In the image, the segments inside the square go from a vertex to the middle point of the opposite side. If the length of the sides of the square is $1$, the area of $ABCD$ is?
Any hints?
I tried some things matching the bigger triangles but i can't get the area of $ABCD$.
On
As N74 suggested, you can rotate the small triangles on to the trapezoids that are on each side of the square $ABCD$ to get a plus sign shape made of five congruent squares.
Since all of the translations we have done are distance preserving, we can safely say that the plus sign made of squares has an area equal to the original big square which is $1$. Thus the area of $ABCD$ is one fifth of the total area, or $\frac{1}{5}$ square units.
If we move the figure to a $xy$ axis, we can solve this problem by finding the line equations and intersection points. Consider this image:
Since we know the point coordinates in the image, we can easily find the line equation of lines 1, 2 and 3 (we can substitute the point values in the slope-intercept form equation $y = ax + b$ and find the equation of each line).
For line 1, the line equation is $y = 2x$.
For line 2, the line equation is $y = \frac{2 - x}{2}$
For line 3, the line equation is $y = \frac{1 - x}{2}$
To find point A, we have to find the intersection point of lines 1 and 3:
$$ 2x = \cfrac{1 - x}{2} \Rightarrow x = \cfrac{1}{5} \;and \; y = \cfrac{2}{5} \Rightarrow A = \left(\cfrac{1}{5}, \cfrac{2}{5}\right) $$
We can do the same with lines 1 and 2 in order to find point B:
$$ 2x = \cfrac{2 - x}{2} \Rightarrow x = \cfrac{2}{5} \; and \; y = \cfrac{4}{5} \Rightarrow B = \left(\cfrac{2}{5}, \cfrac{4}{5}\right) $$
The side of the square (we can call it $L$) will then be the distance $\overline{AB}$:
$$ L = \overline{AB} = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2 } = \sqrt{\cfrac{1}{25} + \cfrac{4}{25}} = \sqrt{\cfrac{1}{5}} $$
Finally, the area of the square will be:
$$ Area = L^2 = \cfrac{1}{5} $$