Pythagorean Thm: A piece of wire is cut into two pieces....

[Sophie]

It is definatly a day for silly mistakes today here is my next involving Pythag. I just need help with the start, I have a worked example, but it does not show how it got the height of the triangle, I assume using pythag and I just can not figure it out.

Question: A piece of wire 10m long is cut into two pieces. One peice is bent into a sqaure and the other is bent into an equiliateral triangle. How should the wire be cut so that the total area enclosed is a. maximum b. minimum

Here is how I started:

The length of 10m wire is split into x and (10-x)

Lenth of 1 side of of the sqaure x/4

length of 1 side of equaliateral triangle (10-x)/3

Now my problem. I can not figure out the height of the triangle the following is what I tryed:
I put a line down the middle of the equaliateral creating 2 right angled triangles to work out the height and then used pythag

Length of Hypotemuse ((10-x)/3))

Length of base of new right angled triangle ((10-x)/3))/2 = (10-x)/6

Length of height = a

((10-x)/3)^2) = (((10-x)/6)^2) + (a^2)

a = sqrt( ((10-x)/3)^2) - (((10-x)/6)^2) )

a = ((10-x)/3) - ((10-x)/6)

which is wrong

The answer book says the height should be ((Sqrt(3))/2) ((10-x)/3)
I have tryed numeriouse ways to get the but can not!!!

I would be very very greatful if someone could show me the error of my ways

Thanks Sophie

 

[galactus]

Hey Sophie.

I don't think you have to use the height of the triangle.

The area of an equilateral triangle is \(\displaystyle \L\\\frac{\sqrt{3}}{4}r^{2}\), where r is the length of a side.


Let's make a diagram like so:

 *******************************************************
|<-------  x  -------->|<----------- y  ---------------->|
              |
              |                                  |
              |                                  |
             \/                                 \/
            /\                             ----------  
          /   \   r                       |          |  
        /       \                         |          |  s
        --------                          -----------

Since x is the length of the portion of the wire that makes the perimter of the triangle, then r=x/3.
Since y is the length of the wire that makes the perimeter of the square, then s=y/4.

\(\displaystyle \L\\A=\frac{\sqrt{3}}{4}r^{2}+s^{2}\)

Therefore, \(\displaystyle A=\frac{\sqrt{3}}{4}(\frac{x}{3})^{2}+(\frac{y}{4})^{2}\)

But x+y=10, so y=10-x

\(\displaystyle \L\\A=\frac{\sqrt{3}}{4}\cdot\frac{x^{2}}{9}+\frac{(10-x)^{2}}{16}=
\frac{\sqrt{3}}{36}x^{2}+\frac{x^{2}}{16}-\frac{5}{4}x+\frac{25}{4}\)

\(\displaystyle \L\\A'(x)=\frac{4\sqrt{3}+9}{72}x-\frac{5}{4}\)

Finish now?.

 

[Sophie]

Thanks Galactus

I can not belive I have got this far in math and have not seen the area of an equalateral triangle. Oh well it all makes sence now.

Thanks Sophie