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I encountered the following question in GMAT Prep Exam and was unable to solve it. The answer that is selected (image below) is correct. Though I did select the right answer but it was a fluke.

Q-1) In the figure above (below here) , the radius of circle with center O is 1 and BC = 1. What is the area of triangular region ABC ?

Re: Geometry ques...tricky [#permalink]
07 Jun 2011, 10:28

1

This post received KUDOS

Expert's post

bblast wrote:

I did not apply pythagoras.

I solved this pretty quickly, as I interpreted this as a right triangle inscribed in a circle as 30-60-90 triangle.

so longer leg = 1/2*hypotenuse*root3.

area = 1/2 *root 3 * 1

hope I am correct ?

Yes, you are. There are multiple ways of arriving at the value of AB. You see that Cos C = 1/2 so C must be 60 degrees \(Sin 60 = \sqrt{3}/2\) so \(AB = \sqrt{3}\) (Let me point out here that you are not expected to know trigonometry in GMAT.)

Or since the sides are 1 and 2, the third side must be \(\sqrt{3}\) by Pythagorean theorem. _________________

Re: Geometry ques...tricky [#permalink]
08 Jun 2011, 07:11

thanks karishma

so whenever I see a triangle in a circle and one arm of triangle as diameter if circle, I can quickly solve it using 30-60-90 formulas ? _________________

Cheers !!

Quant 47-Striving for 50 Verbal 34-Striving for 40

Re: Geometry ques...tricky [#permalink]
08 Jun 2011, 07:26

3

This post received KUDOS

Expert's post

bblast wrote:

thanks karishma

so whenever I see a triangle in a circle and one arm of triangle as diameter if circle, I can quickly solve it using 30-60-90 formulas ?

Actually no. Here we know that the diameter is 2 units and one side is 1 unit so the triangle is 30-60-90. It may not always be the case. Say, if the diameter is 2 and one side is given as root 2, it will be a 45-45-90 triangle. GMAT generally questions you on one of 30-60-90 and 45-45-90... _________________

Re: Geometry ques...tricky [#permalink]
08 Jun 2011, 07:36

Thank u . +1 Kudos

But it makes sense that if one triangle-circle holds this relation. Any triangle having the diameter as the hypotenuse should. As the angles for it in a circle will not change. Only the sides will grow or shrink. But ya maybe the ratio 1:\sqrt{3} :2 occurs at this particular size of the 2 geometric figures !!

So I will put ur words in my flashcard. _________________

Cheers !!

Quant 47-Striving for 50 Verbal 34-Striving for 40

Re: Geometry ques...tricky [#permalink]
08 Jun 2011, 18:17

2

This post received KUDOS

Expert's post

bblast wrote:

Thank u . +1 Kudos

But it makes sense that if one triangle-circle holds this relation. Any triangle having the diameter as the hypotenuse should. As the angles for it in a circle will not change. Only the sides will grow or shrink. But ya maybe the ratio 1:\sqrt{3} :2 occurs at this particular size of the 2 geometric figures !!

So I will put ur words in my flashcard.

To help you visualize, I will leave you with a diagram.

Re: Geometry ques...tricky [#permalink]
09 Jun 2011, 20:48

B...using Pythagoras theorem and using the property that a triangle with the hypotenuse as the diameter of a circle subtends a 90 degree anywhere on the circle. _________________

My will shall shape the future. Whether I fail or succeed shall be no man's doing but my own.

Re: Geometry ques...tricky [#permalink]
20 Nov 2011, 21:45

VeritasPrepKarishma wrote:

bblast wrote:

thanks karishma

so whenever I see a triangle in a circle and one arm of triangle as diameter if circle, I can quickly solve it using 30-60-90 formulas ?

Actually no. Here we know that the diameter is 2 units and one side is 1 unit so the triangle is 30-60-90. It may not always be the case. Say, if the diameter is 2 and one side is given as root 2, it will be a 45-45-90 triangle. GMAT generally questions you on one of 30-60-90 and 45-45-90...

Hi Karishma

I wanted to clarify my misunderstanding. In 45:45:90 the multiples will be x:x: \(\sqrt{2x}\) So if the diameter is 2 the other 2 sides should be 2 and 2\(\sqrt{2x}\)?

Re: Geometry ques...tricky [#permalink]
21 Nov 2011, 20:35

Expert's post

melguy wrote:

Hi Karishma

I wanted to clarify my misunderstanding. In 45:45:90 the multiples will be x:x: \(\sqrt{2x}\) So if the diameter is 2 the other 2 sides should be 2 and 2\(\sqrt{2x}\)?

Thanks

In 45-45-90, the sides will be in the ratio 1:1:\(\sqrt{2}\) or you can say they will be x, x and \(\sqrt{2}x\) (Mind you, the root is only on 2, not on x). \(\sqrt{2}x\), the longest side, is the hypotenuse.

Here, the diameter is the hypotenuse i.e. the longest side. The right angle is opposite to the diameter. We know that the side opposite to the largest angle is the longest side. Hence the diameter has to be the longest side i.e. \(\sqrt{2}x\). If \(\sqrt{2}x\) = 2 x = \(\sqrt{2}\) So the other two equal sides will be \(\sqrt{2}\) each. _________________

Re: Geometry ques...tricky [#permalink]
28 Nov 2011, 22:59

VeritasPrepKarishma wrote:

bblast wrote:

thanks karishma

so whenever I see a triangle in a circle and one arm of triangle as diameter if circle, I can quickly solve it using 30-60-90 formulas ?

Actually no. Here we know that the diameter is 2 units and one side is 1 unit so the triangle is 30-60-90. It may not always be the case. Say, if the diameter is 2 and one side is given as root 2, it will be a 45-45-90 triangle. GMAT generally questions you on one of 30-60-90 and 45-45-90...

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