2020 AMC 10A Problem 20

Question

Quadrilateral

ABCD

satisfies

\angle ABC = \angle ACD = 90^{\circ},

AC=20,

and

CD=30.

Diagonals

\overline{AC}

and

\overline{BD}

intersect at point

E,

and

AE=5.

What is the area of quadrilateral

ABCD?

330

340

350

360

370

Answer: D

Concepts:coordinate geometry inscribed angle area decomposition

Difficulty rating: 2150

Small hint:

Put

AC

on the

x

-axis and use the known point

E

Big hint:

Point

B

lies both on line

DE

and on the circle with diameter

AC

Video solution:

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Written solution:

Place

A=(0,0)

and

C=(20,0)

. Since

\angle ACD=90^\circ

and

CD=30

, take

D=(20,30)

. The point

E

is

(5,0)

, so line

BD

has equation

y=2(x-5)

.

Because

\angle ABC=90^\circ

, point

B

lies on the circle with diameter

AC

:

(x-10)^2+y^2=100

. Intersecting with

y=2(x-5)

gives

x=2

or

10

. The convex quadrilateral uses

B=(2,-6)

.

Then

[ACD]=\dfrac12\cdot20\cdot30=300

, and

[ABC]=\dfrac12\cdot20\cdot6=60

. The total area is

360

. Thus, D is the correct answer.