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 the radius is $2$ and $BD =3,$ we have $AD = 7.$ Since $ED = 5$ and the angle at $D$ is a right angle, the area of $ADE$ is $\dfrac{5\cdot 7}{2} = \dfrac{35}{2} .$

Also, the value of $AE$ is $\sqrt{5^2+7^2} = \sqrt{74}$ by the Pythagorean Theorem. Also, $\angle ACB$ is a right angle since $AB$ is a diameter. Thus, by angle-angle symmetry, we have $ACB \sim ADE.$

This means the area of $ABC$ is equal to the area of $AED$ times $\dfrac{AB}{AE}^2 = \left(\dfrac{4}{\sqrt{74}}\right)^2 = \dfrac{8}{37} .$ Then, we have an area of $\dfrac{8}{37} \cdot \dfrac{35}2 = \dfrac{140}{37} .$

Thus, the correct answer is D .

2017 AMC 10B Problem 22

Solution:

Problem 22 in Other Years