The dotted diagonal AC (Fig. 42) has the length of twice the radius. Find the area of the emblem only.

The dotted diagonal AC (Fig. 42)  has the length of twice the radius. Find the area of the emblem only.

$$\\A_{circle}=\pi r^2 \\
A_{square}=(r \sqrt{2})^2=2r^2\\
A_{goblet}=\frac{ A_{circle} - A_{square} }{4} = \frac{\pi r^2 - 2r^2}{4}\\\\
A = 4\cdot
\left[
A_{quadrant}-2\cdot A_{goblet}- A_{triangle}
\right]\\\\
A = 4\cdot
\left[
\frac{\pi r^2 }{4}
-2\cdot \left( \frac{\pi r^2 - 2r^2}{4} \right)
- \frac{ \left( r\sqrt{2}-r \right)^2 }{2}
\right]\\\\
A = \pi r^2 -2\cdot \left(\pi r^2 - 2r^2\right)
- 2\cdot \left( r\sqrt{2}-r \right)^2\\\\
A = \pi r^2 -2\pi r^2 + 4r^2
- 2\cdot \left( r\sqrt{2}-r \right)^2\\\\
A = \pi r^2 -2\pi r^2 + 4r^2
- 2\cdot \left(2r^2-2\sqrt{2}r^2 +r^2 \right)\\\\
A = \pi r^2 -2\pi r^2 + 4r^2
- 4r^2 +4\sqrt{2}r^2-2r^2\\\\
A = -\pi r^2 +4\sqrt{2}r^2-2r^2\\\\
A=r^2\cdot (4\sqrt{2}-2-\pi)$$

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The dotted diagonal AC (Fig. 42)  has the length of twice the radius. Find the area of the emblem only.

Radius (r) = 1 

AC = 2r       r = 1

Area of the square is:  (sqrt(2))²  = 2.000u²

Triangles:  (lkC + oAf) = (sqrt(2) -1)² = 0.171572875253809862u²

Half circle area is:  r²pi/2 = 1.5707963267948966u²

(2.000u² - 1.570796326794896u² - 0.17157287525380986u²)*2= 0.515261595902587u²

You don't specify the value of the radius in the question civonamzuk, but if r = 1 then heureka's answer evaluates to the same as the one you give.

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Heureka's answer is correct!

Really?  That's a new one on me!  (Forgive the pun!)

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