2011 AMC 10B Problem 14

Below is the professionally curated solution for Problem 14 of the 2011 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 10B solutions, or check the answer key.

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Concepts:rectanglePythagorean Theoremalgebraic manipulation

Difficulty rating: 1370

14.

A rectangular parking lot has a diagonal of 2525 meters and an area of 168168 square meters. In meters, what is the perimeter of the parking lot?

5252

5858

6262

6868

7070

Solution:

Let the side lengths be l,w.l,w. We wish to find 2(l+w).2(l+w). From the Pythagorean Theorem, we get 25=l2+w225 = \sqrt{l^2+w^2}l2+w2=625l^2+w^2 = 625 We also know lw=168.lw = 168.

As such l2+2lw+w2=(l+w)2=961=312.\begin{align*}l^2+2lw+ w^2 &= (l+w)^2 \\&= 961 \\&= 31^2.\end{align*} This makes l+w=31,l+w = 31, and as such, our answer is 312=62.31\cdot 2=62.

Thus, the correct answer is C .

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