What is the area of the shaded region below? 
I think the solution requires subtracting the area of each of the four triangles from the area of the rectangle. I can calculate the area of triangles a and b since I have a base and height. But I don't see how to calculate the height of triangles c and d.
I've tried drawing a line parallel to the 3 cm width of the rectangle that intersects with the point where triangles c and d touch. The height of that line (call it h) along the 4 cm height of the rectangle would give me the height of both triangles c and d. But I don't see how to derive h.
$\endgroup$4 Answers
$\begingroup$The heights of $c$ and $d$ are in the ratio $1:3$ (equal to the ratio of their bases). So the heights are $1$ and $3$ respectively.
The area is $$3\times 4-\frac{1}{2}\times 1\times 4-\frac{1}{2}\times 1\times 4-\frac{1}{2}\times 1\times 1-\frac{1}{2}\times 3\times 3$$
$\endgroup$10$\begingroup$For each of the shaded triangles, the length of a vertical line from the top corner until it hits the bottom side is $2$. The horizontal distance between the two other corners is $\frac 32$. So the area of each shaded triangle is $\frac12\cdot 2\cdot \frac32$.
$\endgroup$4$\begingroup$Following from @CY Aries answer about c and d ratio. We can see 2 triangles (the grey arreas including the d area in each one).So we find the area of that two(4*3/2 *2) and substracting d area 2 times (3*3) gives as 3
Another approach is given below.
In the following Ax is the area of one of the shaded regions.
The total square give us:
$$Aa + Ab + Ac + Ad + 2\times Ax = 4\times 3$$
$$2 + 2 + \frac{1}{2}\times Hc + \frac{3}{2} \times Hd + 2\times Ax = 12$$
$$\frac{1}{2}\times Hc + \frac{3}{2} \times Hd + 2\times Ax = 8$$
The triangle made up by a, c and one shade region gives us:
$$Aa + Ac + Ax = \frac{1}{2}\times 4\times 2$$
$$2 + \frac{1}{2}\times Hc + Ax = 4 $$
$$\frac{1}{2}\times Hc + Ax = 2 $$
The height of the square give us:
$$Hc + Hd = 4$$
Now we have 3 equations with 3 unknown which is easy to solve, i.e.
$$\frac{1}{2}\times Hc + \frac{3}{2} \times Hd + 2\times Ax = 8$$
$$\frac{1}{2}\times Hc + Ax = 2 $$
$$Hc + Hd = 4$$
will give:
$$Hc = 1$$ $$Hd = 3$$ $$Ax = \frac{3}{2}$$
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