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Triangle ABO is situated within the Circle with center O so
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17 Apr 2013, 07:59
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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
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Re: Triangle ABO is situated within the Circle with center O so
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03 Sep 2013, 05:29
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.
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Re: Triangle ABO is situated within the Circle with center O so
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17 Apr 2013, 11:33
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




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Re: Triangle ABO is situated within the Circle with center O so
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17 Apr 2013, 08:30
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



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Re: Triangle ABO is situated within the Circle with center O so
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23 Apr 2013, 04:08
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



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Re: Triangle ABO is situated within the Circle with center O so
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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



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Re: Triangle ABO is situated within the Circle with center O so
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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.



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Re: Triangle ABO is situated within the Circle with center O so
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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...



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Re: Triangle ABO is situated within the Circle with center O so
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24 Apr 2013, 01:03
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 ?



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Re: Triangle ABO is situated within the Circle with center O so
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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?



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Re: Triangle ABO is situated within the Circle with center O so
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27 Jul 2014, 09:16
Small correction. I believe length of arc is found always from centre angle which should be 90 degrees and not 45 degrees.



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Re: Triangle ABO is situated within the Circle with center O so
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07 Aug 2017, 07:54
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?



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Re: Triangle ABO is situated within the Circle with center O so
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28 May 2020, 19:45
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Re: Triangle ABO is situated within the Circle with center O so
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