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In the figure above, the diameter of the circle is 20 and th

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In the figure above, the diameter of the circle is 20 and th [#permalink] New post 19 May 2017, 06:38
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#GREpracticequestion In the figure above, the diameter of the circle is 20.jpg
#GREpracticequestion In the figure above, the diameter of the circle is 20.jpg [ 13.37 KiB | Viewed 379 times ]


In the figure above, the diameter of the circle is 20 and the sum of the area of the sectors CFAO and BGDO 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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Last edited by Carcass on 30 Jan 2019, 04:03, edited 7 times in total.
Corrected the Q
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Re: In the figure above [#permalink] New post 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] New post 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] New post 03 Apr 2018, 07:26
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E is not the answer and no correct answer is given!

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] New post 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] New post 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] New post 03 Jun 2018, 23:09
Which area is the shaded one?
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Re: In the figure above [#permalink] New post 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.

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Re: In the figure above [#permalink] New post 13 Dec 2018, 18:07
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Carcass wrote:
Unfortunately, the book reports this kind of graph.

It should be the area inside the two triangles.

regards


I would suggest to edit the original post indicating that.
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Re: In the figure above [#permalink] New post 14 Dec 2018, 03:32
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Done.

Thanks.
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Re: In the figure above, the diameter of the circle is 20 and th [#permalink] New post 26 Dec 2018, 11:27
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This was one of the most controversial questions due to the poor quality of the geometry figure.

Now I have updated with a good one figure.

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Re: In the figure above, the diameter of the circle is 20 and th [#permalink] New post 28 Jan 2019, 13:49
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Area of Sector: x/360 * πR^2
Two Sector was shaded so each sector = 80π/2 = 40π

x/360 * πR^2 = 40π
x=144

a+b=c+d=144

Since R=D/2=10 a=b=c=d

a=b=c=d=72

a+b+c+d=72*4=288
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Re: In the figure above, the diameter of the circle is 20 and th [#permalink] New post 28 Jan 2019, 19:53
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Carcass wrote:
This was one of the most controversial questions due to the poor quality of the geometry figure.

Now I have updated with a good one figure.

Regards



Hi carcass,

The question or the figure is still flawed. The answer would be 288 ONLY when you are looking at the area of 80 pi for all the area outside the two sector and not the area outside the triangle.
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Re: In the figure above, the diameter of the circle is 20 and th [#permalink] New post 30 Jan 2019, 04:09
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Thanks, Cethan.

However, the figure, the question, and the answer choices are reported as they are in Barron's book without typos or else. Faithfully.

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Re: In the figure above, the diameter of the circle is 20 and th [#permalink] New post 30 Jan 2019, 04:35
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Carcass wrote:
Thanks, Cethan.

However, the figure, the question, and the answer choices are reported as they are in Barron's book without typos or else. Faithfully.

Regards


I have corrected the figure. Hopefully it is clear and no more ambiguous as given by Barron's.
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Re: In the figure above, the diameter of the circle is 20 and th [#permalink] New post 02 Feb 2019, 11:23
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Carcass wrote:
This was one of the most controversial questions due to the poor quality of the geometry figure.

Now I have updated with a good one figure.

Regards



@Carcass Plz can u look into the diagram attached

When the area of triangle is said to \(10\pi\) are we actually considering the the area of the arc AB and arc AC? I didn't really understood the reasoning

However, this is how I solved, @Carcass kindly provide ur valuable feedback::

Let first find the angle COA and angle DOB

i.e. \(\frac{{\angle COA}}{360} = \frac{{40\pi}}{{100\pi}}\) (each shaded area is 40 pi and the area of the circle is 100 pi)

or \(\angle COA = 144\)

Now \(\angle COD = 180 -144 =36\) (DA is the a straight line and diameter of the circle)

Similarly it can be proved \(\angle AOB = 36\)

Hence for the two \(\triangle\)'s combined

it can be written as \(a + b + 36 + c + d + 36 =360\)

or \(a + b + c + d = 288\)
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#GREpracticequestion In the figure above, the diameter of the circle is 20.jpg
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Re: In the figure above, the diameter of the circle is 20 and th [#permalink] New post 03 Feb 2019, 01:22
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This question clearly has a bit of a flaw in its wording. Certainly is that the letters to pinpoint the shaded area by Chetan came in handy.

However, I always try to look at the bright side of the story and stretch my mental muscles.

If we look at the original graph as it has been provided by the book and using logic, the solution will be easier. Of course, your solution is elegant and fine, though the arcs way to solve it I guess is a bit cumbersome. I always try to go down the road straight.

Attachment:
shot39.jpg
shot39.jpg [ 18.31 KiB | Viewed 245 times ]


Now, few easy steps

  • the stem says the diameter is 20 so the radius must be 10.
  • If the radius is 10 the entire area of the circle is \(100 \pi\)

At this point, the stem also says us that the shaded region is \(80 \pi\). At the time I didn't have the right graph and it looks like the following

Attachment:
shot40.jpg
shot40.jpg [ 11.03 KiB | Viewed 245 times ]


Even having this graph, is impossible to think that the shaded area is inside the isosceles triangles. it must be the outside of them. As it turns out, it is the rest of the circle. The total area is \(100 \pi\) so having \(80 \pi\) the latter represents 80% of the entire area.

From this, the area of the triangles is \(\frac{20}{100} = \frac{1}{5}\) of the entire area.

From this, we do know that inside an isosceles triangle, two sides are equal. The triangles inside the circle have the two major sides = 10. Therefore, the four angles a - b - c - d are the biggest. However, which is their measure ??

We do know that the sum of the angles is 180° if we take into account ONE triangle. Considering that the area of BOTH triangles is \(\frac{1}{5}\) of 360. We must divide 180 not by 5 BUT by 2.5 (considering ONE triangle). \(\frac{180}{2.5} = 72\)

Conclusion, \(a+b+c+d = 72*4 = 288\)
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Re: In the figure above, the diameter of the circle is 20 and th   [#permalink] 03 Feb 2019, 01:22
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