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Solution

--PLEASE DO NOT REMOVE TILL 4/6/14-- --APPEND YOUR CHANGES BEFORE THIS NOTE--

We want to find the area of this figure:

$[asy] path rt,tt, tri; real x, y; y = 1+sqrt(2); x = y+(6/1.7); tt=(0,0)..(y,1)--(y,-1)..cycle; rt=(y,-1)--(x,-1)--(x,1)--(y,1); tri=(y,-1)--(y-1,0)--(y,1); draw(rt); draw(tt); draw(tri); label("1.7", (x, 0), E); label("3", (y+(3/1.7), -1), S); label("C", (y-1, -0.1), S); [/asy]$

We label the circle as circle C. We can break the figure into three parts, shown as the 3/4 circle, the triangle, and the rectangle.

Lets first take a look at the rectangle.

$[asy] path rt; real x, y; y = 1+sqrt(2); x = y+(6/1.7); rt=(y,-1)--(x,-1)--(x,1)--(y,1)--cycle; draw(rt); label("1.7", (x, 0), E); label("3", (y+(3/1.7), -1), S); [/asy]$

It has an area of $3 * 1.7 = 5.1$ .

Lets now take a look at the triangle, after drawing the height.

$[asy] unitsize(0.8inch); path tri, lin; real x, y; y = 1+sqrt(2); x = y+(6/1.7); tri=(y,-1)--(y-1,0)--(y,1)--cycle; lin=(y-1, 0)--(y,0); draw(tri); draw(lin); label("1.7", (y, 0), E); label("C", (y-1, -0.1), S); [/asy]$

We see that both the radii are the two shorter sides of the triangle, making this a isosceles 45-45-90 triangle.

We also see that the height that we drew is half the hypotenuse(note the two smaller 45-45-90 isosceles triangles).

Hence, the area of the triangle is $\frac{1.7 * \frac{1.7}{2}}{2} = 0.7225$ .

Now, let's take a look at the 3/4 circle. We know it is 3/4 because there is a 90 degree triangle cut out of it.

$[asy] unitsize(0.6inch); path tt, tri; real x, y; y = 1+sqrt(2); x = y+(6/1.7); tt=(0,0)..(y,1)--(y-1, 0)--(y,-1)..cycle; tri=(y, 1)--(y,-1); draw(tt); draw(tri); fill(tt, gray(0.6)); label("1.7", (y, 0), E); label("C", (y-1, -0.1), S); [/asy]$

We find the radius using the 45-45-90 triangle. Since the ratios of the sides are 1:1: $\sqrt{2}$ , we can find the radius to be $\frac{1.7}{\sqrt{2}} = \frac{1.7 \sqrt{2}}{2}$ .

Hence, the area of the whole circle is $\pi r^2$ , and the area of the 3/4 circle is $\frac{3}{4} * \pi * (\frac{1.7 \sqrt{2}}{2})^2$ .

Adding it all up, we find the answer to be $\frac{3}{4} \pi (\frac{1.7 \sqrt{2}}{2})^2 + 0.7225 + 5.1$ .

Just plug it into your calculator. And please understand where all the numbers came from before writing the answer down.

We want to find the area of the intersection of the circles in this figure:

$[asy] //import graph; draw(circle((0,0), 2)); draw(circle((2,0), 2)); path inter; inter=(1,-sqrt(3))..(2,0)..(1, sqrt(3))--cycle; draw(inter); fill(inter, gray(0.6)); path inter2; inter2=(1,-sqrt(3))..(0,0)..(1, sqrt(3))--cycle; draw(inter2); fill(inter2, gray(0.6)); [/asy]$

Lets call the radius of each of the circles 1, because we are calculating probability. To do this, we need to divide the intersection area into smaller parts that we can find the area of.

$[asy] //import graph; draw(circle((0,0), 2)); draw(circle((2,0), 2)); path a, b, c; a = (0,0)--(1, sqrt(3))--(2,0)--cycle; b = (0,0)--(1, -sqrt(3))--(2,0); c = (0,0)..(2-sqrt(3), 1)..(1, sqrt(3))--cycle; draw(a); fill(a, gray(0.6)); draw(b); label("A", (0,0), W); label("B", (2,0), E); label("C", (1,sqrt(3)), N); label("D", (1,-sqrt(3)), S); label("1", (1.5,sqrt(3)/2), NE); label("1", (1,0), S); [/asy]$

Lets examine the shaded $\triangle ABC$ . Since each of its sides are radii, it is equilateral.

Hence, each of its angles measure 60 degrees. Namely, $\angle ABC$ measures 60 degrees.

Important note: The area of an equilateral triangle is calculated by $\frac{s^2\sqrt{3}}{4}$ , where $s$ is the side length of a triangle. This is an EXTREMELY important formula to know, and it it also extremely simple to prove and memorize.

After this, we can deduce that the area of triangle ABC with side length 1 is $\frac{1^2\sqrt{3}}{4} = \frac{\sqrt{3}}{4}$

Let us now look at the encompassing sector.

$[asy] //import graph; draw(circle((0,0), 2)); draw(circle((2,0), 2)); path a, b, c; a = (0,0)--(1, sqrt(3))--(2,0)--cycle; b = (0,0)--(1, -sqrt(3))--(2,0); c = (0,0)..(2-sqrt(3), 1)..(1, sqrt(3))--cycle; draw(a); fill(a, gray(0.6)); draw(c); fill(c, gray(0.6)); draw(b); label("A", (0,0), W); label("B", (2,0), E); label("C", (1,sqrt(3)), N); label("D", (1,-sqrt(3)), S); label("1", (1.5,sqrt(3)/2), NE); label("1", (1,0), S); [/asy]$ We then find that the area of the entire sector $ABC$ is $\frac{60}{360} * \pi * 1^2 = \frac{\pi}{6}$ .

Subtracting the area of $\triangle ABC$ from sector $ABC$ gives us the shaded part below:

$[asy] //import graph; draw(circle((0,0), 2)); draw(circle((2,0), 2)); path a, b, c; a = (0,0)--(1, sqrt(3))--(2,0)--cycle; b = (0,0)--(1, -sqrt(3))--(2,0); c = (0,0)..(2-sqrt(3), 1)..(1, sqrt(3))--cycle; draw(a); draw(c); fill(c, gray(0.6)); draw(b); label("A", (0,0), W); label("B", (2,0), E); label("C", (1,sqrt(3)), N); label("D", (1,-sqrt(3)), S); label("1", (1.5,sqrt(3)/2), NE); label("1", (1,0), S); [/asy]$

Plugging in the values we know, we find that this has area $\frac{\pi}{6} - \frac{\sqrt{3}}{4}$ .

We see that there are 4 such small curved areas.

$[asy] //import graph; draw(circle((0,0), 2)); draw(circle((2,0), 2)); path a, b, c, d, e, f; a = (0,0)--(1, sqrt(3))--(2,0)--cycle; b = (0,0)--(1, -sqrt(3))--(2,0); c = (0,0)..(2-sqrt(3), 1)..(1, sqrt(3))--cycle; d = (0,0)..(2-sqrt(3), -1)..(1, -sqrt(3))--cycle; e = (2,0)..(sqrt(3), -1)..(1, -sqrt(3))--cycle; f = (2,0)..(sqrt(3), 1)..(1, sqrt(3))--cycle; draw(a); //draw(c); fill(c, gray(0.6)); //draw(d); fill(d, gray(0.6)); //draw(e); fill(e, gray(0.6)); //draw(f); fill(f, gray(0.6)); draw(b); label("A", (0,0), W); label("B", (2,0), E); label("C", (1,sqrt(3)), N); label("D", (1,-sqrt(3)), S); label("1", (1,0), S); [/asy]$

Multiplying the value we got for an individual small curved part by 4, we find the area of the above shaded region to be $\frac{4\pi}{6} - \frac{4\sqrt{3}}{4}$ .

We then need to add back the areas of the equilateral triangles.

$[asy] //import graph; draw(circle((0,0), 2)); draw(circle((2,0), 2)); path a, b, c, d, e, f; a = (0,0)--(1, sqrt(3))--(2,0)--cycle; b = (0,0)--(1, -sqrt(3))--(2,0)--cycle; c = (0,0)..(2-sqrt(3), 1)..(1, sqrt(3))--cycle; d = (0,0)..(2-sqrt(3), -1)..(1, -sqrt(3))--cycle; e = (2,0)..(sqrt(3), -1)..(1, -sqrt(3))--cycle; f = (2,0)..(sqrt(3), 1)..(1, sqrt(3))--cycle; draw(a); fill(a, gray(0.6)); //draw(c); //fill(c, gray(0.6)); //draw(d); //fill(d, gray(0.6)); //draw(e); //fill(e, gray(0.6)); //draw(f); //fill(f, gray(0.6)); draw(b); fill(b, gray(0.6)); label("A", (0,0), W); label("B", (2,0), E); label("C", (1,sqrt(3)), N); label("D", (1,-sqrt(3)), S); label("1", (1,0), S); [/asy]$

Recall that the area of each equilateral triangle was $\frac{\sqrt{3}}{4}$ .

Hence the area of the above shaded region is $\frac{2\sqrt{3}}{4}$ .

We then look at the total.

$[asy] //import graph; draw(circle((0,0), 2)); draw(circle((2,0), 2)); path a, b, c, d, e, f; a = (0,0)--(1, sqrt(3))--(2,0)--cycle; b = (0,0)--(1, -sqrt(3))--(2,0)--cycle; c = (0,0)..(2-sqrt(3), 1)..(1, sqrt(3))--cycle; d = (0,0)..(2-sqrt(3), -1)..(1, -sqrt(3))--cycle; e = (2,0)..(sqrt(3), -1)..(1, -sqrt(3))--cycle; f = (2,0)..(sqrt(3), 1)..(1, sqrt(3))--cycle; draw(a); fill(a, gray(0.6)); draw(c); fill(c, gray(0.6)); draw(d); fill(d, gray(0.6)); draw(e); fill(e, gray(0.6)); draw(f); fill(f, gray(0.6)); draw(b); fill(b, gray(0.6)); label("A", (0,0), W); label("B", (2,0), E); label("C", (1,sqrt(3)), N); label("D", (1,-sqrt(3)), S); label("1", (1,0), S); [/asy]$

Hence the area of the total shaded region is $\frac{2\sqrt{3}}{4} + \frac{4\pi}{6} - \frac{4\sqrt{3}}{4} = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}$ .

We plug this into the calculator to find the final answer.

In order to find the union(total) of the two circles, we examine the figure.

We want to find the area of the intersection of the circles in this figure:

$[asy] //import graph; fill((circle((2,0), 1.98)), gray(0.6)); fill((circle((0,0), 1.98)), gray(0.6)); draw(circle((0,0), 2)); draw(circle((2,0), 2)); draw((0,0)--(2,0)); label("1", (1,0), S); [/asy]$

Using $\pi r^2$ , the area of the circle on the left is $\pi$ . We also know that the area of the circle on the right is $\pi$ . When we add these together, we are counting the entire area, but we are double-counting the overlap area. We thus need to subtract the overlap from the $2\pi$ in order to find the union(total) of the figure.

Hence, the total is $2\pi - (\frac{2\pi}{3} - \frac{\sqrt{3}}{2}) = \frac{4\pi}{3} + \frac{\sqrt{3}}{2}$ .

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