Mar 27 03:00 PM PDT  04:00 PM PDT Join a free live webinar and learn the winning strategy for a 700+ score on GMAT & the perfect application. Save your spot today! Wednesday, March 27th at 3 pm PST Mar 29 06:00 PM PDT  07:00 PM PDT Join a FREE 1day workshop and learn how to ace the GMAT while keeping your fulltime job. Limited for the first 99 registrants. Mar 29 10:00 PM PDT  11:00 PM PDT Right now, their GMAT prep, GRE prep, and MBA admissions consulting services are up to $1,100 off. GMAT (Save up to $261): SPRINGEXTRAGMAT GRE Prep (Save up to $149): SPRINGEXTRAGRE MBA (Save up to $1,240): SPRINGEXTRAMBA Mar 30 07:00 AM PDT  09:00 AM PDT Attend this webinar and master GMAT SC in 10 days by learning how meaning and logic can help you tackle 700+ level SC questions with ease. Mar 31 07:00 AM PDT  09:00 AM PDT Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes.
Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 29 Mar 2013
Posts: 13

Triangle ABO is situated within the Circle with center O so
[#permalink]
Show Tags
17 Apr 2013, 08:59
Question Stats:
73% (03:07) correct 27% (02:54) wrong based on 238 sessions
HideShow timer Statistics
Attachment:
Untitled1.jpg [ 9.24 KiB  Viewed 16370 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? A. 4pi8 B. 8pi4 C. 2pi2 D. 3pi3 E. 3pi2
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 53831

Re: Triangle ABO is situated within the Circle with center O so
[#permalink]
Show Tags
03 Sep 2013, 06:29
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. 4pi8 B. 8pi4 C. 2pi2 D. 3pi3 E. 3pi2 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 454590 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 \(4\pi8\). Answer: A.
_________________




Manager
Status: Looking to improve
Joined: 15 Jan 2013
Posts: 151
GMAT 1: 530 Q43 V20 GMAT 2: 560 Q42 V25 GMAT 3: 650 Q48 V31

Re: Triangle ABO is situated within the Circle with center O so
[#permalink]
Show Tags
17 Apr 2013, 12:33
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
_________________
KUDOS is a way to say Thank You




Manager
Status: Final Lap
Joined: 25 Oct 2012
Posts: 230
Concentration: General Management, Entrepreneurship
GPA: 3.54
WE: Project Management (Retail Banking)

Re: Triangle ABO is situated within the Circle with center O so
[#permalink]
Show Tags
17 Apr 2013, 09:30
Since OA = OB = r , Perimeter (AOB) = AB + OB + OA = 2OA + AB = 8 + \(\sqrt{32}\) Hence, \(AB = \sqrt{32}\) and OA = OB = r = 4 \(Lenght of arc AB = diameter * Pi * 45/360\) > \(Lenght of arc AB = Pi * \sqrt{2} / 2\) Finally, the area of the portion bounded by line AB and arc AB \(Area = AB* ArcAB  2r\) \(Area = (Pi \sqrt{2} * \sqrt{32} / 2)  8\) Thus, Answer A : 4Pi8
_________________
KUDOS is the good manner to help the entire community.
"If you don't change your life, your life will change you"



Manager
Joined: 14 Jan 2013
Posts: 136
Concentration: Strategy, Technology
GMAT Date: 08012013
GPA: 3.7
WE: Consulting (Consulting)

Re: Triangle ABO is situated within the Circle with center O so
[#permalink]
Show Tags
23 Apr 2013, 05:08
Rock750 wrote: Since OA = OB = r , Perimeter (AOB) = AB + OB + OA = 2OA + AB = 8 + \(\sqrt{32}\)
Hence, \(AB = \sqrt{32}\) and OA = OB = r = 4
\(Lenght of arc AB = diameter * Pi * 45/360\)
> \(Lenght of arc AB = Pi * \sqrt{2} / 2\)
Finally, the area of the portion bounded by line AB and arc AB \(Area = AB* ArcAB  2r\)
\(Area = (Pi \sqrt{2} * \sqrt{32} / 2)  8\)
Thus, Answer A : 4Pi8 Did not understand this one. Probably, I am missing something



Manager
Status: Final Lap
Joined: 25 Oct 2012
Posts: 230
Concentration: General Management, Entrepreneurship
GPA: 3.54
WE: Project Management (Retail Banking)

Re: Triangle ABO is situated within the Circle with center O so
[#permalink]
Show Tags
23 Apr 2013, 06:59
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"



Manager
Joined: 14 Jan 2013
Posts: 136
Concentration: Strategy, Technology
GMAT Date: 08012013
GPA: 3.7
WE: Consulting (Consulting)

Re: Triangle ABO is situated within the Circle with center O so
[#permalink]
Show Tags
23 Apr 2013, 16:54
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.



Manager
Joined: 09 Apr 2013
Posts: 193
Location: United States
Concentration: Finance, Economics
GMAT 1: 710 Q44 V44 GMAT 2: 740 Q48 V44
GPA: 3.1
WE: Sales (Mutual Funds and Brokerage)

Re: Triangle ABO is situated within the Circle with center O so
[#permalink]
Show Tags
23 Apr 2013, 17:43
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...



Manager
Status: Final Lap
Joined: 25 Oct 2012
Posts: 230
Concentration: General Management, Entrepreneurship
GPA: 3.54
WE: Project Management (Retail Banking)

Re: Triangle ABO is situated within the Circle with center O so
[#permalink]
Show Tags
24 Apr 2013, 02:03
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"



Intern
Joined: 04 May 2013
Posts: 44

Re: Triangle ABO is situated within the Circle with center O so
[#permalink]
Show Tags
31 Aug 2013, 17:57
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?



Manager
Joined: 28 May 2014
Posts: 52

Re: Triangle ABO is situated within the Circle with center O so
[#permalink]
Show Tags
27 Jul 2014, 10:16
Small correction. I believe length of arc is found always from centre angle which should be 90 degrees and not 45 degrees.



Intern
Joined: 02 Jul 2017
Posts: 32

Re: Triangle ABO is situated within the Circle with center O so
[#permalink]
Show Tags
07 Aug 2017, 08:54
Bunuel wrote: 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. 4pi8 B. 8pi4 C. 2pi2 D. 3pi3 E. 3pi2 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 454590 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 \(4\pi8\). Answer: A. If an angle in the triangle is 90 degress, shouldn't the hypotenuse be the diameter. How is this scenario possible?



NonHuman User
Joined: 09 Sep 2013
Posts: 10205

Re: Triangle ABO is situated within the Circle with center O so
[#permalink]
Show Tags
21 Feb 2019, 08:14
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.
_________________




Re: Triangle ABO is situated within the Circle with center O so
[#permalink]
21 Feb 2019, 08:14






