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 350,000 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:

Triangle ABO is situated within the Circle with center O so [#permalink]
17 Apr 2013, 07:59

1

This post received KUDOS

12

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

78% (03:37) correct
22% (03:32) wrong based on 246 sessions

Attachment:

Untitled1.jpg [ 9.24 KiB | Viewed 3910 times ]

Triangle ABO is situated within the Circle with center O so that one vertex is at the center of the circle, O, and its other vertices are located on its perimeter. The perimeter of triangle ABO is 8 + √32. What is the area of the hatched portion of the circle, the portion bounded by line AB and arc AB?

Re: Triangle ABO is situated within the Circle with center O so [#permalink]
17 Apr 2013, 11:33

3

This post received KUDOS

Answer is A

The solution is dependent on the formula - Center angle formed by an arc = 2 * Interior angle formed by the same arc.

Does the question provide <C as 45 degrees? The reason I ask is the question does not describe anything about triangle ABC.

Triangle AOB is an isosceles right angle triangle since <O is 90 and AO = OB. Let's assume a as the radius of the circle and which is same as AO or BO.

2a + a 2^1/2 = 8 + 32^1/2. Implies a = 4.

Area of shaded region = area of the sector AOB - area of triangle AOB

pi * 4^2 / area of the sector AOB = 360 / 90 ==> area of sector AOB = 4*pi

Area of shaded region = 4*pi - 8

//kudos please, if this explanation is good _________________

Re: Triangle ABO is situated within the Circle with center O so [#permalink]
23 Apr 2013, 05:59

Hi

First of all, Points A and B lie on the circle with center O : OA = OB = r hence triangle OAB is iscocele triangle . (1)

Second, given that perimeter of triangle AOB is equal to \(AB + OB + OA = 8 + \sqrt{32}\)

From (1) , \(2OA + AB = 8 + \sqrt{32}\) Thus, \(OA = 4\) AND \(AB = \sqrt{32}\) because \(\sqrt{32}\)cannot equal 2OA ( otherwise the triangle OAB wouldn't stand as isocele )

Now, we know that OA = OB = r = 4 and \(AB = \sqrt{32}\) and OAB is isocele

and AB^2 = 32 and OA^2 = OB^2 = 16 Hence OAB is a right isocele triangle. from pythagor (AB^2 = OA^2 + OB^2)

Area of shaded region = area of sector AOB - area of triangle AOB

area of sector AOB = Lenght of Arc(AB) * AB Lenght of Arc (AB) = diameter * Pi * 45/360 = Pi \(\sqrt{2}\) / 2

area of sector AOB = 4Pi area of triangle = 4*4 /2 = 8

Finally: area of shaed region = 4Pi - 8

Hope that helps _________________

KUDOS is the good manner to help the entire community.

"If you don't change your life, your life will change you"

Re: Triangle ABO is situated within the Circle with center O so [#permalink]
23 Apr 2013, 15:54

Rock750 wrote:

Hi

First of all, Points A and B lie on the circle with center O : OA = OB = r hence triangle OAB is iscocele triangle . (1)

Second, given that perimeter of triangle AOB is equal to \(AB + OB + OA = 8 + \sqrt{32}\)

From (1) , \(2OA + AB = 8 + \sqrt{32}\) Thus, \(OA = 4\) AND \(AB = \sqrt{32}\) because \(\sqrt{32}\)cannot equal 2OA ( otherwise the triangle OAB wouldn't stand as isocele )

Now, we know that OA = OB = r = 4 and \(AB = \sqrt{32}\) and OAB is isocele

and AB^2 = 32 and OA^2 = OB^2 = 16 Hence OAB is a right isocele triangle. from pythagor (AB^2 = OA^2 + OB^2)

Area of shaded region = area of sector AOB - area of triangle AOB

area of sector AOB = Lenght of Arc(AB) * AB Lenght of Arc (AB) = diameter * Pi * 45/360 = Pi \(\sqrt{2}\) / 2

area of sector AOB = 4Pi area of triangle = 4*4 /2 = 8

Finally: area of shaed region = 4Pi - 8

Hope that helps

Thanks!. This surely helped me. _________________

"Where are my Kudos" ............ Good Question = kudos

Re: Triangle ABO is situated within the Circle with center O so [#permalink]
23 Apr 2013, 16:43

Rock750 wrote:

From (1) , \(2OA + AB = 8 + \sqrt{32}\) Thus, \(OA = 4\) AND \(AB = \sqrt{32}\) because \(\sqrt{32}\)cannot equal 2OA ( otherwise the triangle OAB wouldn't stand as isocele )

I have a problem with this part..

\(\sqrt{32} = 4 \sqrt{2}\)

Therefore, why couldn't the two equal sides be \(2\sqrt{2}\) each and the third side be 8?

And even then, how are you getting the angle of the triangle? I feel like there are some number properties of triangles i'm missing that I need to memorize here...

Re: Triangle ABO is situated within the Circle with center O so [#permalink]
24 Apr 2013, 01:03

1

This post received KUDOS

dave785 wrote:

\(\sqrt{32} = 4 \sqrt{2}\)

Therefore, why couldn't the two equal sides be \(2\sqrt{2}\) each and the third side be 8?

And even then, how are you getting the angle of the triangle? I feel like there are some number properties of triangles i'm missing that I need to memorize here...

Hi dave785,

Answer to the 1st question: The lenght of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides; Therefore OA and OB must be equal to 4 and AB must be equal to \(\sqrt{32}\)

Now , for the second question, which angle are you talking about ? _________________

KUDOS is the good manner to help the entire community.

"If you don't change your life, your life will change you"

Re: Triangle ABO is situated within the Circle with center O so [#permalink]
31 Aug 2013, 16:57

Rock750 wrote:

Hi

First of all, Points A and B lie on the circle with center O : OA = OB = r hence triangle OAB is iscocele triangle . (1)

Second, given that perimeter of triangle AOB is equal to \(AB + OB + OA = 8 + \sqrt{32}\)

From (1) , \(2OA + AB = 8 + \sqrt{32}\) Thus, \(OA = 4\) AND \(AB = \sqrt{32}\) because \(\sqrt{32}\)cannot equal 2OA ( otherwise the triangle OAB wouldn't stand as isocele )

Now, we know that OA = OB = r = 4 and \(AB = \sqrt{32}\) and OAB is isocele

and AB^2 = 32 and OA^2 = OB^2 = 16 Hence OAB is a right isocele triangle. from pythagor (AB^2 = OA^2 + OB^2)

Area of shaded region = area of sector AOB - area of triangle AOB

area of sector AOB = Lenght of Arc(AB) * AB Lenght of Arc (AB) = diameter * Pi * 45/360 = Pi \(\sqrt{2}\) / 2

area of sector AOB = 4Pi area of triangle = 4*4 /2 = 8

Finally: area of shaed region = 4Pi - 8

Hope that helps

Lenght of Arc (AB) = diameter * Pi * 45/360 = Pi \(\sqrt{2}\) / 2

How is this operation not equal to just pi. Diameter is 8, so 8*45/360 cancels out and the only thing left is Pi. Where do you get sqrt(2)/2 from?

Re: Triangle ABO is situated within the Circle with center O so [#permalink]
03 Sep 2013, 05:29

6

This post received KUDOS

Expert's post

jjack0310 wrote:

Rock750 wrote:

Hi

First of all, Points A and B lie on the circle with center O : OA = OB = r hence triangle OAB is iscocele triangle . (1)

Second, given that perimeter of triangle AOB is equal to \(AB + OB + OA = 8 + \sqrt{32}\)

From (1) , \(2OA + AB = 8 + \sqrt{32}\) Thus, \(OA = 4\) AND \(AB = \sqrt{32}\) because \(\sqrt{32}\)cannot equal 2OA ( otherwise the triangle OAB wouldn't stand as isocele )

Now, we know that OA = OB = r = 4 and \(AB = \sqrt{32}\) and OAB is isocele

and AB^2 = 32 and OA^2 = OB^2 = 16 Hence OAB is a right isocele triangle. from pythagor (AB^2 = OA^2 + OB^2)

Area of shaded region = area of sector AOB - area of triangle AOB

area of sector AOB = Lenght of Arc(AB) * AB Lenght of Arc (AB) = diameter * Pi * 45/360 = Pi \(\sqrt{2}\) / 2

area of sector AOB = 4Pi area of triangle = 4*4 /2 = 8

Finally: area of shaed region = 4Pi - 8

Hope that helps

Lenght of Arc (AB) = diameter * Pi * 45/360 = Pi \(\sqrt{2}\) / 2

How is this operation not equal to just pi. Diameter is 8, so 8*45/360 cancels out and the only thing left is Pi. Where do you get sqrt(2)/2 from?

Triangle ABO is situated within the Circle with center O so that one vertex is at the center of the circle, O, and its other vertices are located on its perimeter. The perimeter of triangle ABO is 8 + √32. What is the area of the hatched portion of the circle, the portion bounded by line AB and arc AB? A. 4pi-8 B. 8pi-4 C. 2pi-2 D. 3pi-3 E. 3pi-2

The Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle.

Therefore the angle O is twice the angle C, or 90 degrees.

Next, since OA=OB=radius, then triangle OAB is isosceles right triangle, or 45-45-90 triangle, and thus its sides are in ratio \(1:1:\sqrt{2}\). Therefore \(r+r+r\sqrt{2}=8+\sqrt{32}\) --> \(r(2+\sqrt{2})=4(2+\sqrt{2})\) --> \(r=4\).

The area of the circle is \(\pi{r^2}=16\pi\) and the area of the sector OAB is 1/4 of that, or \(4\pi\).

The area of triangle OAB is \(\frac{1}{2}*4*4=8\), therefore the area of shaded region is \(4\pi-8\).

Re: Triangle ABO is situated within the Circle with center O so [#permalink]
28 Aug 2015, 02:37

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. _________________

On September 6, 2015, I started my MBA journey at London Business School. I took some pictures on my way from the airport to school, and uploaded them on...

When I was growing up, I read a story about a piccolo player. A master orchestra conductor came to town and he decided to practice with the largest orchestra...

I’ll start off with a quote from another blog post I’ve written : “not all great communicators are great leaders, but all great leaders are great communicators.” Being...