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Difference between revisions of "2002 AMC 10P Problems/Problem 6"

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== Problem ==
 
== Problem ==
−
The perimeter of a rectangle <math>100</math> and its diagonal has length <math>x.</math> What is the area of this rectangle?
+
The perimeter of a rectangle is <math>100</math> and its diagonal has length <math>x.</math> What is the area of this rectangle?
 
<math>
 
<math>
 
\text{(A) }625-x^2
 
\text{(A) }625-x^2
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== Solution 1==
 
== Solution 1==
−
Let <math>l</math> be the length of the rectangle and <math>w</math> be the width of the rectangle. We are given <math>2l+2w=100</math> and <math>l^2+w^2=x^2.</math> We are asked to find <math>lw.</math> Using a bit of algebraic manipulation:
+
Let <math>l</math> be the length of the rectangle and <math>w</math> be the width of the rectangle. We are given <math>2l+2w=100</math> and <math>l^2+w^2=x^2.</math> We are asked to find the area, which is equivalent to <math>lw.</math> Using a bit of algebraic manipulation:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
2l+2w=100 \\
+
2l+2w &= 100 \\
−
l+w=50 \\
+
l+w &= 50 \\
−
(l+w)^2=2500 \\
+
(l+w)^2 &= 2500 \\
−
l^2 + w^2 +2lw = 2500 \\
+
l^2 + w^2 +2lw &= 2500 \\
−
x^2 + 2lw = 2500 \\
+
x^2 + 2lw &= 2500 \\
−
lw=\frac{2500-x^2}{2} \\
+
lw &= \frac{2500-x^2}{2} \\
−
lw=1250 - \frac{x^2}{2}
+
lw &= 1250 - \frac{x^2}{2} \\
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 +
 +
Thus, our answer is <math>\boxed{\textbf{(D) } 1250 - \frac{x^2}{2}}.</math>
  
 
== See also ==
 
== See also ==
 
{{AMC10 box|year=2002|ab=P|num-b=5|num-a=7}}
 
{{AMC10 box|year=2002|ab=P|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:29, 15 July 2024

Problem

The perimeter of a rectangle is $100$ 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 $\boxed{\textbf{(D) } 1250 - \frac{x^2}{2}}.$

See also

2002 AMC 10P (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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