2017 AMC 10B Problems/Problem 22

Problem

The diameter $\overline{AB}$ of a circle of radius $2$ is extended to a point $D$ outside the circle so that $BD=3$ . Point $E$ is chosen so that $ED=5$ and line $ED$ is perpendicular to line $AD$ . Segment $\overline{AE}$ intersects the circle at a point $C$ between $A$ and $E$ . What is the area of $\triangle ABC$ ?

$\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}$

Solution 1

Notice that $ADE$ and $ABC$ are right triangles. Then $AE = \sqrt{7^2+5^2} = \sqrt{74}$ . $\sin{DAE} = \frac{5}{\sqrt{74}} = \sin{BAE} = \sin{BAC} = \frac{BC}{4}$ , so $BC = \frac{20}{\sqrt{74}}$ . We also find that $AC = \frac{28}{\sqrt{74}}$ , and thus the area of $ABC$ is $\frac{\frac{20}{\sqrt{74}}\cdot\frac{28}{\sqrt{74}}}{2} = \frac{\frac{560}{74}}{2} = \boxed{\textbf{(D) } \frac{140}{37}}$ .

Solution 2

We note that $\triangle ACB ~ \triangle ADE$ by $AA$ similarity. Also, since the area of $\triangle ADE = \frac{7 \cdot 5}2 = \frac{35}2$ and $AE = \sqrt{74}$ , $\frac{[ABC]}{[ADE]} = \frac{[ABC]}{\frac{35}2} = (\frac{4}{\sqrt{74}})^2$ , so the area of $\triangle ABC = \boxed{\textbf{(D) } \frac{140}{37}}$ .

2017 AMC 10B (Problems • Answer Key • Resources)
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2017 AMC 10B Problems/Problem 22

Contents

Problem

Solution 1

Solution 2

See Also