base * height = 1/2*1*square root of 3 = (square root of 3)/2
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by MBA.Aspirant » Wed Jun 08, 2011 9:26 am
BC radius = 1 so AC = 2

Triangle ratio 30:60:90

so AB = root 3

area = 1/2 root 3 * 1 = root 3/2
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