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

Thanks,

Ron

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- Dec 15th 2009, 12:49 AMrn5aPerimeter 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 - Dec 15th 2009, 01:05 AMdedust
area of triangel using sine rule : $\displaystyle \frac{1}{2}ab \sin C$.

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

now you have what you need to find the perimeter - Dec 15th 2009, 08:08 PMrn5a
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

- Dec 16th 2009, 12:01 AMdedust
Let $\displaystyle a = $ side of triangel,

then $\displaystyle base = a $

and $\displaystyle height = a \sin \frac{\pi}{3}$

$\displaystyle Area = \frac{1}{2} \times base \times height = \frac{1}{2} \times a \times a\sin \frac{\pi}{3}$ - Dec 16th 2009, 01:59 AMHallsofIvy
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, $\displaystyle x^2+ s^2/4= s^2$ so $\displaystyle x^2= 3s^2/4$ and $\displaystyle x= s\sqrt{3}/2$.

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

Then, of course, the perimeter is 3s. - Dec 16th 2009, 03:46 AMSoroban

Hey, did anyone READ the problem?

It's an old joke . . .

Quote:

How do you find theof an equilateral triangle whose*area*is 4√3 sq.cm?*area*

- Dec 16th 2009, 01:01 PMjgv115
but he's asking how to find the area.

So the answer is

$\displaystyle \frac{b*h}{2} $ - Dec 16th 2009, 01:14 PMBacterius
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 - Dec 17th 2009, 06:52 AMCaptainBlack
The most direct way of doing this is to use Heron's formula for the area of a triangle:

$\displaystyle

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

$

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

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

So:

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

CB