Difference between revisions of "2002 AMC 10P Problems/Problem 6"
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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> | ||
Latest revision as of 21:04, 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 the area, which is equivalent to 
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. 
