Four Surprising Semicircles

Let the center and radius of the large semicircle be $O$ and $1$ , the radius of the small semicircle be $r$ , the diameter of the top small semicircle be $AB$ , $CE$ through $O$ bisects $AB$ perpendicularly at $D$ , and $CD=k$ .

We note that $CO = CD+DO \implies 1 = k+2r \implies k = 1-2r$ . By intersecting chords theorem ,

$\begin{aligned} AD \cdot DB & = CD \cdot DE \\ r^2 & = k(2-k) & \small \blue{\text{Substituting }k=1-2r} \\ r^2 & = (1-2r)(1+2r) \\ & = 1 - 4r^2 \\ \implies r^2 & = \frac 15 \end{aligned}$

The area ratio is $\dfrac {4 \cdot \frac 12 \pi r^2}{\frac 12 \pi \cdot 1^2} = 4r^2 = \boxed {\frac 45}$ .

Just focus on the two smaller semicircles with parallel 'bases'. They can be fitted inside a square with side lengths the same as that base, so the R is the same as the largest V-shape that we can draw in the inscribed square. If we let r = 1, then R² = r² + (2r)² = 1² + 2² = 5. Our ratio is just nr² / R² = (4 × 1²) / (5) = 4/5. All the pies and area formula can be forgotten for a while since all are semicircles (congruency and similarity played their parts) with the important point being area is 2D against length's 1D.

$OA = AB = BC$ hence $OC = \sqrt{2^2+1^2} \cdot OA = \sqrt{5} \cdot OA$ . The area of the large semicircle is $5$ times the one of a small semicircle so the yellow area is $\frac{4}{5}$ of total area.