Singapore Geometry Problem.

$\textit{ First Solution:}$

$\textit{Second Solution}:$

The above solutions are from this source: https://www.youtube.com/watch?v=yb_JUZFKi-A

For excitement, I am gonna leave the actual solving to the reader, for I believe he/she already knows how to solve the area of a circle.

I am just gonna show my steps in a visual way, adding and subtracting areas along the way.

• Step 1: Add

• Step 2: Subtract

• Step 3: Add

• Step 5: Add

• Step 6: Subtract

$\boxed{\text{The problem becomes level 1 when approached this way.}}$

Define the following shape as $\text{M}_k$ , where k is the dimension(s) shown. Let $\text{A}_k$ be the area of $\text{M}_k$ ; then clearly $\text{A}_k$ is $\dfrac{\pi k^2}{4} - \dfrac{k^2}{2}$ .

We draw a diagonal line to split the region (sorry, couldn't figure out how to get Geogebra to shade this) as shown below. Then the area of the part of the region above the diagonal line is $\text{A}_8 - \text{A}_4$ . Similarly, the area of the part of the region below the diagonal line is $\text{A}_{12} - \text{A}_8$ . Thus the total area is $\left(\text{A}_8 - \text{A}_4\right) + \left(\text{A}_{12} - \text{A}_8\right) = \left(\text{A}_{12} - \text{A}_4\right) = \left(\dfrac{\pi (12)^2}{4} - \dfrac{12^2}{2}\right) - \left(\dfrac{\pi (4)^2}{4} - \dfrac{4^2}{2}\right) = (36\pi - 72) - (4\pi - 8) = \boxed{32\pi - 64}$

https://www.youtube.com/watch?v=B_yNI3gAHNk