2017 AMC 10B Problem 22

Draw the segment $BC$. Since $AB$ is a diameter and $C$ is on the circle, the angle $ACB$ is a right angle (any angle that "looks at" a diameter from the circle is $90^\circ$). Now there are two right triangles in the picture that share the corner $A$.

1. The big triangle $ADE$. It has a right angle at $D$. Its sides: $AD = AB + BD = 4 + 3 = 7$ and $DE = 5$. So
$$AE = \sqrt{7^2 + 5^2} = \sqrt{74}, \qquad \text{area of } ADE = \frac12 \cdot 7 \cdot 5 = \frac{35}{2}.$$
2. Same shape. Triangle $ACB$ has a right angle at $C$, triangle $ADE$ has a right angle at $D$, and both have the same angle at $A$ (the segment $AC$ lies along $AE$ and $AB$ lies along $AD$). So the two triangles are similar.
3. Matching sides. The longest sides (opposite the right angles) match: $AB = 4$ in the small triangle and $AE = \sqrt{74}$ in the big one. So every length in the small triangle is $\frac{4}{\sqrt{74}}$ times the matching length in the big one.
4. Areas. Areas scale by the square of that factor, $\frac{16}{74}$:
$$\text{area of } ABC = \frac{35}{2} \cdot \frac{16}{74} = \frac{35 \cdot 8}{74} = \frac{280}{74} = \frac{140}{37}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{140}{37}}$.

A point on a circle together with a diameter always gives a right angle at that point. Here that right angle and the right angle at $D$ make two similar triangles sharing corner $A$, so one ratio of sides, squared, gives the area. We never needed the exact position of $C$.

Use coordinates: $A = (-2, 0)$, $B = (2, 0)$, $D = (5, 0)$, $E = (5, 5)$. Because angle $ACB$ is a right angle, $BC$ is the shortest distance from $B$ to line $AE$. In right triangle $ABC$, $BC = AB \cdot \frac{DE}{AE} = 4 \cdot \frac{5}{\sqrt{74}} = \frac{20}{\sqrt{74}}$ and $AC = AB \cdot \frac{AD}{AE} = \frac{28}{\sqrt{74}}$. The area is $\frac12 \cdot \frac{20}{\sqrt{74}} \cdot \frac{28}{\sqrt{74}} = \frac{280}{74} = \frac{140}{37}$.