# Equilateral Triangles..Finding Sides..Areas...

• Mar 22nd 2008, 08:21 AM
zachcumer
Equilateral Triangles..Finding Sides..Areas...
Hello. Here is the question I have to solve. Its attached. Now, what I came up with was the tick marks are each 6 (whatever unit) (doesn't specify). However I am very confused, on how to solve this.

Thanks
• Mar 22nd 2008, 08:34 AM
Nyoxis
Hope this helps.
• Mar 22nd 2008, 08:42 AM
zachcumer
I thought I was supposed to use the pythagorean therorum. What did you do?
• Mar 22nd 2008, 09:06 AM
Nyoxis
Quote:

Originally Posted by zachcumer
I thought I was supposed to use the pythagorean therorum. What did you do?

The tick marks indicate they are of the same length, so each side is 12.
Because each side is split in the middle we know that each $\displaystyle \frac{1}{2}$ is 6 units.
We also know that the 3 corners of the triangle each have $\displaystyle 60^0$ because all side lengths are the same, meaning it's equilateral.
So for our seperate triangles we have $\displaystyle 30^0$ angles, right angles and 6 unit lengths, which is enough for trigonometry.

Though if you're looking for a solution using Pythagorean theorem then I'll take another look.
• Mar 22nd 2008, 09:09 AM
zachcumer
cool. thanks. any help I could get would be great.
• Mar 22nd 2008, 10:38 AM
Nyoxis
Bah I keep disproving myself, I got x=2.1962...
• Mar 22nd 2008, 11:02 AM
janvdl
Quote:

Originally Posted by zachcumer
Hello. Here is the question I have to solve. Its attached. Now, what I came up with was the tick marks are each 6 (whatever unit) (doesn't specify). However I am very confused, on how to solve this.

Thanks

(See photo2.jpg)

Solve for L using pythagoras. I find $\displaystyle L = 6 \sqrt{3}$

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(See photo3.jpg)

That red line is half the length of the green line.
Therefore the red line has length: $\displaystyle 2 \sqrt{3}$

----

(See photo4.jpg)

We once again have a nice right angled triangle to work with.
We have two sides so solve for the third(green) one.
The green side is $\displaystyle 4 \sqrt{3}$

----

(See photo5.jpg)

A right angled triangle again, solving for the green side, which is finally $\displaystyle x$.

$\displaystyle x = 2 \sqrt{3}$

=============

Man i was stupid. This is an equiliteral triangle. All these lines inside the triangle have length $\displaystyle 6 \sqrt{3}$ !! :D anyway you have the value for $\displaystyle x$ now...
• Mar 22nd 2008, 11:19 AM
Nyoxis
Quote:

Originally Posted by janvdl
(See photo2.jpg)

$\displaystyle x = 2 \sqrt{3}$

=============

Man i was stupid. This is an equiliteral triangle. All these lines inside the triangle have length $\displaystyle 6 \sqrt{3}$ !! :D anyway you have the value for $\displaystyle x$ now...

Ah good, well if $\displaystyle x = 2 \sqrt{3}$ then:

$\displaystyle y^2=(2 \sqrt{3})^2+36 \Rightarrow y^2= 12+36 \Rightarrow y= 4\sqrt{3} = 6.928$
• Mar 22nd 2008, 11:21 AM
janvdl
Quote:

Originally Posted by Nyoxis
Ah good, well if $\displaystyle x = 2 \sqrt{3}$ then:

$\displaystyle y^2=(2 \sqrt{3})^2+36 \Rightarrow y^2= 12+36 \Rightarrow y= 4\sqrt{3} = 6.928$

Why such a long approach??

$\displaystyle y = 2x$ according to that rule... Now what was it called again? Anyway, it does exist :D