areageometry
The big square given below is made up of four equal small squares. The area of the unshaded part is $ 108 $ cm$^2$. What is the area of the shaded part?
What I tried : I tried to see if I can solve this with the theory of equilateral and Isosceles triangles (combination of the shaded and unshaded).
Consider the image below:
Let $R$ be the radius of the big circle and $r$ the radius of the small circles.
If we connect the center of the small circles we get an equilateral triangle with side = $2r$. The center of the triangle is the center of the big circle as well.
The distance between the center of the triangle to the midpoint of one of its sides (aka apothem) is $h/3$, where $h$ is the height of the triangle. In our case, $h = r\sqrt{3}$ (we can easily calculate this with the pythagorean theorem).
As we can see in the image, we can now say that $R = r + \frac{2}{3}r\sqrt{3} = r \left(1 + \frac{2\sqrt{3}}{3}\right) \approx 2.1547r$.
$Area_{shaded} = 3\pi r^2 $
$Area_{bigcircle} = \pi R^2 = \pi (2.1547r)^2 \approx 4.6427\pi r^2$
$Area_{unshaded} = Area_{bigcircle} - Area_{shaded} \approx (4.6427 - 3) \pi r^2 \approx 1.6427 \pi r^2 $
Finally, we can check that $2 (Area_{unshaded}) \approx 3.2854 \pi r^2$ is bigger than $Area_{shaded} = 3 \pi r^2$
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 $$3\times 4-\frac{1}{2}\times 1\times 4-\frac{1}{2}\times 1\times 4-\frac{1}{2}\times 1\times 1-\frac{1}{2}\times 3\times 3$$
Best Answer
The answer is $\frac13 \times 108 = 36$ cm2.
The key is if you slide the vertex $X$ of a triangle $XYZ$ along any line parallel to $YZ$, the area of the triangle remains unchanged. From the picture below, one can see that the area covered by the quadrilateral (in red) is one third of the uncovered area (in light orange).