How would you answer part c) simply. The idea is to work out what fraction of the larger square the shaded area occupies. I know the answer is 47/90, but can't find a simple way of doing it. This question is meant for academically gifted 12-13 year olds and I need to explain it to my student in a way that won't confuse him.
On
A coordinate-free approach is to extend one of the diagonal lines to form the following picture:
The two red triangles whose area we want to find obviously have area $2$ and $3$; the only problem is that they overlap in a small triangle (darker red) whose area we don't know.
Extending the diagonal line forms the blue triangle, which is similar to the dark red triangle, and their bases are in a $4:1$ ratio. Therefore their heights are also in a $4:1$ ratio; since their heights add up to $3$ (the side of the square), this is enough to find the height of the dark red triangle. Then we can find the area of the dark red triangle and finish the problem.
Let the bottom left corner be the origin $(0, 0)$.
The hypotenuse of the left triangle has slope $-1$ and $y$-intercept $2$ , so it has equation $y = -x + 2$. Similarly, the hypotenuse of the right triangle has slope $\frac{3}{2}$, and $y$-intercept $0 + (-1)(\frac{3}{2})$ or $-\frac{3}{2}$, so it has the equation $y = \frac{3}{2}x -\frac{3}{2}$.
The coordinate where the two lines intersect is $\frac{3}{2}x -\frac{3}{2} = -x + 2$, which gives $x = \frac{7}{5}$. Plugging this into $y = -x + 2$ gives $y = \frac{3}{5}$.
Can you find the area of the white region, then finally the grey region?