1. ## Problem finding area

i am having a problem finding the area of the three triangles outside the Pitagoras drawing
you can see what i am meaning in this picture
http://i51.tinypic.com/28gxmyd.jpg
so i need the area of all the triangles in the middle of the squares

2. ## Re: Problem finding area

Hello, leart369!

http://i51.tinypic.com/28gxmyd.jpg
i need the area of all the triangles.

Label the inner right triangle like this:

Code:
      A
*
|   *     c
b |       *
|           *
* - - - - - - - * B
C         a

Note that for any triangle, its area is: .$\displaystyle A \;=\;\tfrac{1}{2}ab\sin C$
. . (One-half the product of two sides and the sine of the included angle)

The right triangle at the lower-left has sides $\displaystyle a$ and $\displaystyle b$.
. . Its area is: .$\displaystyle \boxed{A_1 \:=\:\tfrac{1}{2}ab}$

The triangle at the right has sides $\displaystyle a$ and $\displaystyle c$, and included angle $\displaystyle (180^o - B).$
. . Its area is: .$\displaystyle A_2 \:=\:\tfrac{1}{2}ac\sin(180^o-B) \:=\:\tfrac{1}{2}ac\sin B$
In the above diagram, we see that: .$\displaystyle \sin B \,=\,\tfrac{b}{c}$
. . Hence: .$\displaystyle A_2 \;=\;\tfrac{1}{2}ac\left(\tfrac{b}{c}\right) \quad\Rightarrow\quad \boxed{A_2 \:=\:\tfrac{1}{2}ab}$

The third triangle has sides $\displaystyle b$ and $\displaystyle c$, and included angle $\displaystyle (180^o - A).$
. . Its area is: .$\displaystyle A_3 \:=\:\tfrac{1}{2}bc\sin(180^o-A) \:=\:\tfrac{1}{2}bc\sin A$
In the above diagram, we see that: .$\displaystyle \sin A \,=\,\tfrac{a}{c}$
. . Hence: .$\displaystyle A_3 \;=\;\tfrac{1}{2}bc\left(\tfrac{a}{c}\right) \quad\Rightarrow\quad \boxed{A_3 \:=\:\tfrac{1}{2}ab}$

Fascinating! . . . All four triangles have the same area.

3. ## Re: Problem finding area

does exist any way to find the area off all that drawing?

4. ## Re: Problem finding area

Originally Posted by leart369
does exist any way to find the area off all that drawing?
1. I assume that you mean the area of the hexagon, containing 3 squares, 2 right triangles and 2 obtuse triangles. If so:

2. As Soroban has shown you have:

$\displaystyle \text{complete area} = a^2+b^2+c^2+4 \cdot \frac12 \cdot a \cdot b = c^2 +(a+b)^2$

3. Since $\displaystyle a^2+b^2 = c^2$ there are other possibilities to simplify the sum of areas.

5. ## Re: Problem finding area

thank you very much that was what i was looking for