1994 AMC 12 Problem 2

Attempt Problem 2 of the 1994 AMC 12 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 1994 AMC 12 solutions, or check the answer key.

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A large rectangle is partitioned into four rectangles by two segments parallel to its sides. The areas of three of the resulting rectangles are shown. What is the area of the fourth rectangle?

$6$	$14$
?	$35$

10

15

20

21

25

Answer: B

Concepts:rectangle areas proportionality

Difficulty rating: 960

Small hint:

Represent the two column widths and two row heights by variables

Big hint:

Products of diagonally opposite areas are equal

Solution:

Let the column widths be

u,v

and the row heights be

s,t.

The displayed areas give

us=6,

vs=14,

and

vt=35.

Therefore the missing area is

ut=\frac{(us)(vt)}{vs}=\frac{6\cdot35}{14}=15.

Thus the correct answer is B.