Area of an equilateral triangle

[hospitalteacher]

The problem states that I must find the area of an equilateral triangle with side 5. I k now that it is a regular polygon and the formula is Area is = 1/2ap. How do I find the apothem in order to solve the problem?

 

[Deleted member 4993]

The problem states that I must find the area of an equilateral triangle with side 5. I k now that it is a regular polygon and the formula is Area is = 1/2ap. How do I find the apothem in order to solve the problem?

Draw a picture - draw the apothem. It will be the perpendicular bisector of the base.

Now use Pythagorean theorem to calculate the length of the apothem (height).

 

[masters]

The problem states that I must find the area of an equilateral triangle with side 5. I k now that it is a regular polygon and the formula is Area is = 1/2ap. How do I find the apothem in order to solve the problem?

You don't really need to find the apothem to determine the area.
If you drop a perpendicular bisector from any vertex, you can determine the height of the triangle by the Pythagorean Theorem.

$\displaystyle 5^2=2.5^2+h^2$

Then, all you need to do is apply $\displaystyle A = \frac{1}{2}bh$, where $\displaystyle b = 5$ and $\displaystyle h=2.5\sqrt{3}$

$\displaystyle A=\frac{1}{2}(5)(2.5\sqrt{3})$

$\displaystyle A \approx 10.8$

However, if you have to use the apothem, then you must know that the apothem is the segment drawn from the center of the equilateral triangle perpendicular to a side and thus bisecting the side.

You can then use the 30-60-90 rule to determine that the apothem is $\displaystyle \frac{2.5\sqrt{3}}{3}$

Now, use the area formula which includes the apothem and perimeter.

$\displaystyle A=\frac{1}{2}ap$

$\displaystyle A=\frac{1}{2}\left(\frac{2.5\sqrt{3}}{3}\right)(15)$

$\displaystyle A \approx 10.8$

 

[BigGlenntheHeavy]

$\displaystyle Or \ better \ yet, \ just \ look \ into \ a \ book \ on \ formulas \ for \ geometry \ and \ you'll$

$\displaystyle find \ A_{Equil. \ triangle} \ = \ \frac{\sqrt{3}s^2}{4} \ and \ plug \ in \ 5 \ for \ s.$