2001 AMC 10 Problem 24

Attempt Problem 24 of the 2001 AMC 10 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 2001 AMC 10 solutions, or check the answer key.

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24.

In trapezoid

ABCD,

\overline{AB}

and

\overline{CD}

are perpendicular to

\overline{AD},

with

AB+CD=BC,

AB\lt CD,

and

AD=7.

What is

AB\cdot CD?

12

12.25

12.5

12.75

13

Answer: B

Concepts:trapezoid Pythagorean Theorem difference of squares

Difficulty rating: 1810

Small hint:

Drop a perpendicular from

B

to

CD

at

E;

then

BE=AD=7

and

CE=CD-AB

Big hint:

BC^2=BE^2+CE^2

with

BC=CD+AB

Solution:

Drop a perpendicular from

B

to

CD,

meeting it at

E.

Then

BE=AD=7

and

CE=CD-AB.

By the Pythagorean theorem,

BC^2=BE^2+CE^2.

Since

BC=CD+AB,

\begin{aligned} &(CD+AB)^2 \\ &\quad {}-(CD-AB)^2 \\ &\quad =BE^2=49. \end{aligned}

The left side equals

4\cdot AB\cdot CD,

so

AB\cdot CD=\dfrac{49}{4}=12.25.

Thus, the correct answer is B.