January 17, 2019 January 17, 2019 08:00 AM PST 09:00 AM PST Learn the winning strategy for a high GRE score — what do people who reach a high score do differently? We're going to share insights, tips and strategies from data we've collected from over 50,000 students who used examPAL. January 19, 2019 January 19, 2019 07:00 AM PST 09:00 AM PST Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 52119

In the figure above (not drawn to scale), triangle ABC is inscribed in
[#permalink]
Show Tags
28 Jul 2015, 00:50
Question Stats:
78% (02:13) correct 22% (02:12) wrong based on 148 sessions
HideShow timer Statistics



CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2722
Location: India
GMAT: INSIGHT
WE: Education (Education)

In the figure above (not drawn to scale), triangle ABC is inscribed in
[#permalink]
Show Tags
Updated on: 28 Jul 2015, 03:12
Bunuel wrote: In the figure above (not drawn to scale), triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. Segments AC and OB are equal. If the area of triangle ABC is 8√3, then what is the area of the circle? A. π B. 8π C. 16π D. 48π E. 64π Kudos for a correct solution.Attachment: circle_inscribed_triangle.gif AC = OB = Radius (r) and AB = 2*r ABC is right angle triangle as traingle drawn in Semicircle is always the Right angle triangle with right angle at vertex touching circumference and Hypotenuse at Diameter of circle
Also BC^2 = AB^2  AC^2 = (2r)^2  r^2 = 3r^2 i.e. BC = r√3 Area of ABC = (1/2)*BC*AC = (1/2)*r√3*r = 8√3 i.e. r = 4 Area of Circle = π*r^2 = π*4^2 = 16π Answer: optionC
_________________
Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha email: info@GMATinsight.com I Call us : +919999687183 / 9891333772 Online OneonOne Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html
ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION
Originally posted by GMATinsight on 28 Jul 2015, 03:10.
Last edited by GMATinsight on 28 Jul 2015, 03:12, edited 1 time in total.



Manager
Joined: 14 Mar 2014
Posts: 146

Re: In the figure above (not drawn to scale), triangle ABC is inscribed in
[#permalink]
Show Tags
28 Jul 2015, 03:11
Bunuel wrote: In the figure above (not drawn to scale), triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. Segments AC and OB are equal. If the area of triangle ABC is 8√3, then what is the area of the circle? A. π B. 8π C. 16π D. 48π E. 64π Kudos for a correct solution.Attachment: circle_inscribed_triangle.gif IMO: C Angle in a semi circle = 90. Thus Angle ACB = 90. AC = OB = r 2AC = AB (OB=radius AB = diameter = 2* radius) So angles of the triangle will be ABC = 30 , ACB = 90 , BAC = 60. So sides of the triangle will be AB = 2r AC = r BC = r√3Area of triangle is 8√3 = 1/2 b*h base and heights will be the sides of the triangle since it is right angled triangle 8√3 = 1/2 * r * r√3 16 = r^2 r =4 (r=radius) Area of circle = πr^2 = 16π
_________________
I'm happy, if I make math for you slightly clearer And yes, I like kudos ¯\_(ツ)_/¯



Manager
Joined: 12 Nov 2014
Posts: 63

Re: In the figure above (not drawn to scale), triangle ABC is inscribed in
[#permalink]
Show Tags
28 Jul 2015, 03:21
From the figure it can be seen that the triangle ACB is inscribed within a semicircle. So angle C is 90 degree. Lets take OB = x , so OA =x (both radius) 2x, Also it is given that AC = OB = x. For triangle OAC (draw a line ) all sides are equal (OA = AC= OC) so it is an equilateral triangle. So <A = 60degree The right triangle ACB is thus a 30 60 90 triangle so sides are x, x√3and 2x. Area of triangle ACB = 0.5 * x * x√3 = 8√3 So x = 4 i.e. radius of the circle is 4 Area of circle = π x * x = 16π Answer C
_________________
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Kindly press Kudos if the explanation is clear. Thank you Ambarish



CEO
Joined: 20 Mar 2014
Posts: 2639
Concentration: Finance, Strategy
GPA: 3.7
WE: Engineering (Aerospace and Defense)

Re: In the figure above (not drawn to scale), triangle ABC is inscribed in
[#permalink]
Show Tags
28 Jul 2015, 03:29
Bunuel wrote: In the figure above (not drawn to scale), triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. Segments AC and OB are equal. If the area of triangle ABC is 8√3, then what is the area of the circle? A. π B. 8π C. 16π D. 48π E. 64π Kudos for a correct solution.Attachment: circle_inscribed_triangle.gif Given, AB is the diameter of the circle= 2r and AC = OB =r Let BC = h Thus is right triangle ABC, right angled at C, \(r^2+h^2 = (2r)^2\) and from area of the triangle = \(8\sqrt{3}\) = \(0.5*h*r\) > \(h = \frac{16\sqrt{3}}{r}\) Thus, \(r^2+ (\frac{16\sqrt{3}}{r})^2 = 2r^2\) > \(r^4 = 256\) > \(r = 4\) Thus the area of the circle = \(\pi*4^2\) = \(16\pi\). C is the correct answer.



Current Student
Affiliations: Scrum Alliance
Joined: 09 Feb 2010
Posts: 79
Location: United States (MI)
Concentration: Strategy, General Management
GMAT 1: 600 Q48 V25 GMAT 2: 710 Q48 V38
WE: Information Technology (Retail)

Re: In the figure above (not drawn to scale), triangle ABC is inscribed in
[#permalink]
Show Tags
28 Jul 2015, 12:53
Here, \(AC = OB = OA = r\), we can say that \(AB = 2r\). Given the inscribed triangle property, we can get that triangle ABC is a right triangle and so \(AB^2 = AC^2 + BC^2\). So, \((2r)^2 = r^2 + BC^2\), which means that \(BC = \sqrt{3}r\). Area of the triangle ABC = \(\frac{1}{2}(AC)(BC) = 8\sqrt{3}\), means \(\frac{1}{2}(r)(\sqrt{3}r) = 8\sqrt{3}\), which simplifies to \(r^2 = 16\), so Area of the circle = 16\(\pi\) Answer is C. Kudos please
_________________
kudos please



Math Expert
Joined: 02 Sep 2009
Posts: 52119

Re: In the figure above (not drawn to scale), triangle ABC is inscribed in
[#permalink]
Show Tags
17 Aug 2015, 08:41
Bunuel wrote: In the figure above (not drawn to scale), triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. Segments AC and OB are equal. If the area of triangle ABC is 8√3, then what is the area of the circle? A. π B. 8π C. 16π D. 48π E. 64π Kudos for a correct solution.Attachment: circle_inscribed_triangle.gif 800score Official Solution:There are four main concepts that must be understood in order to solve this problem. The first concept is that any triangle inscribed within a circle such that any one side of the triangle is a diameter of the circle, is a right triangle. The second concept is that the proportions of the sides of a 30⁰60⁰90⁰ triangle are x, x√3, 2x. The third concept is that the area of a right triangle is equal to one half the product of its legs: (1/2)bh. Finally, the fourth concept is that the area of a circle is π × radius². Since the triangle is inscribed in the circle and one of its sides constitutes a diameter of the circle, the triangle must be a right triangle. AB is the diameter so AB = 2 × OB. We know that OB = AC. Therefore AB = 2 × AC. Therefore it must be a 30°60°90° triangle where AC = x, AB = 2x and CB = √3 × x (This can also be calculated using the Pythagorean Theorem). We are now ready to solve this problem. Using the third concept from above, the area must be equal to (1/2) × √3 × x × x = 8√3. The solution of the equation is x = 4. So the radius of the circle is 4. Therefore the area of the circle is π × 4² = 16π. The correct answer is C.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Board of Directors
Joined: 17 Jul 2014
Posts: 2601
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

Re: In the figure above (not drawn to scale), triangle ABC is inscribed in
[#permalink]
Show Tags
06 Feb 2016, 09:31
Bunuel wrote: In the figure above (not drawn to scale), triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. Segments AC and OB are equal. If the area of triangle ABC is 8√3, then what is the area of the circle? A. π B. 8π C. 16π D. 48π E. 64π Kudos for a correct solution.Attachment: circle_inscribed_triangle.gif OB is the radius, thus, AC has the same length as the radius. since we have a triangle inscribed in a circle, and one side is the diagonal, the triangle MUST be a right triangle. since the hypotenuse is 2R,and since one leg is R, it must be true that the triangle is 306090. It must also be true, from the property of 306090 triangle, that the other leg must be R*sqrt(3). we then have the area of the triangle: R*R(sqrt3)/2 = 8*sqrt(3) r^2(sqrt3) = 16*sqrt(3) r=4. Area of the circle: pi*r^2 = 16pi.



NonHuman User
Joined: 09 Sep 2013
Posts: 9405

Re: In the figure above (not drawn to scale), triangle ABC is inscribed in
[#permalink]
Show Tags
21 Dec 2018, 16:08
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: In the figure above (not drawn to scale), triangle ABC is inscribed in &nbs
[#permalink]
21 Dec 2018, 16:08






