Sweet Red Fish

First, we find the point $P$ of intersection of the two circles that correlates with the point where the two shaded regions touch. Letting the lower left corner of the square being the origin and the sides of the squares in alignment with the $x$ and $y$ axes, the larger circle has the equation

$(x - 5)^{2} + (y - 5)^{2} = 25 \Longrightarrow x^{2} - 10x + y^{2} - 10y + 25 = 0$

$\Longrightarrow x^{2} + y^{2} = 10x + 10y - 25,$

and the smaller circle has the equation

$(x - \frac{5}{2})^{2} + y^{2} = \frac{25}{4} \Longrightarrow x^{2} + y^{2} = 5x.$

Thus $5x = 10x + 10y - 25 \Longrightarrow x + 2y = 5 \Longrightarrow x - 5 = -2y.$

Plugging this result back into the equation for the larger circle gives us that

$4y^{2} + y^{2} - 10y + 25 = 25 \Longrightarrow 5y^{2} = 10y \Longrightarrow y = 2,$

where we can discard the solution $y = 0.$ This gives us $P(1,2).$

Looking now at the small square, the area $B$ of the lower white region and the area $A$ of the shaded region on the left add to $25 - \frac{25\pi}{4},$ and the area $C$ of the shaded region on the right along with area $B$ add to $\frac{25\pi}{8}.$ Thus the total area of the shaded regions will be $25 - \frac{25\pi}{8} - 2B.$ So we must now find $B.$

Define the points $O(0,0), Q(5,5), R(5,2), S(5,0), T(\frac{5}{2}, 0), U(1,0).$ Now $\angle PQR = \angle PQS = \tan^{-1}(\frac{4}{3}),$ so the area of sector $PQS$ is $\frac{25}{2}\tan^{-1}(\frac{4}{3}).$ The area of the region bounded by chord $PS$ and arc $PS$ is then the area of sector $PQS$ minus the areas of right triangles $\Delta PQR$ and $\Delta PRS$ , which comes out to $a = \frac{25}{2}\tan^{-1}(\frac{4}{3}) - 10.$

The portion of the area $B$ that lies to the right of $x = 1$ is then the area of $\Delta PSU$ minus $a$ , i.e., $4 - (\frac{25}{2}\tan^{-1}(\frac{4}{3}) - 10) = 14 - \frac{25}{2}\tan^{-1}(\frac{4}{3}).$

The portion of area $B$ that lies to the left of $x = 1$ is the area of sector $PTO$ minus the area of $\Delta PTU.$ Now $\angle PTO = \angle PTU = \tan^{-1}(\frac{2}{\frac{3}{2}}) = \tan^{-1}(\frac{4}{3}).$ Thus the area of sector $PTO$ is $\frac{25\pi}{8}\tan^{-1}(\frac{4}{3}),$ and so the area of the portion of $B$ to the left of $x = 1$ is $\frac{25\pi}{8}\tan^{-1}(\frac{4}{3}) - (\frac{1}{2})(\frac{3}{2})(2) = \frac{25\pi}{8}\tan^{-1}(\frac{4}{3}) - \frac{3}{2}.$ Thus the area of $B$ is

$14 - \frac{25}{2}\tan^{-1}(\frac{4}{3}) + \frac{25\pi}{8}\tan^{-1}(\frac{4}{3}) - \frac{3}{2} = \frac{25}{2} - \frac{75}{8}\tan^{-1}(\frac{4}{3}).$

The area of two shaded regions is then

$25 - \frac{25\pi}{8} - 2*(\frac{25}{2} + \frac{75}{8}\tan^{-1}(\frac{4}{3})) = \dfrac{75}{4}\tan^{-1}\left(\dfrac{4}{3}\right) - \dfrac{25\pi}{8} = \boxed{7.569}$

to 3 decimal places.

Let $A,B,C$ denote the area of orange, purple, and pink regions respectively. So we want to find $A+C$

$\begin{aligned} A+C & = & (A+B-B) + (C+B-B) \\ & = & (A+B) + (B+C) - 2B \\ & = & (A+B) + (B+C) - 2(B+C- C) \\ & = & (A+B) - (B+C) +2 C \\ \end{aligned}$

$A+B$ is the area quarter of the square minus the area of sector with $90^\circ$ of the larger circle. So $A+B = 5^2 - \frac {1}{4} \pi \cdot 5^2 = 25 - \frac {25}{4} \pi$

$B+C$ is the area of semicircle of the smaller circle, which means $B+C = \frac {1}{2} \cdot \pi \left ( \frac 5 2 \right )^2 = \frac {25}{8} \pi$

The radius of the smaller circle is simply $\frac 5 2$

To solve for $C$ , we first need to find the coordinates of the intersection of the two circles. Apply coordinate geometry: $x^2 + y^2 = 5^2, \left (x+ \frac 5 2 \right )^2 + (y + 5)^2 = \left ( \frac 5 2 \right )^2$ solving them simultaneously gives points $D(-4,-3), E(0,-5)$

Let $O_1 , O_2$ denote the point of the center of the larger and smaller circle respectively. And denote $\theta_1, \theta_2$ as the measure $\angle EO_1 D, \angle D O_2 E$ respectively.

Then, $DO_1 = EO_1 = 5, DO_2 = E O_2 = \frac 5 2$ , and $DE = \sqrt{4^2 + 2^2} = \sqrt{20}$

By Cosine Rule, we have $\cos \theta_1 = \frac { 5^2 + 5^2 - 20}{2 \cdot 5 \cdot 5 } = \frac 3 5 \Rightarrow \sin \theta_1 = \frac 4 5$ . Similarly, $\cos \theta_2 = \frac { \left ( \frac 5 2 \right )^2 + \left ( \frac 5 2 \right ) - 20 }{2 \cdot \left ( \frac 5 2 \right ) \cdot \left ( \frac 5 2 \right ) } = - \frac 3 5 \Rightarrow \sin \theta_2 = \frac 4 5, \theta_2 = \pi - \sin^{-1} \left ( \frac 4 5 \right )$

We split $C$ into two parts: one part above line $DE$ , the other below line $DE$ .

For the area above $DE$ denoting as $C_1$ , it is simply the area of sector $D O_2 E$ minus the area of triangle $D O_2 E$

$\begin{aligned} C_1 & = & \frac {1}{2} \cdot r^2 ( \theta_2 - \sin \theta_2 ) \\ & = & \frac 1 2 \cdot \left (\frac 5 2 \right )^2 \left ( \pi - \sin^{-1} \left ( \frac 4 5 \right ) - \frac 4 5 \right ) \\ \end{aligned}$

Likewise, for the area below $DE$ denoting as $C_2$ , it is the area of sector $D O_1 E$ minus thue area of triangle $D O_1 E$

$\begin{aligned} C_2 & = & \frac {1}{2} \cdot R^2 ( \theta_1 - \sin \theta_1 ) \\ & = & \frac 1 2 \cdot 5^2 \left ( \sin^{-1} \left ( \frac 4 5 \right ) - \frac 4 5 \right ) \\ \end{aligned}$

Add all of them up: $(A+B) - (B+C) + 2(C_1 + C_2)$ gives $\boxed{7.569}$

Note: $\color{#3D99F6} {P},\color{teal} {Q}, \color{#D61F06} {R}, \text{ \& } \color{#69047E} {S}$ represents the 4 area inside the smaller square.

First of all, take a look at point O, A, B & C, we'll be using only these 4 points in our whole solution. I'll use coordinate system first to get some value. Assume that the center of the bigger circle is at origin $O(0,0)$ & the center of the smaller circle is $B(-2.5,-5)$ . Then their respective equations will be, $x^2+y^2=5^2$ $(x+2.5)^2+(y+5)^2=2.5^2$ Solving them we get the coordinates of, $A\equiv(-4,-3)$ $C\equiv(0,-5)$ From the slope of line AB we get, $\tan \theta_2=-\frac{2}{1.5}$ $\Rightarrow \theta_2=\tan^{-1}\left(-\frac43\right)=\sin^{-1}\frac45$ Similarly, from slope of line AO we get, $\theta_3=\tan^{-1}\left(\frac34\right)$ $\therefore \theta_1=\frac{\pi}{2}-\theta_3=\frac{\pi} {2}-\tan^{-1}\frac34=\sin^{-1}\frac45$ Now, let's divide area $\color{#69047E} {S}$ by segment AC into 2 parts named $\color{#69047E} {S1}$ & $\color{#69047E} {S2}$ :

Then area of, $\color{#69047E} {S1}=\text {Sector} \, BAGC-\Delta BAC$ $=\frac{(2.5) ^2\sin^{-1}\frac45}{2}-\frac12 \times 2.5\times 2.5\times \sin \left(\sin ^{-1}\frac 45\right)$ $=6.9197-2.5\,\color{#0C6AC7} {\left[\text{In 2nd quadrant,} \sin^{-1}\frac45=2.2143\right] }$ $=4.4197$

Similarly, area of, $\color{#69047E} {S2}=\text {Sector} \, OAHC-\Delta OAC$ $=\frac{5^2\times \sin ^{-1}\frac45}{2}-\frac12 \times 5\times 5\times \sin \left(\sin ^{-1}\frac 45\right)$ $=11.5912-10$ $=1.5912$ Thus, $\color{#69047E} {S} = \color{#69047E} {S1} + \color{#69047E} {S2}$ $=4.4197+1.5912=\boxed{6.0109}$

Now, area of $\color{#D61F06} {R}$ is, $\text{Area of the small semicircle} - \color{#69047E} {S}$ $=\frac{\pi\times(2.5)^2}{2}-6.0109$ $=3.8066$

And area of, $\color{teal} {Q} +\color{#D61F06} {R} =\text{Area of the smaller square} - \text{Area of the big quartercirle}$ $=5^2-\frac{\pi\times 5^2}{4}$ $=5.365$ So, $\color{teal} {Q} =5.365-3.8066=\boxed{1.5584}$

Aah! Finally, we got the answer, $\color{teal} {Q} +\color{#69047E} {S} =1.5584+6.0109=7.5693\approx\color{#3D99F6} {\boxed {7.569}}$

For convenience, we will use inches, letting 1 inch = 2.5 cm.

Let A be the area of the right red region, B the left red region, C the top white region, and D the lower white region (all our work will take place in the small square on the lower left.)

We have the obvious equations $A+B+C+D=4, A+C=\pi,$ $A+D=\pi/2$ so $A+B=4+2A-\frac{3\pi}{2}$ .

Let $\theta=\arcsin(\frac{1}{\sqrt5})=\frac{\arcsin(4/5)}{2}$ . This is the top angle in the right triangle formed by the center of the small circle and the two right corners of the square.

Considering the "kite" formed by the centers of the two circles and the two points of region A, we find A = (sector of large circle) + (sector of small circle) - (area of kite) = $4\theta + (\pi/2-\theta)-2$ .

Thus $A+B=4+2A-\frac{3\pi}{2}=6\theta-\frac{\pi}{2}\approx1.2111$ . Converting the result into square centimetres, we have $(6\theta-\frac{\pi}{2})(\frac{5}{2})^2\approx7.5693$ .

Let O be the center of the smaller circle, A be the upper intersection of both circles, and AH be the line parallel to DB.

△AFB and △ABC are isosceles triangles, which share the same base AB. Thus, the line bisecting these triangles in halves will be perpendicular to AB, also bisecting the segment at point I: that line is OF.

Since AH // OB and AI = IB, △AHI is identical to △OIB as they have all the same angles with at least one equal side. Consequently, △AIO is identical to △HIB, and AO = OB = HB = BA.

Therefore, ⎕AHBC is a rhombus.

Let ∠IBO = 𝜭. △IBO is a right-angled triangle, so ∠IOB = 90 ̊ - 𝜭.

Now considering △FOB, which is also a right-angled triangle, since ∠FOB = 90 ̊ - 𝜭, ∠OFB = 𝜭.

Thus tan 𝜭 = 2.5/5 = ½.

In other words, (sin 𝜭)/(cos 𝜭) = ½.

cos 𝜭 = 2(sin 𝜭)

(cos 𝜭)^2 = 4(sin 𝜭)^2

1-(sin 𝜭)^2 = 4(sin 𝜭)^2

(sin 𝜭)^2 = 1/5.

sin 𝜭 = 1/√5 ; cos 𝜭 = 2/√5

Since ⎕AHBC is a rhombus, its diagonals will bisect all its angles in halves. Thus, ∠IBO = ∠IBH = 𝜭, which means ∠OBH = 2𝜭.

And because AO // HB, ∠AOC = ∠OBH = 2𝜭.

Now applying this to △AOC,

sin ∠AOC = sin 2𝜭 = AC/2.5

sin 2𝜭 = 2(sin 𝜭)(cos𝜭) = 2(1/√5 ) (2/√5) = 4/5

So AC = 2.5(4/5) = 2.

Now we get AG = 3, and using Pythagorean theorem for △AGF, we’ll get:

(AF)^2 = (AG)^2 + (GF)^2

5^2 = 3^2 + (GF)^2

GF = 4

Notice that cos ∠GFA = GF/AF = 4/5 = sin 2𝜭.

That means ∠GFA = 90 ̊ - 2𝜭.

Then we can calculate the total shaded area by working out one portion at a time.

First, we formulate the “fish tail” part: Portion EAD “tail” = ⎕EGDC – Portion EGA – Portion ADC

Then, we formulate the “fish body” part: Portion AB “Body” = Portion GABF + Portion ABC - ⎕GCBF

         = (Portion EFB – Portion EGA) + (Portion ADB – Portion ADC)- ⎕GCBF

Combining the terms,

the total area = ⎕EGDC - ⎕GCBF + Portion EFB – 2(Portion EGA) + Portion ADB – 2(Portion ADC)

⎕EGDC = (1)(5) = 5

⎕GCBF = (4)(5) = 20

Portion EFB = (1/4)(π)(5^2) = 6.25 π ≈19.6349

2(Portion EGA) = 2(Portion EFA - △GAF)

        = 2[((π/2-2ϴ)/2π)(π)(5^2) – (1/2)(3)(4)]

(converting 90 ̊ - 2𝜭 to radian mode)

        = (π/2-2ϴ)(25) – 12
        (sin 2𝜭 = 4/5, so 2𝜭 = arcsin (4/5) ≈0.92729)

        ≈ (π/2-0.92729)(25) – 12 

        ≈4.087625

Portion ADB = (1/2)(π)(2.5^2) = 3.125 π ≈9.8175

2(Portion ADC) = 2(Portion ADO – △ACO)

= 2[(2𝜭/2 π)( π)(2.5^2) – (1/2)(2)(1.5)]

        ≈ (0.92729)(6.25) – 3

        ≈ 2.79556

So the shaded area = 5 – 20 + 19.6349 - 4.087625 + 9.8175 - 2.79556 ≈ 7.569

$\large \color{#D61F06}{ \text{Full geometrical solution I have written has vanished!!!!! }}\\\text{ I am rewriting the solution.} \\~~\\\text{A circle radius R and a segment, central angle } 2 \theta ~has~ an ~area ~A_{seg}. \\A_{seg}=A_{sect}, ~its ~sector~ area-A_{tri}, ~\Delta ~area~ formed ~by~the ~chord~and \\its ~bounding ~radii. ~~\therefore~ A_{seg}=A_{sect}- A_{tri}=R^2(\theta-Sin \theta*Cos \theta)..(1) \\ \text{Here, central distance D, is also the hypotenuse of the right angled}\\\text{ triangle with the two radii as legs. }~~\therefore~D^2=5^2+2.5^2~~and D=2.5\sqrt5.\\\implies~\theta_5=Cos^{-1}\dfrac{2}{\sqrt5},~~\theta_{2.5}=Cos^{-1}\dfrac{1}{\sqrt5} , ~~this~ gives ~Sins ~and~ Coses. \\\therefore \text{ the COMMON area of the intersecting circles, is } A_{com}~from~(1)~is :-\\A_{seg5}+A_{seg2.5} =5^2(\theta_5-Sin \theta_5*Cos\theta_5) +2.5^2(\theta_{2.5}-Sin \theta_{2.5}*Cos\theta_{2.5}) \\\therefore \color{#D61F06}{A_{com}=6.011}..\text{Top half small circle area, } \color{#D61F06}{A_{half}}=\dfrac{\pi*2.5^2}{2}=\color{#D61F06}{9.817}.\\ \text{Area of bottom left corner within the square without big circle,}\\$ $\color{#D61F06}{A_{out}}=\dfrac{10^2-\pi*5^2}{4} =\color{#D61F06}{5.365}...Left~red~area=A_{out}+A_{com}-A_{half}. \\Total =\{A_{out}+A_{com}-A_{half}\}+A_{com}={ \large \color{#3D99F6}{7.569}}$

Soln

Take the lower horizontal side of the large square as X-axis

and take the left vertical side of the large square as Y-axis

The equation of the large circle is :

x^2 + y^2 - 10 x - 10 y + 25 = 0

The equation of the small circle is :

x^2 + y^2 - 5 x = 0

These two circles cut each other at two points

The first point is (5 , 0) on the lower horizontal side of the large square ,

the second point is (1 , 2) which is the contact point of the two red regions

Let the area of the large red region be denoted by X

Let the area of the small red region be denoted by Y

Let the part of the small square (above the red regions) be denoted by Z

Then

X + Z = (1/4)(Pi)(5^2) = 19.63495408 ................................................ (1)

Y + Z = 25 - (1/2)(Pi)(2.5^2) = 15.18252296 .....................................(2)

From (1) , (2) we get

X - y = 4.452431122 ....................................................................................(3)

Region X can be divided into two circular segments A , B

Segment A is of radius 5 cm. and angle of measure 0.92729 (in radian )

Segment B is of radius 2.5 cm. and angle of measure (Pi - 0.92729)

Then A = 1.591125 , B = 4.4196875

So

X = A + B = 6.0108

Substituting in (3) we get

Y = 1.558368878

Then

The area of the red regions = X + Y = 7.569

We need to find the area in common between the 2 circles. To do this we think in terms of area of the 2 minor segments. We need the angle so that we can find the area of the 2 segments (0.5(r^2)(x-sin(x)), (x is in radians). Now by the cosine rule we can equate the distance of the chord shared by the 2 circles. root(2.5^2+2.5^2-2(6.25)cos(x))=root(5^2+5^2-2(25)cos(y)). We can then show that y = π - x such that x > y by using theorems from geometry. Using these facts we can show that cos(x) = 3/5. Thus sin(x) = 4/5 (sin(x)>0). Now we can find the area of the intersection by applying the formula twice to get the intersection area as 6.0108697. Now Let A be the set of area contained in the smaller square and B be the set of area of the larger circle such that B is a subset of A and C be the set of area of the smaller circle such that C is a subset of A. We have notB∩C = C - B∩C = (π((5/2)^2)/2)-6.0108697=3.80660733. Now notB∩notC = 25 - 25(π/4)-3.80660733. Hence notB∩notC + B∩C = 25 - 25(π/4)-3.80660733 + 6.0108697 = 7.569

Equation of the large of the large circle +> y=5-sqrt(10x-x^2)

Equation of the small circle= y=sqrt(5x-x^2)

They intersect at point (1,2) and (5,0)

Calculate the area under large circle between x=1 and x=5 Integrate 5-sqrt(10x-x^2) from 1 to 5 = 2.4088

Calculate area under small circle from x=0 to x=1 Integrate sqrt(5x-x^2) from 0 to 1 = 1.3978

Read area (Big circle n Large Circle) = Semi circle -( the two regions calculated above)= ((2.5^2*pi)/2)-3.8066= 6.01087

Second red area= (Square - large circle)/4- two regions calculated above= (100-(pi*25))/4 - 3.8066= 1.55846

Lastly, two red regions area= 6.01087+1.55846=7.56933...