Dianthus

Let's assume the radius of the big semi-circle is $\boxed{1}$

• We label BD as $\boxed{h}$
• By the pythagorean theorem we can then label DC as $\boxed{\sqrt{4-h²}}$
• We label the diameter of the two big circles as $\boxed{x}$
• We label the diameter of the small circle as $\boxed{y}$

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• Let's solve for the diameter $\boxed{x}$ of the big circles

  • By the product of chords theorem we can write the following equation : $\boxed{x(2-x)=\frac{\sqrt{(4-h²)²}}{2²}}$
  • The diameter of a circle inscribed in a triangle is four times the area of the triangle divided by it's perimeter.
  • This give us the second equation: $\boxed{x=\frac{2h\sqrt{4-h²}}{2+h+\sqrt{4-h²}}}$
  • We then solve this system of equations and get $\boxed{x=\frac{8}{13}}$ and $\boxed{h=\frac{10}{13}}$

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Now we need to solve for the radius of the small circle

• We again use the product of chords theorem to get the following equation : $\boxed{y(2-y)=\frac{h²}{2²}}$
• We solve the equation and find $\boxed{y=\frac{1}{13}}$

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Since the value of y need to be 1 we calculate the value of x and the value of the diameter of the big semi-circle (D) using proportional relationships. We get $\boxed{x=8}$ and $\boxed{D=26}$
Finally : $\boxed{8 \times 26=208}$