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# Triangle ABO is situated within the Circle with center O so

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Triangle ABO is situated within the Circle with center O so  [#permalink]

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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. 4pi-8
B. 8pi-4
C. 2pi-2
D. 3pi-3
E. 3pi-2
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Re: Triangle ABO is situated within the Circle with center O so  [#permalink]

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03 Sep 2013, 05:29
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3
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 $$4\pi-8$$.

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Re: Triangle ABO is situated within the Circle with center O so  [#permalink]

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17 Apr 2013, 11:33
6

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  [#permalink]

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17 Apr 2013, 08:30
2
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 $$Area = AB* ArcAB - 2r$$

$$Area = (Pi \sqrt{2} * \sqrt{32} / 2) - 8$$

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Re: Triangle ABO is situated within the Circle with center O so  [#permalink]

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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 $$Area = AB* ArcAB - 2r$$

$$Area = (Pi \sqrt{2} * \sqrt{32} / 2) - 8$$

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  [#permalink]

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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  [#permalink]

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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  [#permalink]

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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  [#permalink]

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24 Apr 2013, 01:03
1
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  [#permalink]

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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  [#permalink]

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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  [#permalink]

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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. 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 $$4\pi-8$$.

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  [#permalink]

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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   [#permalink] 28 May 2020, 19:45