# Find the area of the shade area plus the middle curved square, regions - 1, 2, 3, 4 and 5? All four circles are equal with radius r?

Nov 7, 2016

see explanation.

#### Explanation:

I have to assume that $r = 3$ units and ${O}_{1} {O}_{2} = {O}_{2} {O}_{3} = 5$ units.
${O}_{1} B = 3 - 0.5 = 2.5$
$\implies \cos a = \frac{2.5}{3} , \implies a = {\cos}^{-} 1 \left(\frac{2.5}{3}\right) = {33.557}^{\circ}$

Yellow area $\left({A}_{Y}\right)$ = Area 2 = Area 3 =Area 4 = Area 5
${A}_{Y} = 2 \cdot \left(\pi {r}^{2} \left(\frac{2 a}{360}\right) - \frac{1}{2} {r}^{2} \sin 2 a\right)$
$= 2 \cdot 9 \left(\pi \cdot \frac{67.115}{360} - \frac{1}{2} \sin 67.115\right) = 2.2508$

${O}_{1} {O}_{2} = {O}_{2} {O}_{3} = 5 , \angle {O}_{2} = {90}^{\circ} , \implies {O}_{1} {O}_{3} = 5 \sqrt{2}$
$\implies G H = 5 \sqrt{2} - 2 r = 5 \sqrt{2} - 2 \times 3 = 1.0711$
$b = \angle C {O}_{1} {O}_{3} = 45 - a = 45 - 33.557 = {11.443}^{\circ}$
${O}_{1} G = r \cos b = 3 \cdot \cos 11.443 = 2.9404$
$G H = 5 \sqrt{2} - 2 \cdot {O}_{1} G = 5 \sqrt{2} - 2 \cdot 2.9404 = 1.1903$
$G H = C D = D E = E F = F C$
Area of square $C D E F \left({A}_{s q}\right) = {1.1903}^{2} = 1.4168$

Let the curved square area be ${A}_{c s}$
Let the red area $\left(C \to F \to C\right)$ be ${A}_{R}$
${A}_{R} = \pi {r}^{2} \cdot \left(\frac{2 b}{360}\right) - \frac{1}{2} {r}^{2} \sin 2 b$
${A}_{R} = \pi \cdot {3}^{2} \cdot \frac{2 \cdot 11.443}{360} - \frac{1}{2} \cdot {3}^{2} \sin \left(2 \cdot 11.443\right) = 0.0475$

$\implies {A}_{c s} = {A}_{s q} - 4 \cdot {A}_{R}$
$\implies {A}_{c s} = 1.4168 - 4 \cdot 0.0475 = 1.227$

Finally, let (Area 1 + Area 2 +Area 3 + Area 4 + Area 5) $= {A}_{T}$

${A}_{T} = {A}_{c s} + 4 \cdot {A}_{Y} = 1.227 + 4 \cdot 2.2508 = 10.23$ (unit^2)