2017 AMC 10B Problem 22

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}

Use the semicircle angle to get a right triangle

Compare

\triangle ABC

and

\triangle AED

by similarity

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 .

Problem 22 in other years