Thread: Pythagorean Theorem - College Algebra

An equilateral triangle is shown below.



a.) Find an expression for its height h in terms of a.


Ok, I have the solution, but I don't understand it. This is how the solutions manual shows it is solved:

step #1: h^2 + (a/2)^2 = a^2
a^2 + b^2 = c^2 Pathagorean formula, ok I understand this.

step #2: h^2 + [(a^2) / 4] = a^2
a divided by 2 squared equals a squared over 4. Got it.

step #3: h^2 = (3a^2) / 4
Ok this is where I'm lost. Where did the 3 times a squared come from?

step #4: h = [(3a^2) /4]^(1/2)
I understand this except I still don't know where the 3a^2 came from. h equals the square root of, 3 times a squared divided by 4.

step #5: h = (a/2)(3^(1/2))
I understand this as well. a divided by 2, times the square root of 3.

For step 3, they divided the $\displaystyle a^2$ on the right of the equation by 4 to get a common denominator with the other $\displaystyle a^2$.
They then moved the $\displaystyle \frac{a^2}{4}$ to the side with $\displaystyle \frac{4a^2}{4}$.
4a^2 - 1a^2= 3a^2

Thanks a lot; this makes perfect sense now.