Resolve this problem about area of circle

Question

I spent a few days thinking about how to solve this problem without using the formula of the circle, anyone have any ideas?

Here you can find a picture with the problem

Thank you!

Answer 1

The ratios of the areas of the semi-circles with radius $AB, BD, AD$ is $1:4:9$. Let $S_{AB}$ denote the area of the semicircle with diameter $AB$. Then, we can see that the shaded area can be derived by $$S_{shaded} = \frac{S_{AD}-S_{BD}+S_{AB}}{S_{AD}}\cdot S_{AD} = \frac{9-4+1}{9}\cdot \frac{1}{2}\cdot S_{circle} = 27\pi$$ where we used the fact that $S_{AD} = \frac{1}{2}S_{circle}$.

Answer 2

Hint : Compute first the area of the shadded region in the upper semi circle. You know that the total area of the upper semi-circle is $\frac{81}{2} \pi$. And you can compute how much you need to remove ($\frac{36}{2} \pi$) to get the exact answer .

Then compute the shadded region in the lower semi circle. ($9 \pi$)

Answer 3

Maybe the task wants you to use ratios?

If you put $S=81\pi$ the area of circle with diameter AD, then for the areas of circles with diameters BD and AB we have:

$$ S_{BD} = \bigl( \frac{2}{3}\bigr)^2S$$ $$ S_{AB} = \bigl( \frac{1}{3}\bigr)^2S$$

Since $BD=\frac{2}{3}AD$ and $AB=\frac{1}{3}AD$ .

Then you can compute area of the shaded region this way: $$ S_{shaded}=\frac{1}{2}S_{AB}+(\frac{1}{2}S-\frac{1}{2}S_{BD})$$

Answer 4

We will divide the whole figure into 6 regions as shown below:-

Let n(A) = [2] + [3] + [3’] and n(B) = [2’] + [3’] + [3].

Note that [1] = [1’]; [2] = [2’] and [3 = [3’]. By symmetry, n(A) = n(B).

$ n(A \cup B) = [Whole] – [Green]$

$n(A \cap B) = [3’] + [3] = [Red]$

By inclusion-exclusion, we have $2 \times n(A) = n(A) + n(B) = n(A \cup B) + n(A \cap B)$

The required $= \dfrac 12 ([Whole] – [Green] – [Red])$