Sample Question 22:

The diameter \(AB\) 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 \(AE\) intersects the circle at a point \(C\) between \(A\) and \(E\). What is the area of \(\triangle ABC\)?

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

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Answer Keys

Question 22: D

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Solutions

Question 22
Step 1: ADE and ABC are both right triangles. You can solve for AE using the Pythagorean theorem and get \(AE = \sqrt{7^2+5^2} = \sqrt{74}\).

Step 2: As the triangles ADE and ABC are similar, the sine of angle DAE is the same as the sine of angle BAC. We have \(\sin{DAE} = \frac{5}{\sqrt{74}} = \sin{BAE} = \sin{BAC} = \frac{BC}{4}\), and hence, we can find BC as \(BC = \frac{20}{\sqrt{74}}\).

Step 3: Similarly, we find that \(AC = \frac{28}{\sqrt{74}}\).

Step 4: The area of triangle ABC is given by the half the product of its base and height. So, the area of triangle ABC is \(\frac{\frac{20}{\sqrt{74}}\cdot\frac{28}{\sqrt{74}}}{2} = \frac{\frac{560}{74}}{2} = \frac{140}{37}\).

Therefore, the area of \(\triangle ABC\) is \(\frac{140}{37}\) which corresponds to option \(\textbf{(D) } \frac{140}{37}\).