1951 AMC 12 Problem 2

Attempt Problem 2 of the 1951 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 1951 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

A rectangular field is half as wide as it is long and is completely enclosed by

x

yards of fencing. The area in terms of

x

is:

\dfrac{x^2}{2}

2x^2

\dfrac{2x^2}{9}

\dfrac{x^2}{18}

\dfrac{x^2}{72}

Let the width be

w

, so the length is

2w

Use the perimeter equation

2(w+2w)=x

before finding the area

Let the width be

w

, so the length is

2w.

The fencing gives

2(w+2w)=6w=x,

hence

w=\frac{x}{6}.

Therefore

A=w(2w)=2\left(\frac{x}{6}\right)^2=\frac{x^2}{18}.

Thus, the correct answer is D.

Problem 2 in other years