# Problem of the Week Problem D and Solution They’re Blue

## Problem

In rectangle $$ABCD$$, the length of side $$AB$$ is $$7$$ m and the length of side $$BC$$ is $$5$$ m. Four points, $$W$$, $$X$$, $$Y$$, and $$Z$$, lie on diagonal $$BD$$, dividing it into five equal segments. Triangles $$AWX$$, $$AYZ$$, $$CWX$$, and $$CYZ$$ are then painted blue, as shown.

Determine the area of the painted region.

## Solution

Solution 1

Using the formula for area of a triangle, $$\text{area} = \frac{\text{base}\times\text{height}}{2}$$, we have area $$\triangle ABD = \frac{7\times 5}{2} = \frac{35}{2} \text{ m}^2$$.

The five triangles $$\triangle ADW$$, $$\triangle AWX$$, $$\triangle AXY$$, $$\triangle AYZ$$, and $$\triangle ABZ$$ have the same height, which is equal to the perpendicular distance between $$BD$$ and $$A$$. Since $$DW = WX = XY = YZ = ZB$$, it follows that the five triangles also have equal bases. Therefore, the area of each of these five triangles is equal to $$\frac{1}{5} (\text{area }\triangle ABD) = \frac{1}{5}\left(\frac{35}{2}\right) = \frac{7}{2} \text{ m}^2$$.

Similarly, the area of $$\triangle BCD$$ is equal to $$\frac{7\times 5}{2} = \frac{35}{2} \text{ m}^2$$. The five triangles $$\triangle CDW$$, $$\triangle CWX$$, $$\triangle CXY$$, $$\triangle CYZ$$, and $$\triangle CBZ$$ also have the same height and equal bases. Therefore, the area of each of these five triangles is equal to $$\frac{1}{5} (\text{area }\triangle BCD) = \frac{1}{5}\left(\frac{35}{2}\right) = \frac{7}{2} \text{ m}^2$$.

Therefore, the area of the painted region is $$4\left(\frac{7}{2}\right) = 14 \text{ m}^2$$.

Solution 2

Since $$ABCD$$ is a rectangle, $$\angle DAB=90\degree$$, so $$\triangle ABD$$ is a right-angled triangle. We can then use the Pythagorean Theorem to calculate $$BD^2 = AB^2 + AD^2 = 7^2 + 5^2 = 49 + 25 = 74$$, and so $$BD = \sqrt{74}$$, since $$BD>0$$. Therefore, $$DW = WX = XY = YZ = ZB = \frac{1}{5}(BD) = \frac{1}{5}\sqrt{74}$$.

Using the formula for area of a triangle, $$\text{area} = \frac{\text{base}\times\text{height}}{2}$$, we have area $$\triangle ABD = \frac{7\times 5}{2} = \frac{35}{2} \text{ m}^2$$.

Let’s treat $$BD=\sqrt{74}$$ as the base of $$\triangle ABD$$ and let $$h$$ be the corresponding height. Since the area of $$\triangle ABD$$ is $$\frac{35}{2}$$, then we have $$\frac{\sqrt{74}\times h}{2} = \frac{35}{2}$$ and so $$\sqrt{74}\times h= 35$$, thus $$h = \frac{35}{\sqrt{74}}$$.

$$\triangle AWX$$ and $$\triangle AYZ$$ both have height $$h = \frac{35}{\sqrt{74}}$$ and base $$\frac{\sqrt{74}}{5}$$, so $$\text{area }\triangle AWX = \text{area }\triangle AYZ = \frac{1}{2}\left(\frac{\sqrt{74}}{5}\right)\left(\frac{35}{\sqrt{74}}\right) = \frac{7}{2} \text{ m}^2$$.

Similarly, $$\triangle CWX$$ and $$\triangle CYZ$$ both have height $$h=\frac{35}{\sqrt{74}}$$ and base $$\frac{\sqrt{74}}{5}$$, so $$\text{area }\triangle CWX = \text{area }\triangle CYZ = \frac{1}{2}\left(\frac{\sqrt{74}}{5}\right)\left(\frac{35}{\sqrt{74}}\right) = \frac{7}{2} \text{ m}^2$$.

Therefore, the area of the painted region is $$4\left(\frac{7}{2}\right) = 14 \text{ m}^2$$.