[phpBB Debug] PHP Notice: in file /viewtopic.php on line 1524: Undefined array key "topic_tags_visibility"
[phpBB Debug] PHP Notice: in file /viewtopic.php on line 1525: Undefined array key "timer_statistics_visibility"
AB is the Diameter and point O is the center of the circle shown above : Quantitative Comparison Questions

Last visit was: 20 Jul 2024, 07:58 It is currently 20 Jul 2024, 07:58

Close

GRE Prep Club Daily Prep

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 GRE 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.

Close

Request Expert Reply

Confirm Cancel
avatar
Manager
Manager
Joined: 13 Aug 2016
Posts: 77
Own Kudos [?]: 142 [33]
Given Kudos: 0
GRE 1: Q158 V154
Send PM
Most Helpful Expert Reply
User avatar
Retired Moderator
Joined: 07 Jun 2014
Posts: 4810
Own Kudos [?]: 10776 [5]
Given Kudos: 0
GRE 1: Q167 V156
WE:Business Development (Energy and Utilities)
Send PM
General Discussion
avatar
Manager
Manager
Joined: 13 Aug 2016
Posts: 77
Own Kudos [?]: 142 [2]
Given Kudos: 0
GRE 1: Q158 V154
Send PM
avatar
Director
Director
Joined: 09 Nov 2018
Posts: 505
Own Kudos [?]: 131 [0]
Given Kudos: 0
Send PM
Re: AB is the Diameter and point O is the center of the circle shown above [#permalink]
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 30-60-90 triangle and we may use properties of 30-60-90 triangle to solve this question.Sides of a 30-60-90 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


how is it possible if 2x=60, the triangle is equilateral. how do we calculate angle ODB and DBO ? Please, answer.
avatar
Supreme Moderator
Joined: 01 Nov 2017
Posts: 371
Own Kudos [?]: 451 [0]
Given Kudos: 0
Send PM
Re: AB is the Diameter and point O is the center of the circle shown above [#permalink]
Expert Reply
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 30-60-90 triangle and we may use properties of 30-60-90 triangle to solve this question.Sides of a 30-60-90 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


how 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=180-60=120, so both are 120/2=60, as both are equal
avatar
Manager
Manager
Joined: 07 Aug 2016
Posts: 59
Own Kudos [?]: 67 [3]
Given Kudos: 0
GRE 1: Q166 V156
Send PM
Re: AB is the Diameter and point O is the center of the circle shown above [#permalink]
3
yasir9909 wrote:
Image

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.html

Where 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/30-60-90-triangle.htm

The 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.
avatar
Manager
Manager
Joined: 22 Feb 2018
Posts: 163
Own Kudos [?]: 212 [0]
Given Kudos: 0
Send PM
Re: AB is the Diameter and point O is the center of the circle shown above [#permalink]
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 (180-2x)/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)
avatar
Supreme Moderator
Joined: 01 Nov 2017
Posts: 371
Own Kudos [?]: 451 [2]
Given Kudos: 0
Send PM
Re: AB is the Diameter and point O is the center of the circle shown above [#permalink]
2
Expert Reply
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 (180-2x)/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
avatar
Manager
Manager
Joined: 27 Feb 2017
Posts: 188
Own Kudos [?]: 142 [0]
Given Kudos: 0
Send PM
Re: AB is the Diameter and point O is the center of the circle shown above [#permalink]
Can someone please explain this?
avatar
Supreme Moderator
Joined: 01 Nov 2017
Posts: 371
Own Kudos [?]: 451 [3]
Given Kudos: 0
Send PM
Re: AB is the Diameter and point O is the center of the circle shown above [#permalink]
3
Expert Reply
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 30-60-90 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 (180-60)/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_12-02-09.44.19.jpg
PicsArt_12-02-09.44.19.jpg [ 11.06 KiB | Viewed 14893 times ]

avatar
Intern
Intern
Joined: 01 Feb 2019
Posts: 37
Own Kudos [?]: 12 [0]
Given Kudos: 0
Send PM
Re: AB is the Diameter and point O is the center of the circle shown above [#permalink]
chetan2u wrote:
kruttikaaggarwal wrote:
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..


I get that (1/2)(√3/2*r^2)=(1/2) will be half of the area of ABC but how do we then establish that the area of the equilateral and the 30-60-90 are the same?
Manager
Manager
Joined: 05 Sep 2023
Posts: 63
Own Kudos [?]: 66 [1]
Given Kudos: 11
Send PM
Re: AB is the Diameter and point O is the center of the circle shown above [#permalink]
1
Angle ACB is an inscribed angle, so you know that it's half of the opposing angle. The opposing angle, the diameter, is 180 degrees, meaning angle ACB is 90 degrees.

Since x + 2x + 90 = 180, x = 30.
This makes triangle ACB 30-60-90 triangle.
To make things easy, let's say that side AB is 4, AC is 2 root 3, and CB is 2.
Then we find triangle ACB's area --> (b*h)/2 = (2*2root3)/2 = 2 root 3

Since 2x = 60, angle BOD is 60. Sides OB and OD are the radius, so you know they're equal. This means angle obd and angle odb need to be the same.
180-60 = 120/2 = 60
Triangle OBD is an equilateral triangle.
From the first part, AB = 4 so OB = 2.
Then, find the area of OBD. This creates another 30-60-90 triangle to find the height. The height of the triangle is root 3. The area of OBD = b * h / 2 = 2 * root 3/2 = root 3

Since QB is 2 times area of OBD, root 3 times 2 = 2 root 3.

QA = QB meaning Choice C is the correct answer.
Prep Club for GRE Bot
Re: AB is the Diameter and point O is the center of the circle shown above [#permalink]
Moderators:
GRE Instructor
51 posts
GRE Forum Moderator
28 posts
Moderator
1092 posts
GRE Instructor
218 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne