What is the area of the shaded region in this rectangle?

Question

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.

Answer 1

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 $\frac12\cdot 2\cdot \frac32$.

Answer 2

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 $$3\times 4-\frac{1}{2}\times 1\times 4-\frac{1}{2}\times 1\times 4-\frac{1}{2}\times 1\times 1-\frac{1}{2}\times 3\times 3$$

Answer 3

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

Answer 4

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:

$$Hc = 1$$ $$Hd = 3$$ $$Ax = \frac{3}{2}$$