Question

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 $\triangle ABC?$

$\dfrac{120}{37}$

$\dfrac{140}{39}$

$\dfrac{145}{39}$

$\dfrac{140}{37}$

$\dfrac{120}{31}$

Solution:

Since $\angle ACB$ is inscribed in a semicircle, it is a right angle, so $\triangle ABC \sim \triangle AED$ (both right-angled and sharing angle $A$ ). Their areas are in ratio $AB^2 : AE^2.$ Here $AB = 4,$ so $AB^2 = 16,$ and $AD = AB + BD = 7,$ so $AE^2 = AD^2 + ED^2 = 49 + 25 = 74.$ The area of $\triangle AED$ is $\tfrac12 \cdot 7 \cdot 5 = \tfrac{35}{2}.$ Thus $[\triangle ABC] = \frac{16}{74} \cdot \frac{35}{2} = \frac{140}{37}.$

Thus, the correct answer is D.

2017 AMC 12B Problem 18

Solution:

Problem 18 in Other Years