# An equilateral triangle and a circle have the same center....

• Mar 4th 2008, 01:57 PM
help1
An equilateral triangle and a circle have the same center....
An equilateral triangle and a circle have the same center. The area of the triangle not in the circle equals the area of the circle not in the triangle. If the radius of the circle is 1, find the length of the side of the traingle to the nearest hundredth.

• Mar 4th 2008, 09:49 PM
earboth
Quote:

Originally Posted by help1
An equilateral triangle and a circle have the same center. The area of the triangle not in the circle equals the area of the circle not in the triangle. If the radius of the circle is 1, find the length of the side of the traingle to the nearest hundredth.

1. Calculate the area of the circle:

$\displaystyle a_c = \pi \cdot 1^2 = \pi$

2. Let s denote the side of the equilateral triangle. Then the area of the triangle can be calculated by:

$\displaystyle a_t = \frac14 \cdot s^2 \cdot \sqrt{3}$

3. Both areas have the same value:

$\displaystyle \frac14 \cdot s^2 \cdot \sqrt{3} = \pi ~\iff~ s^2 = \frac{4 \pi}{\sqrt{3}}$ That means:

$\displaystyle s \approx 2.6935...$
• Mar 4th 2008, 09:56 PM
Mathnasium
I'll take a crack at it - it'd be much easier with a picture, but oh well.

Draw your triangle with your circle concentric to it. Parts of your circle, obviously, are outside, and parts are inside. We're going to find the area of the circle that is NOT inside the triangle by finding the area of the circle that IS inside the triangle and then subtracting that from $\displaystyle \pi$, the total area of the circle (since it has a radius of 1).

From the center, draw six lines - one line to each point of intersection of the circle and the triangle. These six lines are all radii, so they all have length 1. Inside the circle, your triangle is now divided up into 6 parts. 3 of these parts are congruent to one another, and the other 3 are congruent to one another. We'll find the area of each of the two kinds of pieces and then multiply them each by 3. This will give us the area of the circle that is inside the triangle.

The central angle in each piece is 60 degrees - you've divided the circle up using 6 equally-spaced radii, so the 360 degrees in the circle is divided evenly.

The first set of 3 parts: look at the parts of the triangle that are triangles themselves - 3 of your pieces should have 3 straight edges, and 3 will have a third "side" that is an arc of the circle. So we have three triangles, each of which is equilateral - the central angle is 60, and we know it is an isosceles triangle since two of the legs are both radii of the circle and thus congruent. This means the base angles are also congruent, giving us 60-60-60. The area of an isosceles triangle is $\displaystyle \frac{\sqrt{3}}{4}l^2$, so each of our triangles has an area of $\displaystyle \frac{\sqrt{3}}{4}$, giving a total area for these 3 pieces of:

$\displaystyle \frac{3\sqrt{3}}{4}$.

Now let's look at the other three pieces of our triangle that are inside the circle - these are really sectors of the circle, each again with a central angle of 60 degrees. Put them all together and they form a single 180-degree sector of the circle - in other words, half the circle. The circle has area $\displaystyle \pi$, so these sectors have area:

$\displaystyle \frac{\pi}{2}$.

Thus, the area of the circle inside the triangle is:

$\displaystyle \frac{3\sqrt{3}}{4} + \frac{\pi}{2} = \frac{3\sqrt{3}+2{\pi}}{4}$.

Then the area of the circle not in the triangle is:

$\displaystyle \pi - \frac{3\sqrt{3}+2{\pi}}{4} = \frac{2 \pi - {3\sqrt{3}}}{4}$.

The problem says this is the same as the area of the triangle not in the circle.

Let the total area of the triangle be $\displaystyle \frac{\sqrt{3}}{4}l^2$. We already know the area of the triangle in the circle:

$\displaystyle \frac{3\sqrt{3}+2{\pi}}{4}$

so the area of the triangle NOT in the circle is the total area less this amount:

$\displaystyle \frac{\sqrt{3}}{4}l^2 - \frac{3\sqrt{3}+2{\pi}}{4} = \frac{l^2 \sqrt{3} - 3 \sqrt{3} - 2 \pi}{4}$.

This is equal to the area of the circle not in the triangle, which gives:

$\displaystyle \frac{l^2 \sqrt{3} - 3 \sqrt{3} - 2 \pi}{4} = \frac{2\pi - {3\sqrt{3}}}{4}$

Get rid of the denominators, and also get rid of a $\displaystyle -3\sqrt{3}$ on each side. This gives:

$\displaystyle l^2 \sqrt{3} = 4 \pi$.

Solve for l - I got 2.70 rounded to the nearest hundredth.

Hopefully this makes sense. Start with a clear picture.
• Mar 4th 2008, 10:07 PM
earboth
Quote:

Originally Posted by Mathnasium
I'll take a crack at it - it'd be much easier with a picture, but oh well.

To my great surprise I've got the same result even though I misunderstood the question completely.

But I can provide you with a drawing.
• Aug 29th 2008, 02:14 PM
alesianlarken
Area of the triangle
That was tricky question,i am not that much good in mathematics,even i am facing lot problems with quantitative aptitude.Same question i got earlier in my competitive exam where i failed to answer that one.If any one know the answer please send it to us.
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Alesianlarken

WideCircles
• Aug 29th 2008, 03:09 PM
Soroban
Hello, help1!

Did you make a sketch?

Quote:

An equilateral triangle and a circle have the same center.
The area of the triangle not in the circle equals the area of the circle not in the triangle.
If the radius of the circle is 1, find the length of the side of the traingle (nearest hundredth).

Code:

                *               /:\               /:::\             /:::::\             / * * * \           */          *         *:/          \:*       *:/            \:*       :/              \:       */                \*       *        *        *     /*                  *\     /:.                  .:\   /:::*                *:::\   *-----*---------------*-----*           *:::::::::::*               * * *

"The area of the triangle not in the circle" is the sum of the three "triangular" regions.

"The area of the circle not in the triangle" is the sum of the three segments.

These two areas can be expressed, using the inscribed regular hexagon
. . (which I did not try to draw) and a variety of area formulas.

Good luck!

Too slow again . . . Great diagram, Earboth!
.
• Aug 29th 2008, 03:19 PM
Jhevon
Quote:

Originally Posted by Soroban
Hello, help1!

Did you make a sketch?

(well, earboth made a diagram :p)
Quote:

Code:

                *               /:\               /:::\             /:::::\             / * * * \           */          *         *:/          \:*       *:/            \:*       :/              \:       */                \*       *        *        *     /*                  *\     /:.                  .:\   /:::*                *:::\   *-----*---------------*-----*           *:::::::::::*               * * *

"The area of the triangle not in the circle" is the sum of the three "triangular" regions.

"The area of the circle not in the triangle" is the sum of the three segments.

These two areas can be expressed, using the inscribed regular hexagon
. . (which I did not try to draw) and a variety of area formulas.

Good luck!

did you take a course in making these kinds of "sketches"?!
• Aug 29th 2008, 11:49 PM
Moo
Quote:

Too slow again . . . Great diagram, Earboth!
That was 6 months ago ! (Rofl)