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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ps - triangle in a circle from GMAT Prep Test
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jainnikhil02
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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
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
Nikhil K Jain
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Frankenstein
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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
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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ccassel
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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.
Thanks.
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A central angle is formed by two radii.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.
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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cans
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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
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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kruthika
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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
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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