I want to draw a Moon

I've labeled a few points of interest in the figure, which will be showcased here:

Since $\overline{BC}$ is the diameter of the small circumference, and A is a point within the circumference itself, then $\overline{ABC}$ is a right triangle. Furthermore, $\overline{AH}$ is perpendicular to $\overline{BC}$ , thus $\overline{AH}$ is the altitude of the triangle with respect to the hypotenuse.

Let $R$ be the radius of the large circumference. Thus, it is obvious that the following segments measure, respectively:

$\overline{AH} = R - 4$

$\overline{BH} = R - 6$

$\overline{CH} = R$

From the right angle altitude theorem, we can write: $\overline{AH}^{2} = \overline{BH}*\overline{CH}$

Then:

$(R - 4)^{2} = R*(R - 6)$

$R^{2} - 8R + 16 = R^{2} - 6R$

$2R = 16$

$R = 8$

Since the side of the square is equal to the diameter of the large circle, then its side is $16$ ; thus, the area equals $16^{2} = 256$

Let $x$ radii of the biggest circunference.

We have that $CB=x-6$ and $AB=x-4$ . Now $\triangle ACD$ and $AB$ is its heigth so $\triangle ABC \sim \triangle DBA$ .

$\frac{CB}{AB}=\frac{AB}{BD}$

$\frac{x-6}{x-4}=\frac{x-4}{x}$

$x^2-6x=x^2-8x+16$

$2x=16$

$x=8$

Side of the square is equal to $2x \therefore$ side longitude is $16 \Rightarrow$ area is $\boxed{256}$

Let the center of the large circle be O and its radius be r

Let the contact point of the two circles be A

The horizontal diameter of the large circle cuts the small circle at C

The vertical diameter of the large circle cuts the small circle at B above the

horizontal diameter

In the small circle

AC is a diameter

Then

angle ABC is right

In the right triangle ABC

Since BO is perpendicular to AC

Then

(BO)^2 =OA X OC

(r - 4)^2 = r (r - 6)

r = 8

The area of the large square = 16^2 = 256

Let x be the radius of the large circle. Then connect the three points shown to make a triangle. By using the Pythagorean theorem we have the following equality. [(x-6)² + (x-4)²] + [(x-4)² + x²] = [(x-6)+x]² => x = 8 so area of large square is (2x)² = 16² = 256 sq units

$r+y=\dfrac{2r+6}{2}$

$2r+2y=2r+6$

$2y=6$

$y+3$

$2r+6=8+2x$

$2r+2+2x$

$r=1+x$ $\color{#D61F06}(1)$

By the pythagorean theorem, we have

$r^2=x^2+y^2$ $\color{#D61F06}(2)$

$\implies$ $\text{Substitute y=3 and equation} \color{#D61F06}(1)$ in $\color{#D61F06}(2)$

$(1+x)^2=x^2+3^2$

$1+2x+x^2=x^2+9$

$1+2x=9$

$2x=8$

$x=4$

Therefore the area of the big square is four times the area of the small square, we have

$A=4(4+x)^2=4(4+4)^2=4(64)=256$

The diagram on the left shows the problem as stated. The large circle is irrelevant and was omitted. Point O is at the center of the square. Point A is where the circle intersects the y-axis above and point B is where the circle intersects the x-axis to the left. The diagram on the right shows the circle re-positioned such that it is now centered on point O. Points C and D represent the upper and lower intersections between the circle and the y-axis. Line CD also represents the diameter of the circle. Points A' and B' represent the same points on the circle which now lie in new locations within the square. Triangle CAA' is a right triangle with legs of CA=1 and AA'=3. Since line CD is the diameter of the circle, angle CA'D must be a right angle. Thus, by the Pythagorean theorem, DA * AC = 3^2, so DA * 1 = 9, thus DA = 9. Therefore the diameter of the circle (line CD) is 10. Thus, the length of a side of the square is 16 and the area of the square is 256. In my initial solution, I defined a triangle with vertexes at C, O, and the midpoint of CA'. Then I used trig to find the radius (CO) of the circle. As I was writing that up, I noticed I could use Pythagoras instead.

I didn't include the larger circle in the image because it is irrelevant. Let the square be of side: $2a$ , with its surface area being $4a^2$ .

The radius of the circle is: $\frac{2a-6}{2} = \scriptstyle a-3 \Rightarrow OB = OA = a-3$ .

$\scriptstyle OC= OB - BC = (a-3) - (a-6) = 3$

$\scriptstyle OA^2= AC^2 + OC^2 \space \Rightarrow \space (a-3)^2 = (a-4)^2 + 9$

$\scriptstyle a^2 -6a + 4 = a^2 - 8a + 16 + 9 \space \Rightarrow \space a = 8 \space \Rightarrow \space A = 256$

By looking at the diagram, you can see that the line of length 4 divides the height of the small square into 2. So, the height of the side of the small square is 8. So the area of the small square is 64. As there are 4 such squares, the total area is 256. It really is a lot about intuition. Not a very recommended method but fast.

Since the line of 4 is half of the smaller square therefore 8 is the total height and base if the smaller square. 8 * 8= 64 and since there are 4 squares 64 * 4= 256

Using notation as below and assuming r as the radius of the small circle: Diameter of large circle = CB + 8 = 2r + 6. So radius of large circle = CH = r + 3. BH = (r + 3) - 6 = r - 3. AH = HD = (r + 3) - 4 = r - 1 (assuming D is the intersection of AH produced on the small circle). But AH HD = BH HC. So (r + 3) (r - 3) = (r - 1) (r - 1) Solving for r, we have r = 5. So diameter of the large circle = 2r + 6 = 16 and area of the large square = 256.

The diameter of the large circle is 2s where s is the unknown side half-length of the large square. This makes the diameter of the small circle 2s -6 and then the radius s - 3 and if the central lines cross at the origin of a Cartesian coordinate system, the small circle has two x-intercepts: (s, 0) and (-s + 6, 0) and thus has center ( (s + -s + 6)/2, 0 ) or (3,0)

This makes the equation of the smaller circle

(x - 3)^2 + y^2 = (s -3)^2

We know some points on the circle, like the uppermost y-intercept (0, s-4)

Plug that in for x = 0 and y = 4 and solve for s = 8

The square has sides 2*s = 16 and area s^2 = 256.

Nice puzzle!

$ABC$ is a right triangle . Therefore $\overline{AB}^2+\overline{BC}^2=\overline{AC}^2$ .

Express the Pythagorean identity in terms of $x,y$ and $r$ : $(x^2+y^2)+(y^2+r^2)=(x+r)^2$ .

Simplify to get $y^2=x\cdot r$

Substitute $x=r-6$ and $y=r-4$ to get $(r-4)^2=r(r-6) \implies r^2+16-8r=r^2-6r \implies r=8$ .

The area of the large square is then $4r^2=\boxed{256}$ .

I’m probably the only person who solved it with pixel measurements lol

$6+AB=4+BD$ $\implies$ $BD=2+AB$ or $AB=BD-2$

$BC=\dfrac{2FD+6}{2}=FD+3$

$FD+3=4+BD$ $\implies$ $FD=BD+1$

$BF=FD-AB=BD+1-(BD-2)=BD+1-BD+2=3$

Using pythagorean theorem, we have

$FD^2=BD^2+BF^2$

$(BD+1)^2=BD^2+9$

$BD^2+2BD+1=BD^2+9$

$2BD=8$

$BD=4$

So the radius of the big circle is $4+4=8$ .

Thus, the area of the square is $16^2=$ $\boxed{256}$ .

The figure looks to scale. So what I did was look at the smaller inside squares instead of the circles. It looks like they have side lengths of 8, making the bigger squares side lengths 16. A = l X w. 16 X 16 = 256.

If you compare 4 and 6 you end up 8 as your small square size double to 16 multiply by 2 and there's your answer of 256 without all the head racking