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I am being asked to calculate the area of a triangle with a side-length of 15.5 inches.

The formula for calculating a regular polygon's area is 1/2Pa

Where P is the perimeter of the polygon, and a is the apothem. I am completely lost.

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Well, the area of a triangle is 1/2bh, if I remember correctly. – Dustin L. Jul 9 '12 at 4:06
up vote 1 down vote accepted

enter image description hereHint: If you cut the equilateral triangle into two by bisecting an angle, you make two right triangles. Can you identify the base and height?

Added: Look at the right triangles and try to find x

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I assume they're both 15.5, right? – Dustin L. Jul 9 '12 at 4:07
No, the bisector is shorter than that. I suggest you draw a picture. – Ross Millikan Jul 9 '12 at 4:09
Okay but then how would I find the length of the bisector? – Dustin L. Jul 9 '12 at 4:12
Thanks so much! – Dustin L. Jul 9 '12 at 4:22

It is best not to use a general formula that one has not great control over, when the problem is a very concrete one.

However, one can compute the apothem. Recall the apothem is the (length of) a perpendicular from the centre of the equilateral triangle to any one of the sides. So let our equilateral triangle be $ABC$, and let $P$ be the centre. Draw the line through $P$ which is perpendicular to $AB$. Suppose it meets $AB$ at $Q$. The apothem is the length of $PQ$.

Since $\angle CAB=60^\circ$, by symmetry we have $\angle PAQ=30^\circ$. Note that $AQ=15.5/2$. So if $x$ is the apothem, then $$\frac{x}{(15.5/2)}=\tan(30^\circ)=\frac{1}{\sqrt{3}}.$$ It follows that $$x=\frac{15.5}{2}\cdot \frac{1}{\sqrt{3}}.$$

If you do not know the "special angles" fact that $\tan(30^\circ)=\frac{1}{\sqrt{3}}$, you can use the calculator to evaluate $\tan(30^\circ)$ to high precision.

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