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

(Created page with "== Solution 1== == See also == {{AMC10 box|year=2002|ab=P|num-b=5|num-a=7}} {{MAA Notice}}")
 
Line 1: Line 1:
 +
== Problem 6 ==
 +
The perimeter of a rectangle <math>100</math> and its diagonal has length <math>x.</math> What is the area of this rectangle?
 +
<math>
 +
\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}
 +
</math>
 +
 
== Solution 1==
 
== Solution 1==
  

Revision as of 17:41, 14 July 2024

Problem 6

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

See also

2002 AMC 10P (ProblemsAnswer KeyResources)
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. AMC logo.png