Thread: Equilateral Triangle & Circle

A wire 10 meters long is to be cut into two pieces. One piece will be shaped as an equilateral triangle, and the other piece will be shaped like a circle.

(a) Express the total area A enclosed by the pieces of wire as a function of the length x of a side of the equilateral triangle.

(b) What is the domain of A?

Originally Posted by symmetry

A wire 10 meters long is to be cut into two pieces. One piece will be shaped as an equilateral triangle, and the other piece will be shaped like a circle.

(a) Express the total area A enclosed by the pieces of wire as a function of the length x of a side of the equilateral triangle.

(b) What is the domain of A?

We know in an equilateral triangle all the sides are equal; further we know the area of a triangle is (1/2)*b*h.

Similarly, we know the area of a circle is Pi*r^2 and the circumference is 2*Pi*r.

Consider the equilateral triangle. Since we have x on each side, we can find the length of the base by drawing a perpendicular from the vertex opposite the base down to the base. Two right triangles are formed, and using the pythag. theorem we are able to determine the base is equal to:

(sqrt(3)/2)*x = h, where h is the height.

And thus, the area of the triangle is:

(1/2)*x*[(sqrt(3)/2)*x]

Now, solve A = Pi*r^2. But, we know:

10 - 3x = 2*Pi*r

(Since we have three equal sides in the equilateral triangle);

Solve for r;

r = (-3x + 10)/(2*Pi)

And thus the area of the circle is:

A = Pi*((-3x + 10)/(2*Pi))^2

= Pi*[(3x - 10)^2/(4*Pi^2)]

= [(3x - 10)^2/(4*Pi)]

The total area is:

[(3x - 10)^2/(4*Pi)] + (1/2)*x*[(sqrt(3)/2)*x]

= [(3x - 10)^2/(4*Pi)] + (x^2*sqrt(3))/4

Now we can find the domain;

The easiest thing to do now is to graph it. You will get a parabola and it's clear that the domain is (-inf, inf).

Thanks for the math notes. I will review this stuff a bid more carefully.