2017 AMC 10B Problem 22

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 $\dfrac{5\cdot 7}{2} = \dfrac{35}{2} .$ By the Pythagorean Theorem, $AE=\sqrt{5^2+7^2}=\sqrt{74}.$ Also, $\angle ACB$ is a right angle because $AB$ is a diameter. The triangles share the angle at $A,$ so $\triangle ABC\sim\triangle AED$ by angle-angle similarity. Their corresponding hypotenuses are $AB=4$ and $AE=\sqrt{74},$ so their area ratio is $$\dfrac{[ABC]}{[AED]}=\left(\dfrac4{\sqrt{74}}\right)^2=\dfrac8{37}.$$ Therefore, $$[ABC]=\dfrac8{37}\cdot\dfrac{35}{2}=\dfrac{140}{37}.$$ Thus, the correct answer is D .