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Circle inside equilateral triangle

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by sunnyjohn » Mon Nov 23, 2009 6:26 pm
Sorry, I edited my post...!

IMO : 256*Sqrt(3)

How:

the sides of triangle will be tangent to circle. Means,

angle between radius and side of triangle = 90
two tangent will meet at vertex of triangle --> draw a line from any vertex to center of circle.

so you will see - another triangle with sides as ( radius, tangent, line from vertex to center of circle)
this is actually a 30-60-90 triangle. [the line from vertex will bisect the vertex angle]

so the ratio of side will be : 1:sqrt(3):2
so : 8: 8(sqrt(3): 16

this will give us side of triangle : 32
Area : sqrt(3)/4 * 32*32 ==> 256*Sqrt(3)
Last edited by sunnyjohn on Mon Nov 23, 2009 9:51 pm, edited 1 time in total.
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by adam15 » Mon Nov 23, 2009 8:05 pm
I think the answer is 192
I will take some of result sunnyjohn
we have 8:18sqr(3):16
16 is half the segment of one of the sidde the triangle ABC
the height of the triangle passes through the center of the circl "which is barycenter" and the radius is 1/3 of the height
then h=24
thus the surface of the triangle is 16*24/2=192
any comments
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by heshamelaziry » Mon Nov 23, 2009 9:05 pm
IMO the side of the equelitarel triangle is16 root3. Area of equilateral triangle is (s^2 *root3)/4 = 192 root3
Last edited by heshamelaziry on Mon Nov 23, 2009 10:13 pm, edited 1 time in total.
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by papgust » Mon Nov 23, 2009 10:07 pm
sunnyjohn wrote:
so the ratio of side will be : 1:sqrt(3):2
so : 8: 8(sqrt(3): 16

this will give us side of triangle : 32
Area : sqrt(3)/4 * 32*32 ==> 256*Sqrt(3)
Hi Sunny,

How did you get 32 as a side?

This is my calculation (Correct me if i'm wrong):
16 is the side opp to 90 degree angle (angle b/w tangent and radius). 30 degrees is the angle formed in the vertex of triangle (Opp of which is radius 8). so the partial side of the vertex should be 8 root(3). So the side of a triangle must be 16 root(3).

Area of triangle = root(3)/4 * [16 root(3)]^2 = 192 root(3)

IMO it should be 192 root(3)
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by sunnyjohn » Tue Nov 24, 2009 12:39 am
papgust wrote:
sunnyjohn wrote:
so the ratio of side will be : 1:sqrt(3):2
so : 8: 8(sqrt(3): 16

this will give us side of triangle : 32
Area : sqrt(3)/4 * 32*32 ==> 256*Sqrt(3)
Hi Sunny,

How did you get 32 as a side?

This is my calculation (Correct me if i'm wrong):
16 is the side opp to 90 degree angle (angle b/w tangent and radius). 30 degrees is the angle formed in the vertex of triangle (Opp of which is radius 8). so the partial side of the vertex should be 8 root(3). So the side of a triangle must be 16 root(3).

Area of triangle = root(3)/4 * [16 root(3)]^2 = 192 root(3)

IMO it should be 192 root(3)
Yup, you are correct...~ I made a silly mistake at the end of calculation...Ans should be 192(sqrt(3)).

Thank you Papgust..!
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by gmatmachoman » Tue Apr 27, 2010 11:35 am
Handy formula :

r =a *(sqrt 3)/6

where r is the radius of circle & a is the side of triangle.
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by ronsom » Tue Apr 27, 2010 12:25 pm
Another way of solving this problem:-

When a circle is inscribed inside an equilateral triangle, the center of the circle coincides with the centroid.
The centroid divides the median in a 2:1 ratio... Since radius = 8, therefore, the median ( = height of the equi. triangle) = 24.
BY applying pyth. theorem we get 24^2 = side^2 - (side/2)^2
=> side = 16 sqrt(3)
Area = (sqrt(3)*16(sqrt(3))^2)/4 = 192 sqrt(3)
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by tata » Wed Apr 28, 2010 1:07 pm
gen3hatch wrote:If a circle is inscribed in an equilateral triangle and the radius of the circle is 8 what is the area of the triangle?

Thanks
Can someone please explain this using a diagram? I am getting confused here a little but so a diagramatic explanation would really be helpful.
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by goyalsau » Thu Nov 25, 2010 12:35 pm
Great Post Guys,

I want to know if the Circle in inscribed in a right angle triangle ( 30, 60, 90 ) Then that will be the area of the triangle
Saurabh Goyal
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by sumansana88 » Sat Sep 15, 2012 2:55 am
thanks dear........this is the correct one
..
heshamelaziry wrote:IMO the side of the equelitarel triangle is16 root3. Area of equilateral triangle is (s^2 *root3)/4 = 192 root3
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by benjiboo » Sat Sep 15, 2012 9:11 pm
Attached picture was my thought process. This whole process should take maybe 30 seconds to set up... But I tried to lay it out in steps for you. Hope im right... but anyway lmk if this helps.

may have to download pic to see the whole thing... it appears cut off in my browser when viewing on beatthegmat
Attachments
tricirc.jpg
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