2020 AMC 12A Problem 18

Attempt Problem 18 of the 2020 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 12A solutions, or check the answer key.

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18.

Quadrilateral

ABCD

satisfies

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

AC = 20,

and

CD = 30.

Diagonals

AC

and

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 circle area decomposition

Difficulty rating: 1800

Small hint:

Place

A = (0,0),

C = (20, 0),

and

D = (20, 30),

so

E = (5, 0)

Big hint:

Since

\angle ABC = 90^\circ,

B

lies on the circle with diameter

AC;

it also lies on line

DE

Solution:

Place

A = (0,0)

and

C = (20, 0).

Since

\angle ACD = 90^\circ,

D = (20, 30),

and

E = (5, 0)

because

AE = 5.

Since

\angle ABC = 90^\circ,

B

lies on the circle of radius

10

centered at

(10, 0).

Line

DE

is

(5 + t,\, 2t);

substituting gives

t^2 - 2t - 15 = 0,

so

t = 5

or

t = -3.

For

E

to lie between

B

and

D,

take

t = -3,

giving

B = (2, -6),

a distance

6

below line

AC.

Then

[ACD] = \tfrac12 \cdot 20 \cdot 30 = 300

and

[ABC] = \tfrac12 \cdot 20 \cdot 6 = 60,

so the total area is

360.

Thus, D is the correct answer.

Problem 18 in other years