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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?

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

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17 Apr 2013, 11:33

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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 [#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.
_________________

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

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...

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

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24 Apr 2013, 01:03

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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?

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]

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28 Aug 2015, 02:37

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

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19 Jan 2017, 09:48

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).

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