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

Intern
Joined: 05 Jul 2016
Posts: 16
Location: Brazil
Concentration: Finance, Entrepreneurship
WE: Analyst (Investment Banking)

In the figure above, the triangle ABC is inscribed in a semicircle, wi
[#permalink]
Show Tags
30 May 2017, 05:39
Question Stats:
76% (02:23) correct 24% (02:41) wrong based on 173 sessions
HideShow timer Statistics
In the figure above, the triangle ABC is inscribed in a semicircle, with center O. If CÔA = 120° and AC = 3√3, what is the radius of the circle? (a) (4√3)/5 (b) 3 (c) 2√3 (d) 4 (e) 9
Official Answer and Stats are available only to registered users. Register/ Login.
Attachments
gmat_question.png [ 44.86 KiB  Viewed 3860 times ]



Director
Joined: 13 Mar 2017
Posts: 696
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)

In the figure above, the triangle ABC is inscribed in a semicircle, wi
[#permalink]
Show Tags
Updated on: 30 May 2017, 19:29
guireif wrote: In the figure above, the triangle ABC is inscribed in a semicircle, with center O. If CÔA = 120° and AC = 3√3, what is the radius of the circle?
(a) (4√3)/5 (b) 3 (c) 2√3 (d) 4 (e) 9 Lets join O to C in the diagram. Let radius of semicircle = x . So , OA= OB = OC = x Also angle BOC = 180°  angle COA = 60° Since OC = OB angle BCO = angle COB = (180  angle BOC)/2 = 60° So triangle BOC is an equilateral triangle . Hence BC = x Now since ACB is a triangle inside semicircle. So Triangle ACB is right angled triangle. > AC^2 + BC^2 = AB^2 > (3sqrt(3))^2 + x^2 = (2x)^2 > 3x^2 = 27 > x^2 = 9 > x =3 Hence radius = 3.
_________________
CAT 2017 (98.95) & 2018 (98.91) : 99th percentiler UPSC Aspirants : Get my app UPSC Important News Reader from Play store.
MBA Social Network : WebMaggu
Appreciate by Clicking +1 Kudos ( Lets be more generous friends.) What I believe is : "Nothing is Impossible, Even Impossible says I'm Possible" : "Stay Hungry, Stay Foolish".
Originally posted by shashankism on 30 May 2017, 05:57.
Last edited by shashankism on 30 May 2017, 19:29, edited 1 time in total.



Director
Joined: 13 Mar 2017
Posts: 696
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)

Re: In the figure above, the triangle ABC is inscribed in a semicircle, wi
[#permalink]
Show Tags
30 May 2017, 06:00
guireif wrote: In the figure above, the triangle ABC is inscribed in a semicircle, with center O. If CÔA = 120° and AC = 3√3, what is the radius of the circle?
(a) (4√3)/5 (b) 3 (c) 2√3 (d) 4 (e) 9 I forgot to attach the diagram.. I have edited the diagram provided by you for the solution..
Attachments
gmat_question.png [ 45.17 KiB  Viewed 3821 times ]
_________________
CAT 2017 (98.95) & 2018 (98.91) : 99th percentiler UPSC Aspirants : Get my app UPSC Important News Reader from Play store.
MBA Social Network : WebMaggu
Appreciate by Clicking +1 Kudos ( Lets be more generous friends.) What I believe is : "Nothing is Impossible, Even Impossible says I'm Possible" : "Stay Hungry, Stay Foolish".



Intern
Joined: 05 Jul 2016
Posts: 16
Location: Brazil
Concentration: Finance, Entrepreneurship
WE: Analyst (Investment Banking)

Re: In the figure above, the triangle ABC is inscribed in a semicircle, wi
[#permalink]
Show Tags
30 May 2017, 06:17
shashankism wrote: guireif wrote: In the figure above, the triangle ABC is inscribed in a semicircle, with center O. If CÔA = 120° and AC = 3√3, what is the radius of the circle?
(a) (4√3)/5 (b) 3 (c) 2√3 (d) 4 (e) 9 Lest join O to C in the diagram. Let radius of semicircle = x . So , OA= OB = OC = x Also angle BOC = 180°  angle COA = 60° Since OC = OB angle BCO = angle COB = (180  angle BOC)/2 = 60° So triangle BOC is an equilateral triangle . Hence BC = x Now since ACB is a triangle inside semicircle. So Triangle ACB is right angled triangle. > AC^2 + BC^2 = AB^2 > (3sqrt(3))^2 + x^2 = (2x)^2 > 3x^2 = 27 > x^2 = 9 > x =3 Hence radius = 3. Since OC = OB angle BCO = angle COB = (180  angle BOC)/2 = 60°That's what I was missing! You're a genius. Thank you so much. Best, Gui.



Director
Joined: 04 Dec 2015
Posts: 738
Location: India
Concentration: Technology, Strategy
WE: Information Technology (Consulting)

Re: In the figure above, the triangle ABC is inscribed in a semicircle, wi
[#permalink]
Show Tags
30 May 2017, 18:32
guireif wrote: In the figure above, the triangle ABC is inscribed in a semicircle, with center O. If CÔA = 120° and AC = 3√3, what is the radius of the circle?
(a) (4√3)/5 (b) 3 (c) 2√3 (d) 4 (e) 9 Triangle inscribed in a semicircle will always have one angle as \(90^{\circ}\).
From the diagram \(\angle\)ACB = \(90^{\circ}\)
Connect OC. Given; \(\angle\)CÔA = 120° \(\angle\)COB + \(\angle\)COA = 180 Therefore \(\angle\)COB = \(60^{\circ}\) OB = OC (Both are radius). Hence \(\angle\)OCB = \(\angle\)OBC = \(60^{\circ}\)
\(\triangle\) ABC is 30:60:90 Triangle. Sides of 30:60:90 triangle are in ratio = x : x\(\sqrt{3}\) : 2x
BC : AC : AB = x : x\(\sqrt{3}\) : 2x
AC = 3\(\sqrt{3}\)
Therefore BC = 3 and AB = 6
AB is diameter of semicircle. Radius is half of diameter. Therefore Radius = 3. Answer B...



Manager
Joined: 03 Feb 2017
Posts: 75
Location: Australia

In the figure above, the triangle ABC is inscribed in a semicircle, wi
[#permalink]
Show Tags
31 May 2017, 05:19
I tried solving it, but since I was taking more than 1 minute to figure it, I guessed right answer assuming that it has to be a 30:60:90 triangle. Only the 30:60:90 triangle has a √3 component x : x√3 : 2x.
So, I guessed the diameter should be 2x=6 and hence the radius 3.



VP
Joined: 09 Mar 2016
Posts: 1287

In the figure above, the triangle ABC is inscribed in a semicircle, wi
[#permalink]
Show Tags
10 Mar 2018, 04:31
shashankism wrote: guireif wrote: In the figure above, the triangle ABC is inscribed in a semicircle, with center O. If CÔA = 120° and AC = 3√3, what is the radius of the circle?
(a) (4√3)/5 (b) 3 (c) 2√3 (d) 4 (e) 9 Lets join O to C in the diagram. Let radius of semicircle = x . So , OA= OB = OC = x Also angle BOC = 180°  angle COA = 60° Since OC = OB angle BCO = angle COB = (180  angle BOC)/2 = 60° So triangle BOC is an equilateral triangle . Hence BC = x Now since ACB is a triangle inside semicircle. So Triangle ACB is right angled triangle. > AC^2 + BC^2 = AB^2 > (3sqrt(3))^2 + x^2 = (2x)^2 > 3x^2 = 27 > x^2 = 9 > x =3 Hence radius = 3. shashankism why do you asssume that point O divides diameter of circle equally ? it could divide not equally ? no ?:)



Manager
Joined: 27 Jul 2017
Posts: 50

Re: In the figure above, the triangle ABC is inscribed in a semicircle, wi
[#permalink]
Show Tags
10 Mar 2018, 07:37
dave13 wrote: shashankism wrote: guireif wrote: In the figure above, the triangle ABC is inscribed in a semicircle, with center O. If CÔA = 120° and AC = 3√3, what is the radius of the circle?
(a) (4√3)/5 (b) 3 (c) 2√3 (d) 4 (e) 9 Lets join O to C in the diagram. Let radius of semicircle = x . So , OA= OB = OC = x Also angle BOC = 180°  angle COA = 60° Since OC = OB angle BCO = angle COB = (180  angle BOC)/2 = 60° So triangle BOC is an equilateral triangle . Hence BC = x Now since ACB is a triangle inside semicircle. So Triangle ACB is right angled triangle. > AC^2 + BC^2 = AB^2 > (3sqrt(3))^2 + x^2 = (2x)^2 > 3x^2 = 27 > x^2 = 9 > x =3 Hence radius = 3. shashankism why do you asssume that point O divides diameter of circle equally ? it could divide not equally ? no ?:) dave13 Since the question states "In the figure above, the triangle ABC is inscribed in a semicircle, with center O" therefore point O has to divide the diameter of the circle equal as OA and OB are radii of the semicircle shown. Hope it clears!
_________________
Ujjwal Sharing is Gaining!



VP
Joined: 09 Mar 2016
Posts: 1287

Re: In the figure above, the triangle ABC is inscribed in a semicircle, wi
[#permalink]
Show Tags
11 Mar 2018, 04:44
Hi ujjwal80, thanks for an explanation:) i have a couple of questions can you please explan. see below. have a great weekend Question 1. Based on which rule do we assume that the length of OC equals OA and OB? OC could be of any length, no? Question 2. regarding this  > "Angle BCO = angle COB = (180  angle BOC)/2 = 60° " How can we assume that angle BCO is 60° ? it could be any degree… no ?



Board of Directors
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3627

Re: In the figure above, the triangle ABC is inscribed in a semicircle, wi
[#permalink]
Show Tags
11 Mar 2018, 05:03
dave13 wrote: Question 1.
Based on which rule do we assume that the length of OC equals OA and OB? OC could be of any length, no?
Question 2. regarding this  > "Angle BCO = angle COB = (180  angle BOC)/2 = 60° "
How can we assume that angle BCO is 60° ? it could be any degree… no ? Hey dave13 , Here are the answers: Based on which rule do we assume that the length of OC equals OA and OB? > We are not assuming. It is the rule. If the given figure is a semi circle with center as O, all the line segments joining the center and any point on the circle are equal and are also known as radii. Hence, OC = OA = OB How can we assume that angle BCO is 60° ? > Since OC = OA, that means triangle COA is isosceles. We are given that angle COA = 120. => the sum of angles OAC and OCA = 180120 = 60. => angle OCA = 30 (since triangle COA is isosceles). Now, every triangle drawn on the semi circle is right angled triangle. This angle ACB = 90. Therefore, BCO = ACN  OCA = 9030 = 60 Does that make sense?
_________________
My GMAT Story: From V21 to V40 My MBA Journey: My 10 years long MBA Dream My Secret Hacks: Best way to use GMATClub  Importance of an Error Log! Verbal Resources: All SC Resources at one place  All CR Resources at one place Blog: Subscribe to Question of the Day Blog GMAT Club Inbuilt Error Log Functionality  View More. New Visa Forum  Ask all your Visa Related Questions  here. New! Best Reply Functionality on GMAT Club! Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for free Check our new About Us Page here.



Manager
Joined: 27 Jul 2017
Posts: 50

Re: In the figure above, the triangle ABC is inscribed in a semicircle, wi
[#permalink]
Show Tags
11 Mar 2018, 18:13
abhimahna wrote: dave13 wrote: Question 1.
Based on which rule do we assume that the length of OC equals OA and OB? OC could be of any length, no?
Question 2. regarding this  > "Angle BCO = angle COB = (180  angle BOC)/2 = 60° "
How can we assume that angle BCO is 60° ? it could be any degree… no ? Hey dave13 , Here are the answers: Based on which rule do we assume that the length of OC equals OA and OB? > We are not assuming. It is the rule. If the given figure is a semi circle with center as O, all the line segments joining the center and any point on the circle are equal and are also known as radii. Hence, OC = OA = OB How can we assume that angle BCO is 60° ? > Since OC = OA, that means triangle COA is isosceles. We are given that angle COA = 120. => the sum of angles OAC and OCA = 180120 = 60. => angle OCA = 30 (since triangle COA is isosceles). Now, every triangle drawn on the semi circle is right angled triangle. This angle ACB = 90. Therefore, BCO = ACN  OCA = 9030 = 60 Does that make sense? Thanks abhimahna for the explanation, dave13 refer to triangle concepts https://gmatclub.com/forum/mathtriangles87197.html if you have more doubts feel free to post.
_________________
Ujjwal Sharing is Gaining!




Re: In the figure above, the triangle ABC is inscribed in a semicircle, wi &nbs
[#permalink]
11 Mar 2018, 18:13






