Difference between revisions of "2002 AMC 10P Problems/Problem 6"
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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}} | ||
Revision as of 20:03, 14 July 2024
Problem
The perimeter of a rectangle 
and its diagonal has length 
What is the area of this rectangle?
Solution 1
Let 
be the length of the rectangle and 
be the width of the rectangle. We are given 
and 
We are asked to find 
Using a bit of algebraic manipulation:
Thus, our answer is 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
