Question

using the Pythagorean thereom, find the are of an equilateral triangle whose sides measure 5 units. Find the are to the nearest tenth of a square unit.

Answer 1

An equilateral triangle isn't a right triangle, so the Pythagorean Theorem doesn't apply directly.  

But if we divide an equilateral triangle with side s in half along an altitude, we get two congruent right triangles each with hypotenuse s and one leg s/2.

The other leg is the altitude h we divided along.  By the Pythagorean Theorem

$( \frac s 2)^2 + h^2 = s^2$

$\frac{s^2}{4} + h^2 = s^2$

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

$h = \dfrac{\sqrt{3}}{2} s$

The area A of our original equilateral triangle is now

$A= \frac 1 2 sh = \frac 1 2 s \left( \dfrac{ \sqrt{3}}{2} s \right) = \dfrac{\sqrt{3}}{4} s^2$

So the area of our equilateral triangle of side 5 is

$A = \dfrac{\sqrt{3}}{4} (5)^2 = \dfrac{25}{4} \sqrt{3} \approx 10.8$

Answer: 10.8

I don't like ruining a nice exact answer with an approximation, but the question asks for one.

We could have done the above steps with 5 in particular instead of s but I prefer to do the general solution and substitute at the end.

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Since you read this far, I will give you the secret formula for the area S of a triangle given squared sides A, B, C.  Your teachers won't even know this one.

$16S^2 = 4AB-(A+B-C)^2$

Let's try it out for $A=B=C=s^2,$ an equilateral triangle:

$16S^2 = 4A^2 -A^2 = 3A^2$

$S = \sqrt{\dfrac{3}{16}} A = \dfrac{\sqrt{3}}{4} s^2 \qquad\checkmark$