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What is the area of the circle within the triangle

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What is the area of the circle within the triangle [#permalink] New post 10 Dec 2006, 08:06
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What is the area of the circle within the triangle?
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 [#permalink] New post 10 Dec 2006, 08:37
The triangle is equilateral.


Its height/2 is the radius of the circle = sqrt(3)

Area of sector with 60 at vertex = Angle/360 * PI * r^2 = PI/2
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 [#permalink] New post 11 Dec 2006, 06:54
trivikram wrote:
The triangle is equilateral.


Its height/2 is the radius of the circle = sqrt(3)

Area of sector with 60 at vertex = Angle/360 * PI * r^2 = PI/2


I'm afraid we could use this formula only when the angle is at the center. Any one?
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 [#permalink] New post 11 Dec 2006, 12:15
The triangle is equilateral.Apply pythorus theorem,

4^2=(4/2)^2+h^2.

h^2=12

h=diameter of circle.

Area of circle = PI*d^2/4=PI*12/4=3*PI???
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 [#permalink] New post 11 Dec 2006, 12:27
ncprasad wrote:
The triangle is equilateral.Apply pythorus theorem,

4^2=(4/2)^2+h^2.

h^2=12

h=diameter of circle.

Area of circle = PI*d^2/4=PI*12/4=3*PI???

Area of circle is correct but the question asked is about the area of the circle that is inside the triangle.
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 [#permalink] New post 11 Dec 2006, 12:27
Getting pi + 3 (sqrt 3)/2
But adopted fairly long process, takes around 4 to 5 min
any simple approach.. ??
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 [#permalink] New post 11 Dec 2006, 12:30
yogeshsheth wrote:
ncprasad wrote:
The triangle is equilateral.Apply pythorus theorem,

4^2=(4/2)^2+h^2.

h^2=12

h=diameter of circle.

Area of circle = PI*d^2/4=PI*12/4=3*PI???

Area of circle is correct but the question asked is about the area of the circle that is inside the triangle.


Drat! :oops:

Lesson learnt.
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 [#permalink] New post 11 Dec 2006, 12:54
anindyat wrote:
Getting pi + 3 (sqrt 3)/2
But adopted fairly long process, takes around 4 to 5 min
any simple approach.. ??


If I guessed it correctly you divided the area in one sector and two triangles.
The area of sector is is pi and what did u get as the area of other two traingles. I am getting the area as r^2/ for a single triangle?
Am i doing something wrong here?
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 [#permalink] New post 11 Dec 2006, 13:23
Yea ....fairly long process

Join the intersecting points of circle and triangle to the center of the circle.
So, the angle at center should be 120

Part of the circle constitutes 120 degree at the center so the circle, so the area is [pi * (sqrt3) ^ 2] * 120 /360 = pi

The remaining portion has 2 isosceles triangle, having two sides equal to radius = (sqrt 3)
And the angles are 30 and 30 and 120.

I am sure someone will come up with a better approach
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 [#permalink] New post 11 Dec 2006, 13:26
anindyat wrote:
Yea ....fairly long process

Join the intersecting points of circle and triangle to the center of the circle.
So, the angle at center should be 120

Part of the circle constitutes 120 degree at the center so the circle, so the area is [pi * (sqrt3) ^ 2] * 120 /360 = pi

The remaining portion has 2 isosceles triangle, having two sides equal to radius = (sqrt 3)
And the angles are 30 and 30 and 120.

I am sure someone will come up with a better approach


The height of these triangles will be r/2 and the area will be r^2/4 right?
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 [#permalink] New post 11 Dec 2006, 13:37
Yes Height = r /2 but the base is not r. Chord of the circle is base
I think you are considering base = r.
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 [#permalink] New post 11 Dec 2006, 13:43
anindyat wrote:
Yea ....fairly long process

Join the intersecting points of circle and triangle to the center of the circle.
So, the angle at center should be 120

Part of the circle constitutes 120 degree at the center so the circle, so the area is [pi * (sqrt3) ^ 2] * 120 /360 = pi

The remaining portion has 2 isosceles triangle, having two sides equal to radius = (sqrt 3)
And the angles are 30 and 30 and 120.

I am sure someone will come up with a better approach


Same approach... same result :)

Pi*r^2 - 2* (120/360*Pi*r^2 - sqrt(3)/4*r^2)

With:
> r = radius of circle = sqrt(3)/4*4 = sqrt(3)
> Pi*r^2 = the area of the circle
> -2* = the 2 identical areas above each side of the triangle
> 120/360*Pi*r^2 = the area of the sector defined by angle of 120° (iscole triangle with r as equal sides)
> - sqrt(3)/4*r^2 = the aera of the isocele triangle of 120°

Finally,

r^2*(Pi/3+sqrt(3)/2)= Pi + 3*sqrt(3)/2
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 [#permalink] New post 11 Dec 2006, 14:39
SOOOOO SORRRY GUYS :oops:

Maybe I should have phrased the question "what is the area of the circle in the diagram?"
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 [#permalink] New post 11 Dec 2006, 17:29
Sumithra wrote:
trivikram wrote:
The triangle is equilateral.


Its height/2 is the radius of the circle = sqrt(3)

Area of sector with 60 at vertex = Angle/360 * PI * r^2 = PI/2


I'm afraid we could use this formula only when the angle is at the center. Any one?


Can you elucidate this please? :)
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 [#permalink] New post 11 Dec 2006, 19:11
Phew! Thanks guys! Great job! I understood your approach. Few minutes to figure out and few more minutes to calculate... Can we afford this many minutes in the real exam? Hope we don't get such complicated questions.
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 [#permalink] New post 11 Dec 2006, 19:23
trivikram wrote:
Sumithra wrote:
trivikram wrote:
The triangle is equilateral.


Its height/2 is the radius of the circle = sqrt(3)

Area of sector with 60 at vertex = Angle/360 * PI * r^2 = PI/2


I'm afraid we could use this formula only when the angle is at the center. Any one?


Can you elucidate this please? :)


If the angle of the sector is at the centre, we can determine the ratio of the sector to the circle by dividing its angle by 360. Hence, this fraction multiplied by the area of the circle gives the area of the sector.
But the ratio of the sector to the circle cannot be determined if the angle is not at the centre. Hope I'm answering your question.
  [#permalink] 11 Dec 2006, 19:23
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