Difference between revisions of "2011 AMC 10A Problems/Problem 9"

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== Solution ==
 
== Solution ==
  
We have a rectangle of side lengths <math>a-(-b)=a+b</math> and <math>d-(-c)=c+d.</math> Thus the area of this rectangle is  
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We have a rectangle of side lengths <math>a-(-b)=a+b</math> and <math>d-(-c)=c+d.</math> Thus the area of this rectangle is <math>(a + b)(c + d) = ac + ad + bc + bd.</math>
<cmath>
 
(a + b)(c + d) = ac + ad + bc + bd.
 
</cmath>
 
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2011|ab=A|num-b=8|num-a=10}}
 
{{AMC10 box|year=2011|ab=A|num-b=8|num-a=10}}

Revision as of 20:42, 6 February 2012

Problem 9

A rectangular region is bounded by the graphs of the equations $y=a, y=-b, x=-c,$ and $x=d$, where $a,b,c,$ and $d$ are all positive numbers. Which of the following represents the area of this region?

$\textbf{(A)}\ ac+ad+bc+bd\qquad\textbf{(B)}\ ac-ad+bc-bd\qquad\textbf{(C)}\ ac+ad-bc-bd   \quad\quad\qquad\textbf{(D)}\ -ac-ad+bc+bd\qquad\textbf{(E)}\ ac-ad-bc+bd$

Solution

We have a rectangle of side lengths $a-(-b)=a+b$ and $d-(-c)=c+d.$ Thus the area of this rectangle is $(a + b)(c + d) = ac + ad + bc + bd.$

See Also

2011 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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