This has to do with a formula for the perimeter of a triangle, but I am not having any luck finding it anywhere. Maybe there are some math geniuses here that love a challenge.

Here is the question:

Given that an equilateral triangle is inscribed in a circle with a radius of 30 inches, what is the perimeter of the triangle?

This is an old math problem that I never found a formula to a while back. Still I just have to know how to get to the solution of 156 inches.

Does anyone know how to solve this?

 

http://mathworld.wolfram.com/EquilateralTriangle.html

rather than explain it, this will be better because you need the picture for reference.

 

Best I could figure the formula is...

a = r / ((square root of 3) / 6) where...

a = length of 1 side of the triangle

r = radius of inscribed circle

giving us...

a = 30 / (1.732050808 / 6) = 103.923

Perimeter = 3a = 311.769

You can check my math though...

 

Draw lines from the centre of the circle to each corner of the triangle. You now have three identical isocoles trianlges (two sides equal in each). The three peaks at the centre make 360-degrees, therefore you know that the angle of each apex at the centre is 360/3 = 120-degrees. But since the sum of all three angles of a triangle must be 180-deg, and the other two are equal to each other, that means that the other two corners of each trinagle is (180 - 120)/2 = 30-deg.

Now draw a line from the circle centre to cross one of triangle sides perpendicularly. Because triangle you're cutting in half is an isoloces triangle, you know that this perpendicular is cutting the line it crosses exactly in half.

From the previous, you also know that each of the two smaller triangles have angles 60-deg, 90-deg and 30-deg - a right-angled triangle and you know the hypotenuse's legnth is R, the radius of the circle.

If L is the length of one-half of the original triangle's sides - that is, the side that you cut in half with your perpendicular, and theta is the angle between L and R, then from grade 12 trig you know that

L/R = cosine(theta)

So, rearranging

L = R * cosine(theta)

but theta is ALWAYS 60% no matter what the size of the circle,

AND it just so happens that the cosine of (30-deg) is 0.866

And since L is one-sixth of the perimetre, P, then :

P = 6 * L = 6 * (R * cosine(60-deg)) = 6 * (R * .866) = 5.196R.

No matter what size your circle is, the perimitre of the equilateral triangle inscribed in side it will be 5.196 x the radius.

 

Oops. I gave the answer for a circle inscribed in the triangle. Do I at least get partial credit for showing my work?

 

OK, I think I have enough information to do it now. I think the isosceles triangle is the key to this formula. I'll have to work these numbers awhile and get an answer to compare to the 156 inches given.

Oh, I almost forgot. The inscribed circle formula will come in handy later, so thanks for that and I'll give credit for it.

 

1) Draw it with a pencil on a sheet of cardboard.
2) Use the cardboard as a template on a piece of sheet metal.
3) Carve it out with an air saw. *Make 2, they small
4) Use a wire-feed welder to tack them into the exhaust tips of one of the buddies you work with that drives a Chevolet
5) Watch that sucker whine and accuse the day after...

*Practical applications have always been a specialty of mine
**Yes, I am vindictive at times... The bigger the mouth, the more likely justice is a-comin'

 

Greywolf, I'm tempted to do that very thing, but I'm not able to keep a straight face when I lie. I wouldn't make a good politician.

Ok, I figured this thing out. I had devide the equilateral into three isoscelese triangles and then one of them into a right triangle of the 30 60 90 type. Then used sine of the 30 degree angle times the length of the known side of 30 inches which turns out to be 15. Now use Pythagorean's theorem.

I used it in the form a^2 = c^2 - b^2. This turns out to be 25.98076211. Since there are two of these right triangles that makes up one side of the equilateral, you have to add this twice then multiply the sum by three to get the perimeter of 155.8845727. This can be rounded off to 155.8846 or even 156 since the original given lengths have no decimals.

Thanks for all of the replies!

 

Ouch, all this math makes my head hurt. Its no wonder I never took trig. Funny thing is that even though I never took trig, I tested right into Applied Calculus in college. hmm