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# In the figure above

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In the figure above [#permalink]  19 May 2017, 06:38
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Question Stats:

55% (01:17) correct 44% (01:59) wrong based on 29 sessions

In the figure above, the diameter of the circle is 20 and the area of the shaded region is $$80 \pi$$. What is the value of a + b + c + d ?

A) 144

B) 216

C) 240

D) 270

E) 288
[Reveal] Spoiler: OA

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Re: In the figure above [#permalink]  20 May 2017, 04:13
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Damm bro, I think the question looks a bit confusing at the beginning... What is missing is the information that the area of the circle WITHOUT the two triangles is 80. It´s not possible in a other way, since the radius is 10, the area is 100*pi.

Given that, we can make an approximation for a, b, c and d using circle segments. It is only an approximation because we assume the area of the triangle is the same as the area of the respective circle segment. This is fair enough for small triangles and since we do not need to calculate the value. We just have to come near to a provided solution.

So let us say the inner angles of the triangles are called e and f respectively. Since the total are is 100 pi and the area without the two triangles is 80 pi, the two triangles have the area 20 pi. And one triangle has the area of 10 pi, which is exactly 10% of the total area.
Again, assuming the triangle is a circle segment, the inner angle (e and f) must be 360/10=36 degrees. Since a=b=c=d, a+b=180-36 => a+b=144. Therefore a+b+c+d= 144*2=288.

This is the value provided as OA.
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Re: In the figure above [#permalink]  25 Jan 2018, 02:56
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D=20 so the r=10. So the circle is 100pi.
Given that, the circle area without the two triangles is 80pi. So the area of the two triangles is 20pi.
Now simply, considering a triangle as a sector of the circle we can use the formula (theta/360)=(Sector area/circle area).
By solving, theta=36 degrees. So a sector creates an angle of 36 degrees in the center and leaving 144 degrees for other two angles of a triangle.
So for two triangles' we can get 144+144=288 degrees.
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Re: In the figure above [#permalink]  03 Apr 2018, 07:26
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The question read the area of the circle without the two triangles is 80 pi. Not the area of the circle without the sector, both answers above are incorrect.

The area of the two triangles is 2 * ( 1/2 * r^2 * Sin ( Q)) where Q is the angle of the triangle's apex at the center of the circle.

So 20pi = r^2 * Sin (Q) .... Q = asin ( 0.2) = 11.537... thus a + b + c + d = 2* (180-11.537) = 336.926.
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Re: In the figure above [#permalink]  03 Apr 2018, 13:25
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YMAkib wrote:
D=20 so the r=10. So the circle is 100pi.
Given that, the circle area without the two triangles is 80pi. So the area of the two triangles is 20pi.
Now simply, considering a triangle as a sector of the circle we can use the formula (theta/360)=(Sector area/circle area).
By solving, theta=36 degrees. So a sector creates an angle of 36 degrees in the center and leaving 144 degrees for other two angles of a triangle.
So for two triangles' we can get 144+144=288 degrees.

I am not getting 36 degrees for theta/360 = 20pi/100pi. Can you please explain how you got to 36?
Thank you
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Re: In the figure above [#permalink]  03 Apr 2018, 13:38
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There are two versions of this question. I edited the question above changing is the shaded region that I suppose is the area inside the circle but OUTSIDE the two triangles.

Now, the radius is 10 and the area is 100 \pi. The area of the shadow region is $$\frac{80}{100} = \frac{4}{5}$$ of the circle and the rest is $$\frac{1}{5}$$ of the circle.

We do know that the central angle in a circle is always $$360$$ and $$\frac{1}{5}$$ of 360 is 72°.

$$a + b + c + d + 72 = 360$$

$$a + b + c + d = 288$$
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Re: In the figure above [#permalink]  03 Jun 2018, 23:09
Which area is the shaded one?
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Re: In the figure above [#permalink]  04 Jun 2018, 11:10
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Unfortunately, the book reports this kind of graph.

It should be the area inside the two triangles.

regards
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Re: In the figure above   [#permalink] 04 Jun 2018, 11:10
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