Aug 23 08:00 PM PDT  10:00 PM PDT Get Magoosh's App for Free Questions, Free Video lessons, and more. Aug 25 09:00 PM PDT  10:00 PM PDT Target Test Prep is giving 10 lucky winners 4 months of FREE access to our toprated GRE Quant course. Our giveaway contest is open for 10 days only, so enter today for your chance to levelup your prep! Aug 26 08:00 PM PDT  09:00 PM PDT We have a growing team of trained online experts, but much of our online GRE tutoring is delivered by a true GRE expert, MyGuru's Director of Online Tutoring, Stefan. Aug 26 08:00 PM PDT  11:00 PM PDT Learn how to evaluate your profile, skills, and experiences to determine if, when, and where you should apply to graduate school. Aug 28 08:00 PM PDT  11:00 PM PDT Free Video Modules Within each of the 16 learning modules (Geometry, Statistics, Sentence Equivalence, etc.) that comprise the course, you'll find plenty of free videos to help you make an informed purchase. Aug 30 08:00 PM PDT  09:00 PM PDT 7+ POINT GRE SCORE IMPROVEMENT GUARANTEE Sep 01 08:00 PM PDT  09:30 PM PDT Whether you want to take a free practice test, attend a deepdive session about the test itself, or experience the power of live instruction, we’ve got you covered.
Author 
Message 
TAGS:


Manager
Joined: 13 Aug 2016
Posts: 77
Followers: 0
Kudos [?]:
39
[4]
, given: 21

Comparison of Area of Triangles inside a Circle [#permalink]
05 Oct 2016, 06:05
4
This post received KUDOS
Question Stats:
50% (01:30) correct
50% (02:31) wrong based on 42 sessions
AB is the Diameter and point O is the center of the circle shown above
Quantity A 
Quantity B 
Area of Triangle ABC 
Twice the Area of Triangle OBD 
A : Quantity A is greater B : Quantity B is greater C : Both quantities are equal D : Answer Cannot be determined




GRE Prep Club Legend
Joined: 07 Jun 2014
Posts: 4809
WE: Business Development (Energy and Utilities)
Followers: 128
Kudos [?]:
2045
[1]
, given: 397

Re: Comparison of Area of Triangles inside a Circle [#permalink]
05 Oct 2016, 12:20
1
This post received KUDOS
Hey, Let radius of the triangle be r. Now since AB is the diameter of the circle, triangle ABC is a right angle triangle. And from triangle properties we know value of angle \(x = 30\) degrees. Thus triangle OBD is a equilateral triangle. Now AC= \(2*r*cos(x)\)=\(2*r*cos(30)\)=\(2*r*\sqrt{3}*\frac{1}{2}\)=\(\sqrt{3}r\). And BC= \(2*r*sin(x)\)=\(2*r*sine(30)\)=\(2*r*\frac{1}{2}\)=\(r\) Area of traingle ABC= \(\frac{1}{2}\)*AC*BC=\(\frac{1}{2}\sqrt{3}r^2\). Now Area of the equilateral triangle OBD= \(\frac{1}{4}\sqrt{3}r^2\). Clearly 2* Area of OBD= Area of ABC. Hence C is the correct option.
_________________
Sandy If you found this post useful, please let me know by pressing the Kudos Button
Try our free Online GRE Test



Manager
Joined: 13 Aug 2016
Posts: 77
Followers: 0
Kudos [?]:
39
[1]
, given: 21

Re: Comparison of Area of Triangles inside a Circle [#permalink]
09 Oct 2016, 01:20
1
This post received KUDOS
I personally recommend to avoid using Sine,Cosine and other trigonometric function to solve geometry related problems as other easier ways seem more handy and quick.Here is my solution: <BCA is an inscribe angle and AB is the diameter,so <BCA = 90 degree.So as ABC is a right triangle so we can use Pythagorean Theorem on this triangle.Since sum of the interior angles of any triangle is equal to 180 degrees therefore m<BCA+x+2x = 180 => 90+3x = 180 => x = 30 => Triangle ABC is a 306090 triangle and we may use properties of 306090 triangle to solve this question.Sides of a 306090 triangle are in the ratio 1:\(sqrt3\):2.Since hypotenuse of Triangle ABC is equal to Diameter AB = 2r of the circle;where r is radius of the circle.So the sides of Triangle ABc are r,r\sqrt{3} and 2r. => Quantity A : Area of Triangle ABC = (1/2)*(base)*(height) = (1/2)*(BC)*(AC) = = (1/2)*(r)*(r\(sqrt3\)) = \((sqrt3/2)*r^2\) Since 2x = 2*30 = 60 so Triangle OBD is an equilateral triangle and formula for area of an equilateral triangle is \((sqrt3/4)*s^2\),where s is the side of the equilateral triangle.Side of equilateral Triangle OBD is equal to radius r of the circle Quantity B : Twice the Area of Triangle ABC = \(2*(sqrt3/4)*r^2\) = \((sqrt3/2)*r^2\) which is equal to Quantity => Quantity A = Quantity B.Hence answer is C. This question has been taken from tests of Greatest Prep.Here is the link to the website of Greatest Prep : http://www.greatestprep.com



Director
Joined: 09 Nov 2018
Posts: 509
Followers: 0
Kudos [?]:
39
[0], given: 1

Re: Comparison of Area of Triangles inside a Circle [#permalink]
15 Nov 2018, 07:24
yasir9909 wrote: I personally recommend to avoid using Sine,Cosine and other trigonometric function to solve geometry related problems as other easier ways seem more handy and quick.Here is my solution: <BCA is an inscribe angle and AB is the diameter,so <BCA = 90 degree.So as ABC is a right triangle so we can use Pythagorean Theorem on this triangle.Since sum of the interior angles of any triangle is equal to 180 degrees therefore m<BCA+x+2x = 180 => 90+3x = 180 => x = 30 => Triangle ABC is a 306090 triangle and we may use properties of 306090 triangle to solve this question.Sides of a 306090 triangle are in the ratio 1:\(sqrt3\):2.Since hypotenuse of Triangle ABC is equal to Diameter AB = 2r of the circle;where r is radius of the circle.So the sides of Triangle ABc are r,r\sqrt{3} and 2r. => Quantity A : Area of Triangle ABC = (1/2)*(base)*(height) = (1/2)*(BC)*(AC) = = (1/2)*(r)*(r\(sqrt3\)) = \((sqrt3/2)*r^2\) Since 2x = 2*30 = 60 so Triangle OBD is an equilateral triangle and formula for area of an equilateral triangle is \((sqrt3/4)*s^2\),where s is the side of the equilateral triangle.Side of equilateral Triangle OBD is equal to radius r of the circle Quantity B : Twice the Area of Triangle ABC = \(2*(sqrt3/4)*r^2\) = \((sqrt3/2)*r^2\) which is equal to Quantity => Quantity A = Quantity B.Hence answer is C. This question has been taken from tests of Greatest Prep.Here is the link to the website of Greatest Prep : http://www.greatestprep.comhow is it possible if 2x=60, the triangle is equilateral. how do we calculate angle ODB and DBO ? Please, answer.



Supreme Moderator
Joined: 01 Nov 2017
Posts: 370
Followers: 5
Kudos [?]:
123
[0], given: 4

Re: Comparison of Area of Triangles inside a Circle [#permalink]
15 Nov 2018, 08:37
AE wrote: yasir9909 wrote: I personally recommend to avoid using Sine,Cosine and other trigonometric function to solve geometry related problems as other easier ways seem more handy and quick.Here is my solution: <BCA is an inscribe angle and AB is the diameter,so <BCA = 90 degree.So as ABC is a right triangle so we can use Pythagorean Theorem on this triangle.Since sum of the interior angles of any triangle is equal to 180 degrees therefore m<BCA+x+2x = 180 => 90+3x = 180 => x = 30 => Triangle ABC is a 306090 triangle and we may use properties of 306090 triangle to solve this question.Sides of a 306090 triangle are in the ratio 1:\(sqrt3\):2.Since hypotenuse of Triangle ABC is equal to Diameter AB = 2r of the circle;where r is radius of the circle.So the sides of Triangle ABc are r,r\sqrt{3} and 2r. => Quantity A : Area of Triangle ABC = (1/2)*(base)*(height) = (1/2)*(BC)*(AC) = = (1/2)*(r)*(r\(sqrt3\)) = \((sqrt3/2)*r^2\) Since 2x = 2*30 = 60 so Triangle OBD is an equilateral triangle and formula for area of an equilateral triangle is \((sqrt3/4)*s^2\),where s is the side of the equilateral triangle.Side of equilateral Triangle OBD is equal to radius r of the circle Quantity B : Twice the Area of Triangle ABC = \(2*(sqrt3/4)*r^2\) = \((sqrt3/2)*r^2\) which is equal to Quantity => Quantity A = Quantity B.Hence answer is C. This question has been taken from tests of Greatest Prep.Here is the link to the website of Greatest Prep : http://www.greatestprep.comhow is it possible if 2x=60, the triangle is equilateral. how do we calculate angle ODB and DBO ? Please, answer. In ∆OBD, O is 2x=60.. Now OB=OD =radius, so angle OBD =angle ODB OBD+ODB=18060=120, so both are 120/2=60, as both are equal
_________________
Some useful Theory. 1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressionsarithmeticgeometricandharmonic11574.html#p27048 2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effectsofarithmeticoperationsonfractions11573.html?sid=d570445335a783891cd4d48a17db9825 3. Remainders : https://greprepclub.com/forum/remainderswhatyoushouldknow11524.html 4. Number properties : https://greprepclub.com/forum/numberpropertyallyourequire11518.html 5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolutemodulusabetterunderstanding11281.html



Intern
Joined: 07 Aug 2016
Posts: 42
Followers: 0
Kudos [?]:
17
[0], given: 0

Re: Comparison of Area of Triangles inside a Circle [#permalink]
20 Nov 2018, 07:54
yasir9909 wrote: AB is the Diameter and point O is the center of the circle shown above
Quantity A 
Quantity B 
Area of Triangle ABC 
Twice the Area of Triangle OBD 
A : Quantity A is greater B : Quantity B is greater C : Both quantities are equal D : Answer Cannot be determined For a triangle where the side is the diameter the opposing angle will be 90 degree. Given that AB's angle is 90. Then we have 90 + x + 2x = 180 so 3x= 90 and x = 30. Given that 2x = 60 And OB = OD (both are radius) then they have the same corresponding angle. As a result it becomes an equilateral triangle. The area of an equilateral = (s^2 *sqrt of 3)/4 https://www.mathopenref.com/triangleequ ... larea.htmlWhere the side is the radius. For ABC the triangle is 30:60:90 so it has the following values s:s sqrt of 3:2s https://www.themathpage.com/aTrig/306090triangle.htmThe area of ABC = (s * s sqrt of 3)/2 = (s^2 * sqrt of 3)/2 2 times the area of equilateral equals = 2 * (s^2 *sqrt of 3)/4 = (s^2 *sqrt of 3)/2 Hence both are equal.



Manager
Joined: 22 Feb 2018
Posts: 163
Followers: 2
Kudos [?]:
117
[0], given: 22

Re: Comparison of Area of Triangles inside a Circle [#permalink]
20 Nov 2018, 20:36
Answer: A (I need to double check, Not sure about the answer)
AB is the diameter Point O is the center of the circle A: Area of triangle ABC? B: twice the Area of triangle OBD? Consider point C, the environment of circle which it faces (I think you name it arc) equals 180, why? Because AB is diameter and AB arc equals 180, so it’s facing angle which is C equals half of it, so C is 90 degrees. A: Now area of ABC = AC*BC/2 We have: A+B+C = 180 degrees x+2x+90 = 180 > x = 30degrees So angle O is 2x=60degrees. And as OB and OD are radiuses of circle, the OBD triangle is a equilateral. And thus, angles B and D equal (1802x)/2 = 60degrees B: The area of OBD*2 = 2* BD*H/2= BD*H How much is H? we have all angles in OBD sin B = H/OB > H = sinB * OB = sin60 * OB = RADICAL3/2 *OB A = AC*BC/2 sinB(2x)= AC/AB > √3/2 = AC/AB > AC = √3/2*AB > A= √3/2*AB*BC/2 B = √3/2 * OB A= √3/2*AB*BC/2 B = √3/2 * OB So we omit √3/2 from both. A > AB*BC/2 B > OB We know OB = AB/2, so we substitute it in B: A > AB*BC/2 B > AB/2 We omit AB/2 from both And we will have A > BC B > 1 BC is bigger than 1, so A is bigger than B. (Not sure about the answer)
_________________
Follow your heart



Supreme Moderator
Joined: 01 Nov 2017
Posts: 370
Followers: 5
Kudos [?]:
123
[1]
, given: 4

Re: Comparison of Area of Triangles inside a Circle [#permalink]
20 Nov 2018, 20:42
1
This post received KUDOS
FatemehAsgarinejad wrote: Answer: A (I need to double check, Not sure about the answer) AB is the diameter Point O is the center of the circle A: Area of triangle ABC? B: twice the Area of triangle OBD?
Consider point C, the environment of circle which it faces (I think you name it arc) equals 180, why? Because AB is diameter and AB arc equals 180, so it’s facing angle which is C equals half of it, so C is 90 degrees. A: Now area of ABC = AC*BC/2
We have: A+B+C = 180 degrees x+2x+90 = 180 > x = 30degrees So angle O is 2x=60degrees. And as OB and OD are radiuses of circle, the OBD triangle is a equilateral. And thus, angles B and D equal (1802x)/2 = 60degrees
B: The area of OBD*2 = 2* BD*H/2= BD*H
How much is H? we have all angles in OBD sin B = H/OB > H = sinB * OB = sin60 * OB = RADICAL3/2 *OB
A = AC*BC/2 sinB(2x)= AC/AB > √3/2 = AC/AB > AC = √3/2*AB > A= √3/2*AB*BC/2 B = √3/2 * OB
A= √3/2*AB*BC/2 B = √3/2 * OB So we omit √3/2 from both. A > AB*BC/2 B > OB
We know OB = AB/2, so we substitute it in B: A > AB*BC/2 B > AB/2 We omit AB/2 from both And we will have A > BC B > 1 BC is bigger than 1, so A is bigger than B. (Not sure about the answer)
you have forgotten to put BD back in the calculations as shown in coloured portion. so finally A > BC B > BD and BC=BD as they both are equal to the radius.. therefore A=B answer is C
_________________
Some useful Theory. 1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressionsarithmeticgeometricandharmonic11574.html#p27048 2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effectsofarithmeticoperationsonfractions11573.html?sid=d570445335a783891cd4d48a17db9825 3. Remainders : https://greprepclub.com/forum/remainderswhatyoushouldknow11524.html 4. Number properties : https://greprepclub.com/forum/numberpropertyallyourequire11518.html 5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolutemodulusabetterunderstanding11281.html



Manager
Joined: 27 Feb 2017
Posts: 189
Followers: 1
Kudos [?]:
59
[0], given: 15

Re: Comparison of Area of Triangles inside a Circle [#permalink]
01 Dec 2018, 16:07
Can someone please explain this?



Supreme Moderator
Joined: 01 Nov 2017
Posts: 370
Followers: 5
Kudos [?]:
123
[1]
, given: 4

Re: Comparison of Area of Triangles inside a Circle [#permalink]
01 Dec 2018, 20:24
1
This post received KUDOS
kruttikaaggarwal wrote: Can someone please explain this? Take triangle ABC.. As the hypotenuse is diameter, the triangle is right angled at C. Therefore x+2x=90....3x=90....x=30 So ABC is 306090 triangle with sides 1:√3:2 or a:√3a:2a... The hypotenuse AB is diameter and equal to 2r, so 2r=2a...a=r So area of ABC = 1/2 * AC * BC =r*√3r/2=√3r^2/2 Now let's see OBD.. Angle at centre is 2x=2*30=60.. The other two angles will be equal as the sides are equal to radius.. Therefore the other two angles are (18060)/2=60 each. So OBD is an equilateral triangle with sides r, so area = (√3/4)*r^2= (1/2)(√3/2*r^2)=(1/2) area of ABC.. So A=B C
Attachments
PicsArt_120209.44.19.jpg [ 11.06 KiB  Viewed 873 times ]
_________________
Some useful Theory. 1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressionsarithmeticgeometricandharmonic11574.html#p27048 2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effectsofarithmeticoperationsonfractions11573.html?sid=d570445335a783891cd4d48a17db9825 3. Remainders : https://greprepclub.com/forum/remainderswhatyoushouldknow11524.html 4. Number properties : https://greprepclub.com/forum/numberpropertyallyourequire11518.html 5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolutemodulusabetterunderstanding11281.html




Re: Comparison of Area of Triangles inside a Circle
[#permalink]
01 Dec 2018, 20:24





