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

Printable View

- March 22nd 2008, 08:21 AMzachcumerEquilateral 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 - March 22nd 2008, 08:34 AMNyoxis
Hope this helps.

- March 22nd 2008, 08:42 AMzachcumer
I thought I was supposed to use the pythagorean therorum. What did you do?

- March 22nd 2008, 09:06 AMNyoxis
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 is 6 units.

We also know that the 3 corners of the triangle each have because all side lengths are the same, meaning it's equilateral.

So for our seperate triangles we have 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. - March 22nd 2008, 09:09 AMzachcumer
cool. thanks. any help I could get would be great.

- March 22nd 2008, 10:38 AMNyoxis
Bah I keep disproving myself, I got x=2.1962...

- March 22nd 2008, 11:02 AMjanvdl
(See photo2.jpg)

Solve for L using pythagoras. I find

----

(See photo3.jpg)

That red line is half the length of the green line.

Therefore the red line has length:

----

(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

----

(See photo5.jpg)

A right angled triangle again, solving for the green side, which is finally .

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

Man i was stupid. This is an equiliteral triangle. All these lines inside the triangle have length !! :D anyway you have the value for now... - March 22nd 2008, 11:19 AMNyoxis
- March 22nd 2008, 11:21 AMjanvdl