2021 AMC 12A Fall Problem 21

Question

Let

ABCD

be an isosceles trapezoid with

\overline{BC} \parallel \overline{AD}

and

AB = CD.

Points

X

and

Y

lie on diagonal

\overline{AC}

with

X

between

A

and

Y,

as shown in the figure. Suppose

\angle AXD = \angle BYC = 90^\circ,

AX = 3,

XY = 1,

and

YC = 2.

What is the area of

ABCD?

15

5\sqrt{11}

3\sqrt{35}

18

7\sqrt{7}

Place

A, X, Y, C

on a horizontal line with

AX = 3,

XY = 1,

YC = 2.

Then

D

is vertically below

X

and

B

vertically above

Y

Use

\overline{BC}\parallel\overline{AD}

and

AB = CD

to relate the heights of

B

and

D,

then apply the shoelace formula

Put

A = (0,0),

X = (3,0),

Y = (4,0),

C = (6,0).

The right angles give

D = (3, t)

and

B = (4, s)

on opposite sides of

AC.

Parallelism

\overline{AD}\parallel\overline{BC}

forces

t = -\tfrac{3}{2}s,

and

AB = CD

gives

16 + s^2 = 9 + t^2,

so

t^2 - s^2 = 7.

Substituting yields

s^2 = \tfrac{28}{5}.

The shoelace formula gives area

= 3\,|t - s|

= 3\cdot\tfrac{5}{2}s

= \tfrac{15}{2}s

= \tfrac{15}{2}\sqrt{\tfrac{28}{5}}

= 3\sqrt{35}.

Thus, the correct answer is C.

Problem 21 in other years