 Posers and Puzzles

1. 05 Jun '09 00:34
I just got to a question on my homework tonight that I can't make heads or tails of. Its probably quite simple, but I'm not getting it. Here it is:

What is the area of the space trapped between a circle radius 10 and a equalateral triangle inscribed inside it?

Obviously, the circle has an area of 100pi, but how do you find a side lenght of the triangle? could somebody help me out?
2. 05 Jun '09 03:01
Originally posted by clandarkfire
I just got to a question on my homework tonight that I can't make heads or tails of. Its probably quite simple, but I'm not getting it. Here it is:

What is the area of the space trapped between a circle radius 10 and a equalateral triangle inscribed inside it?

Obviously, the circle has an area of 100pi, but how do you find a side lenght of the triangle? could somebody help me out?
I solved it in a very un-elegant way...It hinges on you understanding triginometry, and Cartesian graphing...so it may or may not be the method your looking for...

Ok

On a cartesian graph start with the equation of a circle of raduis 10 which is defined by the following equation:

x^2 + y^2 = 10^2

now place the vertex of the equallateral triangle at an arbitrary point "p" on the circumference of the circle. (chosing the point on one of the axises simplifies the math but is not necessary

I chose the point (0,10) on the y-axis

Now, the triangle in question has 3 interior angles of 60 degrees each.

next find the equation of the line that passes through point "p" at a 30 degree angle refrenced from the y-axis.

here is where some trig comes in handy

give the sides of your triangle the variable lenght "L"

slope of the line in question = (y-y1)/(x-x1)

slope = (L*cos 30 deg.)/(L*sin30deg.) = (sqrt(3)/2) / (1/2) = sqrt(3)

the equation of my line is

y = -sqrt(3)*x + 10

we are trying to find the the side lenght "L", so first we need to find the points of intersection of the circle equation and the line equation just developed.

so solve the system of equations

(1) x^2 + y^2 = 10^2

(2) y = -sqrt(3)*x +10

and i get

x= 5*sqrt(3) & y = -5

from the distance formula

d(0,10),(5*sqrt(3) , -5) = SQRT [(5*sqrt(3) - 0)^2 + (-5-10)^2] = 10*sqrt(3) = L

now that we have "L"

we calculate the area to be

100*pi - (1/2)*(10*sqrt(3))*(10*sqrt(3)*(cos 30 deg.)

A = 100*pi - 75*sqrt(3)

approx = 184.255 units^2

like i said, not elegant, but a solution.

hope this helps, but if not, im sure others will simplify the solution...

😉
3. 05 Jun '09 03:20
Originally posted by joe shmo
I solved it in a very un-elegant way...It hinges on you understanding triginometry, and Cartesian graphing...so it may or may not be the method your looking for...

Ok

On a cartesian graph start with the equation of a circle of raduis 10 which is defined by the following equation:

x^2 + y^2 = 10^2

now place the vertex of the equallateral triangle ...[text shortened]... .

hope this helps, but if not, im sure others will simplify the solution...

😉
Thank you, that was very helpful.

I get that for the most part, I was just lost at getting an equation for the line that passes through point p at a 30 degree angle. Thanks
4. 05 Jun '09 03:221 edit
Originally posted by clandarkfire
Thank you, that was very helpful.

I get that for the most part, I was just lost at getting an equation for the line that passes through point p at a 30 degree angle. Thanks
What is the class that this is for?....you see, Ive never had geometry, and I have a suspicion that is where the simple solution lies...

and if you are still having questions about the methods I used, i will do my best to explain...
5. 05 Jun '09 09:471 edit
Originally posted by clandarkfire
I just got to a question on my homework tonight that I can't make heads or tails of. Its probably quite simple, but I'm not getting it. Here it is:

What is the area of the space trapped between a circle radius 10 and a equalateral triangle inscribed inside it?

Obviously, the circle has an area of 100pi, but how do you find a side lenght of the triangle? could somebody help me out?
Draw lines from the vertex of the triangle to the center of the circle. You're left with 3 isosceles triangles, each with angles 30,30,120.

Use the law of sines to calculate the length of the base of each triangle:

base = sin(120)/sin(30)*10 = 10*sqrt(3)

height is then: height=sqrt(10^2-1/4*base^2) = sqrt(100-300/4) = 5

the area of the triangle is then 10*sqrt(3)*(10+5)/2 = 75*sqrt(3)

and so on.
6. 05 Jun '09 16:584 edits
Originally posted by Palynka
Draw lines from the vertex of the triangle to the center of the circle. You're left with 3 isosceles triangles, each with angles 30,30,120.

Use the law of sines to calculate the length of the base of each triangle:

base = sin(120)/sin(30)*10 = 10*sqrt(3)

height is then: height=sqrt(10^2-1/4*base^2) = sqrt(100-300/4) = 5

the area of the triangle is then 10*sqrt(3)*(10+5)/2 = 75*sqrt(3)

and so on.
another cute way:

call equilateral triangle ABC, oriented with point A at "12 o'clock" and points B and C at "4 o'clock" and "8 o'clock," respectively. now draw diameter AD, which is the perpendicular bisector of BC through its midpoint (we will call it point E). *edit* [this may not be entirely obvious, and i will leave it to the reader to prove if you do not know this already... or if you need a proof just ask! 🙂]

now, to make the algebra of the problem a little simpler, let BE = x. Then, clearly, EC = x. By the 30-60-90 triangles you've created, AE = x*sqrt(3). And since the diameter of the circle is 20, DE = 20 - [x*sqrt(3)].

this is useful because one fact about circles is that any two chords that intersect in a circle have a special property concerning the products of the lengths of the segments created by their intersection. that is, if any two chords AD and BC intersect at point E, it necessarily follows that: AE*ED = BE*EC.

so, in this example: [x*sqrt(3)] * [20 - x*sqrt(3)] = [x] * [x].

distributing, reorganizing, and factoring, we get x = 5*sqrt(3) (i'll leave the algebra to you), and so the base of the triangle = 2x = 10sqrt(3), and the height is x*sqrt(3) = 15. so A = 1/2*bh = 1/2*[5*sqrt(3)]* = 75*sqrt(3).

nice problem! (and p.s. i really like Palynka's law of sines solution too! always good to look for hidden radii and diameters that you can use in a circle diagram.) 🙂
7. 06 Jun '09 12:04
Here's another alternative:
Draw a regular hexagon in your circle, such that three of the vertices of the hexagon are vertices of your equilateral triangle.
The hexagon has double the area of the triangle. [Proof left to the reader ...]
A regular hexagon can be split into 6 identical equilateral triangles. The side-length of these triangles is the radius of the circle.
As long as you know how to find the area of an equilateral triangle (eg using (a*b*sinC)/2 then this does the job.

Dn
8. 06 Jun '09 20:18
Going at this problem with only geometry, your triange has three interior angles at 60 deg each (adding interior angles of any triangle equals 180 degrees and your triangle has three equal sides, so the angle must be 180/3 or 60 deg.).

Bisecting each angle with lines from each angle to the opposite base creates 6 identical triangles with measures of 30,60 and 90 degrees on each angle and the hypotenuse of each right triangle is 10 units (your original radius value).

Since the sum of the squares of the two shorter legs of a right traingle equal the square of the hypotenuse, what two numbers squared equal 100. The answer is 6 and 8 (36 plus 64 equal 100).

You now know the measure of each leg of your six triangles and the rest just falls into place.

That was fun.
9. 06 Jun '09 23:54
Originally posted by olegson
Bisecting each angle with lines from each angle to the opposite base creates 6 identical triangles with measures of 30,60 and 90 degrees on each angle and the hypotenuse of each right triangle is 10 units (your original radius value).
Since the sum of the squares of the two shorter legs of a right traingle equal the square of the hypotenuse, what two numbers squared equal 100. The answer is 6 and 8 (36 plus 64 equal 100).
6 and 8 are the only integer values, but there is an infinite set of other possibilities - eg 5 and sqrt(75).

To show easily why 6 and 8 are wrong, consider the perpendicular height of the original triangle. This is 10 (the radius) added to 6 - a total of 16. The side-length of the original triangle is 8 + 8, which is also 16. The diagonal side-length must be greater than the vertical height.