# Perimeter of equilateral triangle

• December 15th 2009, 01:49 AM
rn5a
Perimeter of equilateral triangle
How do you find the area of an equilateral triangle whose area is 4√3 sq.cm? Please show the steps.

Thanks,

Ron
• December 15th 2009, 02:05 AM
dedust
area of triangel using sine rule : $\frac{1}{2}ab \sin C$.

for your case (equilateral triangle), $a=b$ and $C=\frac{\pi}{3}$

now you have what you need to find the perimeter
• December 15th 2009, 09:08 PM
rn5a
Actually I haven't been taught sin & all those things....so I couldn't understand what you have suggested. Can you please show me how to do it using 1/2 basexheight formula?

Thanks,

Ron
• December 16th 2009, 01:01 AM
dedust
Let $a =$ side of triangel,
then $base = a$
and $height = a \sin \frac{\pi}{3}$
$Area = \frac{1}{2} \times base \times height = \frac{1}{2} \times a \times a\sin \frac{\pi}{3}$
• December 16th 2009, 02:59 AM
HallsofIvy
Let "s" be the length of one side of the equilateral triangle. Drop a line from one vertex to the middle of the opposite side. It can be shown that this line is also perpendicular to the opposite side, dividing the equilateral triangle into two right triangles that have hypotenuse of length s and one leg of length s/2.

Taking "x" to be the length of the altitude to the equilateral triangle, which is the other leg of the right triangles, by the Pythagorean theorem, $x^2+ s^2/4= s^2$ so $x^2= 3s^2/4$ and $x= s\sqrt{3}/2$.

Now, the area of the equilateral triangle is "1/2 base times height" or $\frac{1}{2}\left(s\right)\left(\frac{s\sqrt{3}}{2} \right)= \frac{\sqrt{3}}{4}s^2$. Set that equal to $4\sqrt{3}$ and solve for s.

Then, of course, the perimeter is 3s.
• December 16th 2009, 04:46 AM
Soroban

Hey, did anyone READ the problem?

It's an old joke . . .

Quote:

How do you find the area of an equilateral triangle whose area is 4√3 sq.cm?
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . $\begin{array}{c} {\color{red}\uparrow} \\ {\color{red}\text{Here it is!}} \end{array}$

• December 16th 2009, 02:01 PM
jgv115
but he's asking how to find the area.

$\frac{b*h}{2}$
• December 16th 2009, 02:14 PM
Bacterius
He gave the area. I suspect he meant the perimeter ..
Soroban, I was thinking like you, was this some sort of evil plot ? Why didn't anybody notice the mistake in the few first posts ? :D
• December 17th 2009, 07:52 AM
CaptainBlack
Quote:

Originally Posted by rn5a
How do you find the perimiter of an equilateral triangle whose area is 4√3 sq.cm? Please show the steps.

Thanks,

Ron

The most direct way of doing this is to use Heron's formula for the area of a triangle:

$
A=\sqrt{s(s-a)(s-b)(s-c)}
$

where $s$ is the semi-perimiter and $a$, $b$ and $c$ are the sides.

For an equilateral triangle we have $a=b=c$ and $s=3a/2$

So:

$A^2=48=s^4 / 3^3$

CB