Objective Functions

**tiar** · September 28th 2008, 01:58 PM

Express the area of an equilateral triangle as a function of its height h.

Stuck on this quite a bit..I'm terrible at word problems.
A=h ?
h=x

I'm confused...I need to plug "H" into the area..

A=1/2bh...but how do I get "H"? Or should I solve for the base and then plug it in to get my objective function?

**Prove It** · September 28th 2008, 03:51 PM

Originally Posted by tiar

Express the area of an equilateral triangle as a function of its height h.

Stuck on this quite a bit..I'm terrible at word problems.
A=h ?
h=x

I'm confused...I need to plug "H" into the area..

A=1/2bh...but how do I get "H"? Or should I solve for the base and then plug it in to get my objective function?

It's hard to solve without drawing a diagram - so as I explain this to you, draw it.

Say you have an equilateral triangle of side length $l$ . Clearly all the angles are $60^{\circ}$ .

If you're expressing the Area in terms of his height, then we need to find an expression for the height. So draw the height, remembering that the height is perpendicular to the base, and will cut one of the angles in half.

So in other words, an equilateral triangle is the same as two right-angled 30, 60, 90 degree triangles put next to each other.

Just looking at one of the right angled triangles, we know that the hypotenuse must be of length $l$ and one of the side lengths must be $\frac{1}{2}l$ . We can use pythagoras to find the height.

$a^2 + b^2 = c^2$

$\left (\frac{1}{2}l \right)^2 + h^2 = l^2$

$\frac{1}{4}l^2 + h^2 = l^2$

$h^2 = \frac{3}{4}l^2$

$h = \frac{\sqrt{3}}{2}l$ .

Rearranging for $l$ we get $l = \frac{2 \sqrt{3}}{3} h$ .

So now we can say...

$A = \frac{1}{2}bh$ .

Since $b = l = \frac{2\sqrt{3}}{3}h$ we get...

$A = \frac{1}{2} \times \frac{2\sqrt{3}}{3}h \times h$

and so...

$A = \frac{\sqrt{3}}{3}h^2$ .

**tiar** · September 28th 2008, 03:58 PM

Originally Posted by Prove It

It's hard to solve without drawing a diagram - so as I explain this to you, draw it.

Say you have an equilateral triangle of side length $l$ . Clearly all the angles are $60^{\circ}$ .

If you're expressing the Area in terms of his height, then we need to find an expression for the height. So draw the height, remembering that the height is perpendicular to the base, and will cut one of the angles in half.

So in other words, an equilateral triangle is the same as two right-angled 30, 60, 90 degree triangles put next to each other.

Just looking at one of the right angled triangles, we know that the hypotenuse must be of length $l$ and one of the side lengths must be $\frac{1}{2}l$ . We can use pythagoras to find the height.

$a^2 + b^2 = c^2$

$\left (\frac{1}{2}l \right)^2 + h^2 = l^2$

$\frac{1}{4}l^2 + h^2 = l^2$

$h^2 = \frac{3}{4}l^2$

$h = \frac{\sqrt{3}}{2}l$ .

Hi, I am wondering how you went from...
$\frac{1}{4}l^2 + h^2 = l^2$
to...
$h^2 = \frac{3}{4}l^2$

What am I not thinking of...I do not get the $\frac{3}{4}$ ...
Thank you for your response, I greatly appreciate it!

**Plato** · September 28th 2008, 04:19 PM

Tiar
The post by Prove_It is correct.
Go with it.

**tiar** · September 28th 2008, 04:23 PM

Originally Posted by Plato

Tiar
The post by Prove_It is correct.
Go with it.

I know it is..I am merely wondering how he got from the 1/4 to 3/4. I haven't done math in years...

**11rdc11** · September 28th 2008, 04:42 PM

$h^2 = \frac{4l^2}{4} - \frac{1l^2}{4}$

which is

$h^2 = l^2\bigg(\frac{4}{4} - \frac{1}{4}\bigg)$

$h^2 = \frac{3}{4}l^2$

**tiar** · September 28th 2008, 05:25 PM

Originally Posted by 11rdc11

$h^2 = \frac{4l^2}{4} - \frac{1l^2}{4}$

which is

$h^2 = l^2\bigg(\frac{4}{4} - \frac{1}{4}\bigg)$

$h^2 = \frac{3}{4}l^2$

Oh! Thank you! For some reason I thought I was multiplying it. Goes to show how I am too tired to think...
This makes perfect sense now! Thank you!

**Prove It** · September 29th 2008, 02:53 AM

Originally Posted by Plato

Tiar
The post by Prove_It is correct.
Go with it.

Now now Plato, no need to be like that. Everyone has lapses of memory now and then. It's always the easy stuff too that gets forgotten the most easily.

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