2014 AMC 10B Problem 21

Question

Trapezoid

ABCD

has parallel sides

\overline{AB}

of length

33

and

\overline {CD}

of length

21 .

The other two sides are of lengths

10

and

14 .

The angles

A

and

B

are acute. What is the length of the shorter diagonal of

ABCD ?

10\sqrt{6}

25

8\sqrt{10}

18\sqrt{2}

26

Answer: B

Concepts:trapezoid Pythagorean Theorem

Difficulty rating: 1790

Small hint:

Drop perpendiculars from the shorter base to the longer base

Big hint:

Use the two leg lengths to find the horizontal offsets

Solution:

Let the feet of the perpendiculars from

D

and

C

to

AB

be

E

and

F

, respectively. As drawn, take

AD=10

and

BC=14

; interchanging the two legs only reflects the trapezoid.

Let

AE=x

and let the altitude be

h

. Since

EF=CD=21

and

AB=33

, we have

FB=12-x

.

The two right triangles give

10^2=x^2+h^2

and

14^2=(12-x)^2+h^2.

Subtracting yields

96=144-24x

, so

x=2

and

h^2=96

.

The shorter diagonal is

AC

, whose horizontal displacement is

AE+EF=2+21=23

. Therefore

AC=\sqrt{23^2+96}=\sqrt{625}=25.

Thus, the correct answer is B .