10 Mar '09 03:55

I have a little trouble with my homework tonight - any ideas?

The first question is as follows: There are two circles, each with a radius of (x-5), and an area of pi(4x+1). How much more area does the larger circle have than the smaller one?

Ok, heres what I got:

you can just ignore the pi, since area is radius squared times pi. So:

(x-5)^2 = 4x+1

x^2-10x+25 = 4x+1

x^2-10x+24 = 4x

x^2-10x = 4x-24

x^2-14x + 0 = -24

x^2-14x + 24 = 0, and from there you can solve it like any quadratic equation.

Since I cant use special characters, its hard to show my work, but when you solve this quadratic equation, I come up with the two answers, 12 and 2. These are the two possiblilities for x. They work for the equations, so the area of the larger circle is 49pi. With the smaller circle, the area works out to be 9pi, making the final answer 40pi. However, with a radius of x-5, the radius would be -3, which is obviously not possible. Did I mess up somewhere, or is it just not possible?

Anyway, here is my other problem:

Inside an equilateral triangle are three circles,

each tangent to two sides of the triangle and tangent to

each other. The radius of each circle is 1 inch. What is

the area of the triangle?

I have not Idea how to go about solving that one. Help please!

The first question is as follows: There are two circles, each with a radius of (x-5), and an area of pi(4x+1). How much more area does the larger circle have than the smaller one?

Ok, heres what I got:

you can just ignore the pi, since area is radius squared times pi. So:

(x-5)^2 = 4x+1

x^2-10x+25 = 4x+1

x^2-10x+24 = 4x

x^2-10x = 4x-24

x^2-14x + 0 = -24

x^2-14x + 24 = 0, and from there you can solve it like any quadratic equation.

Since I cant use special characters, its hard to show my work, but when you solve this quadratic equation, I come up with the two answers, 12 and 2. These are the two possiblilities for x. They work for the equations, so the area of the larger circle is 49pi. With the smaller circle, the area works out to be 9pi, making the final answer 40pi. However, with a radius of x-5, the radius would be -3, which is obviously not possible. Did I mess up somewhere, or is it just not possible?

Anyway, here is my other problem:

Inside an equilateral triangle are three circles,

each tangent to two sides of the triangle and tangent to

each other. The radius of each circle is 1 inch. What is

the area of the triangle?

I have not Idea how to go about solving that one. Help please!