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> | ||
| + | [[Category: Introductory Geometry Problems]] | ||
== 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 | + | 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>. |
| − | + | ||
| − | (a + b)(c + d) = ac + ad + bc + bd. | + | ==Video Solution== |
| − | + | https://youtu.be/n0qUd4ukEF8 | |
| + | |||
| + | ~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}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 18:56, 17 November 2020
Contents
Problem 9
A rectangular region is bounded by the graphs of the equations 
and 
, where 
and 
are all positive numbers. Which of the following represents the area of this region?
Solution
We have a rectangle of side lengths 
and 
Thus the area of this rectangle is 
.
Video Solution
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
