Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
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...
The “3 golden nuggets” of MBA admission process With ten years of experience helping prospective students with MBA admissions and career progression, I will be writing this blog through...
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...
Wow! MBA life is hectic indeed. Time flies by. It is hard to keep track of the time. Last week was high intense training Yeah, Finance, Accounting, Marketing, Economics...