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2011 AMC 10B Problem 14

2011 AMC 10B Problem 14AMC 10 Step by Step

Name: Date: September 13, 2026

Algebraic Manipulation · difficulty 2 of 5 · about 3 min · position P11-15

A rectangular parking lot has a diagonal of $25$ meters and an area of $168$ square meters. In meters, what is the perimeter of the parking lot?

A)
$52$
B)
$58$
C)
$62$
D)
$68$
E)
$70$

Answer

Key insight

Diagonal gives l^2 + w^2 = 625, area gives lw = 168, so (l + w)^2 = 625 + 336 = 961 and l + w = 31.

Solution

Let the sides be $l$ and $w$ . The diagonal is the hypotenuse of a right triangle with legs $l$ and $w$ , so $l^2 + w^2 = 25^2 = 625$ . The area gives $lw = 168$ .

The perimeter is $2(l + w)$ , and
$(l + w)^2 = l^2 + w^2 + 2lw = 625 + 336 = 961 = 31^2.$
So $l + w = 31$ and the perimeter is $2 \cdot 31 = 62$ .

The answer is $\boxed{\textbf{(C)}\ 62}$ .

Why this works

Sums of squares and products combine into the square of a sum, so you never need the individual sides. This is the standard move whenever a problem hands you $x^2 + y^2$ and $xy$ and asks about $x + y$ (or the reverse).

Alternative approach

Recognize the $7$ - $24$ - $25$ Pythagorean triple: $7 \cdot 24 = 168$ confirms it, so the sides are $7$ and $24$ and the perimeter is $2(7 + 24) = 62$ .

The trap

Trying to solve for l and w individually with a quartic, or adding lw only once when forming (l + w)^2.

Common mistakes

Trying to solve for $l$ and $w$ individually with a quartic, or adding $lw$ only once when forming $(l + w)^2$ .
Reporting $l + w = 31$ instead of the perimeter $62$ .

Techniques

Apply an identity: SFFT, sum of squares, difference of cubes, Vieta