2002 AMC 10P Problems/Problem 6

Problem

The perimeter of a rectangle $100$ and its diagonal has length $x.$ What is the area of this rectangle? $\text{(A) }625-x^2 \qquad \text{(B) }625-\frac{x^2}{2} \qquad \text{(C) }1250-x^2 \qquad \text{(D) }1250-\frac{x^2}{2} \qquad \text{(E) }2500-\frac{x^2}{2}$

Solution 1

Let $l$ be the length of the rectangle and $w$ be the width of the rectangle. We are given $2l+2w=100$ and $l^2+w^2=x^2.$ We are asked to find the area, which is equivalent to $lw.$ Using a bit of algebraic manipulation: $\begin{align*} 2l+2w &= 100 \\ l+w &= 50 \\ (l+w)^2 &= 2500 \\ l^2 + w^2 +2lw &= 2500 \\ x^2 + 2lw &= 2500 \\ lw &= \frac{2500-x^2}{2} \\ lw &= 1250 - \frac{x^2}{2} \\ \end{align*}$

Thus, our answer is $\boxed{\textbf{(D) } 1250 - \frac{x^2}{2}}.$

2002 AMC 10P (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2002 AMC 10P Problems/Problem 6

Problem

Solution 1

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