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

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<math> \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 </math>
 
<math> \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 </math>
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[[Category: Introductory Geometry Problems]]
  
 
== Solution ==
 
== Solution ==
  
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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>
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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) = \boxed{\textbf{(A)}\ ac + ad + bc + bd}</math>.
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==Video Solution==
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https://youtu.be/n0qUd4ukEF8
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~savannahsolver
  
 
== 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}}
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{{MAA Notice}}

Latest revision as of 18:56, 17 November 2020

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 $(a + b)(c + d) = \boxed{\textbf{(A)}\ ac + ad + bc + bd}$.

Video Solution

https://youtu.be/n0qUd4ukEF8

~savannahsolver

See Also

2011 AMC 10A (Problems • Answer Key • Resources)
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

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