equilateral triangle pS

wow, great logitech. please do share if you come across any other such relations.

scoobydooby wrote:wow, great logitech. please do share if you come across any other such relations.

I dont come across, I find them

scoobydooby wrote: using the property of the 30-60-90 with base as 3, we get radius as 2*3^1/2

excellent logic. but something doesn't add up for me. I know it has to work, but it doesnt:

3^2 + (3/2)^2 = 45/4
(45/4)^(1/2) = 3/2*5^(1/2) which is not the radius in this problem! Radius is 2*3^(1/2)! What gives?

adilka wrote:

scoobydooby wrote: using the property of the 30-60-90 with base as 3, we get radius as 2*3^1/2

excellent logic. but something doesn't add up for me. I know it has to work, but it doesnt:

3^2 + (3/2)^2 = 45/4
(45/4)^(1/2) = 3/2*5^(1/2) which is not the radius in this problem! Radius is 2*3^(1/2)! What gives?

say the triangle inscribed be ABC (say B top, AC base). from A and C extend upto O the centre of the circle. you have triangle OAC within triangle ABC. OA=OC=radius=r. let the perpendicular from B to AC be BN. ON perpendicular to AC.

AN=3 (the perpendicular bisects the base)
angle AON:60 (as angle AOC=2*angleABC)
in the 30-60-90 triangle AON, ratio of sides 1:3^1/2:2
AN=3, this would give radius=OA=2*3^1/2

logitechs relation between triangle and circle would be much faster in this case.

Thanks, Scooby. I found my mistake. Was taking 1:2 proportion of the legs, vs. hyp./smaller leg.

scoobydooby wrote:

adilka wrote:

scoobydooby wrote: using the property of the 30-60-90 with base as 3, we get radius as 2*3^1/2

excellent logic. but something doesn't add up for me. I know it has to work, but it doesnt:

3^2 + (3/2)^2 = 45/4
(45/4)^(1/2) = 3/2*5^(1/2) which is not the radius in this problem! Radius is 2*3^(1/2)! What gives?

say the triangle inscribed be ABC (say B top, AC base). from A and C extend upto O the centre of the circle. you have triangle OAC within triangle ABC. OA=OC=radius=r. let the perpendicular from B to AC be BN. ON perpendicular to AC.

AN=3 (the perpendicular bisects the base)
angle AON:60 (as angle AOC=2*angleABC)
in the 30-60-90 triangle AON, ratio of sides 1:3^1/2:2
AN=3, this would give radius=OA=2*3^1/2

logitechs relation between triangle and circle would be much faster in this case.

Can you attach a visual image? I'm not quite sure what you're trying to describe.

logitech wrote:Guys I have seen this same problem so many times over and over again..I carried the whole calculation and found out:

Area of Triangle / Area of Circle = 3xSQRT(3)/4Pi

In our example:

3xSQRT(3)/4Pi = 9SQRT(3)/Area of Circle

Area of Circle : 12 pi

In your face GMAT!!



Put it in a flash card ( if you have time and patience, you can drive the formula )

Thanks logitech!

Does this equation work for a circle inscribed in an equilateral triangle as well or is there another equation?

logitech wrote:Area of Triangle / Area of Circle = 3xSQRT(3)/4Pi

Hi Logitech,

Is this formula applicable only to an equilateral triangle or any triangle?

I would like to derive this formula, can you please share the starting point for this formula?

Thanks,

Vineet

Anyone found an alternative approach to Logitech's formula? Thanks

Hey guys,

I know this might be too late.Here is how I arrived at the answer12 pi

All of you guys solved that side=6 [needs no explanation]

Now here is how i proceeded.

The perpendicular = 3 root 3.

The theory(property of an equilateral triangles inscribed in a circle) is that "if the perpendicular is divided in a ratio 2:1 such that the longer part is from the vertex and the shorter part is towards the base,to which it is perpendicular and which it bisects" then the radius of the circle is the longer of the two parts.

so the radius is 2/3 [3 root 3]= 2 root 3

Area= pi r^2= pi (2 root 3)^2= 12pi

I think we can use two formula here
step1
traditional
area of eq triangle= (root(3))/4) side^2
so
9 (root) 3= (root(3))/4) side^2
side^2=36
side=6(since it is length)

now generalized one,my teacher told me that it can be used as formula
when eq triangle is inscribed inside circle
area of cirle=(pi/3)xside^2

so pi/3 X 36
12 pi
answer

When circle is inscribed inside eq triangle with side x (opposite of above)
formula is
pi/12 X x^2

This does not require a formula

we know Area of Equilateral triangle = ((Sq Root 3) / 4 )* (side) sq

we get side

Also since the triangle is inscribed the center angle is twice 60 = 120

So we have a 30 - 60 -90 with one side as 6/2 = 3 from which we can derive radius.

Once r is found Area circle can be found.

Hope this helps.

Hi,

As we have seen there are multiple ways to solve this question :

1. Use s(s-a)(s-b)(s-c) = (abc/4R)^2

2. Using the concept ((property of an equilateral triangles inscribed in a circle) that the perpendicular is divided in a ratio 2:1 such that the longer part is from the vertex and the shorter part is towards the base,to which it is perpendicular and which it bisects"]


Let me add another way to use a little bit of trigonometry (which also derives the radius, somebody asked how to find it without knowing these rules) :

3.
In the diagram attached :
angle OAD=angle OBD=30
Triangle OAD and triangle OBD are 30-60-90 triangle

So, AD/OA = sqrt3/2; (If you know that D is midpoint of AB, then you may skip the next few lines altogether).
similarly; BD/OB = sqrt3/2;

=> AD+BD = sqrt3/2 * (OA+OB) = sqrt3/2 * 2r
=> 6 = sqrt3/2 * 2r
=> r = 6/sqrt3

Hence , area = pi*r^2 = 12pi.

Thanks

Area of equilateral triangle = (sqrt(3)/4)*(S^2) = 9 sqrt3)

=> side S = 6

Radius of the circle R = S/sqrt(s) = 2* sqrt(3)

Area of the circle = pi*4**3 = 12pi

Answer C

Answer is 12pi.
Here are 2 important properties:
1. Perpendiculars on the sides of equilateral triangles intersect eacho other in the ratio 2:1.
2. If an equilateral triangle is inscribed in a circle the perpendiculars intersect each other ot the center of the circle.

Based on 1 and 2 above : radius (r) = 2h (height of triangle)/3
For equilateral triangle h/b(base) = tan 60 = sqrt(3)/2 => b = 2h/sqrt(3).
Given area of triangle = bh/2 = 9*sqrt(3)=> h^2/sqrt(3) {replaced value of b from above} => h = 3*sqrt(3).
so r = 2h/3 = 2*sqrt(3) and hence area of circle = 12pi.