The diagram shows eight congruent squares inside a circle. Every shaded square has one vertex on the circle. What is the ratio of the shaded area to the area of the circle? Express your answer as a decimal to the nearest hundredth.

Answer 1

[asy]
pair A,O,B,C,D,EE,F,G,H;

A = (0,0);
B = (17.68,8.68);
C = (10,17.68);
D = (-8.68, 10);
EE = (-17.68,-8.68);
F = (-10,-17.68);
G = (8.68,-10);
H = (6.34, 0);

draw(A--B--C--D--EE--F--G--H--cycle,dashed);
draw(Circle((0,0),20));
fill(A--B--H--cycle,gray);
fill(A--H--G--F--cycle,gray);
fill(A--F--EE--D--cycle,gray);
fill(A--D--C--B--cycle,gray);
draw((-40,0)--(40,0),linewidth(0.7),Arrows(6));
draw((0,-40)--(0,40),linewidth(0.7),Arrows(6));
label("$x$",(40,0),SE);
label("$y$",(0,40),NW);

dot((-10,0));dot((0,0));dot((10,0));dot((20,0));
label("$-2$",(0,0),NW);
label("$-1$",(10,0),NW);
label("$0$",(20,0),NW);
label("$1$",(-10,0),NW);
[/asy]

We label the points as shown. The radius of the circle is labelled 20. Let the sidelength of each square be $s$. From the tangent part of the diagram, $2s = 20$, so $s = 10$.

The area of each square is then $s^2 = 10^2 = 100$. At this point, let's compute the area of the shaded region. We can clearly see that it consists of 8 congruent triangles. Each triangle has base $10$, so we only need to find the height.

[asy]
pair A,O,B,C,D,EE,F,G,H;

O = (0,0);
B = (17.68,8.68);
C = (10,17.68);
D = (-8.68, 10);
EE = (-17.68,-8.68);
F = (-10,-17.68);
G = (8.68,-10);
H = (6.34, 0);

draw(Arc(O,20,0,180));

draw(O--B--C--D--F--G--H,dashed);
draw(O--C--H--cycle);
draw(rightanglemark(O,H,C,40));

draw((-40,0)--(40,0),linewidth(0.7),Arrows(6));
draw((0,-40)--(0,40),linewidth(0.7),Arrows(6));
label("$x$",(40,0),SE);
label("$y$",(0,40),NW);

dot((-10,0));dot((0,0));dot((10,0));dot((20,0));
label("$-2$",(0,0),NW);
label("$-1$",(10,0),NW);
label("$0$",(20,0),NW);
label("$1$",(-10,0),NW);

[/asy]

We compute the area of each triangle by finding the area of the rectangle (since it has two right angles) that it is a part of and then divide by $2$. We know that the length of the rectangle is $10$, so we need to find the width and height. Using the Pythagorean Theorem, we know that $x^2 + y^2 = 20^2 = 400$, so $y = \pm \sqrt{400 - x^2}$. Looking at our diagram suggests that $y$ is positive, so $y = \sqrt{400 - x^2}$.

The area of our rectangle is then $10 \cdot \sqrt{400 - x^2}$. Now we must find the height of each triangle. Recall that the height of a right triangle is the side opposite the right angle. So, we have $10-2s = 10 - 2 \cdot 10 = -10$. But our answer must be positive! Don't forget that we subtracted positive $2s$. To get a positive value, let's just take the absolute value of $10 - 2s$, which gives us $|-10| = 10$.

To find the area of each triangle and hence the shaded area, we multiply the base by the height, and divide by 2. So $\frac{10^2 \cdot \sqrt{400 - x^2}}{2} = 50 \sqrt{400 - x^2}$.

Now, let's divide by the area of the circle. The area of the circle is $\pi \cdot 20^2 = 400 \pi$. Thus, the ratio is $\frac{50 \sqrt{400 - x^2}}{400 \pi}$. Furthermore, the ratio is the same for all $x$ because all 8 shaded triangles are congruent, and they comprise the entire shaded area. Hence we can use any value of $x$ in our formula. Let's use $x=2$ because it gives us a clean right triangle.

We plug in $x=2$ to get $\frac{50 \cdot \sqrt{400 - 2^2}}{400 \pi} = \frac{50 \cdot \sqrt{394}}{400 \pi}$. We rationalize the denominator of the square root expression:
$$\frac{50 \cdot \sqrt{394}}{400 \pi} =\frac{50 \cdot \sqrt{394}(20 \sqrt{394})}{400 \pi(20 \sqrt{394})}=\frac{50 \cdot 20 \cdot 394}{400\pi \cdot 20 \sqrt{394}}=\boxed{\frac{197}{40\sqrt{394}\pi}}.$$