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Difference between revisions of "2010 AMC 10A Problems/Problem 2"
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</math>
  
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==Solution==
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==Solution 1==
  
 
Let the length of the small square be <math>x</math>, intuitively, the length of the big square is <math>4x</math>. It can be seen that the width of the rectangle is <math>3x</math>. Thus, the length of the rectangle is <math>4x/3x = 4/3</math> times large as the width. The answer is <math>\boxed{B}</math>.
 
Let the length of the small square be <math>x</math>, intuitively, the length of the big square is <math>4x</math>. It can be seen that the width of the rectangle is <math>3x</math>. Thus, the length of the rectangle is <math>4x/3x = 4/3</math> times large as the width. The answer is <math>\boxed{B}</math>.
  
−
 
+
==Solution 2==
 +
We can say the smaller squares area is <math>x^2</math>, so <math>\dfrac{1}{4} of the area of the larger square is </math>4x^2<math> so the large squares are is </math>16x^2<math>, so each side is </math>4x<math> so length is </math>4x<math> and the width is </math>4x-x=3x<math> so </math>\dfrac{4x}{3x}=\dfrac{4}{3}$
 
== See also ==
 
== See also ==
 
{{AMC10 box|year=2010|ab=A|num-b=1|num-a=3}}
 
{{AMC10 box|year=2010|ab=A|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:57, 13 January 2018

Problem 2

Four identical squares and one rectangle are placed together to form one large square as shown. The length of the rectangle is how many times as large as its width?

[asy] unitsize(8mm); defaultpen(linewidth(.8pt)); draw((0,0)--(4,0)--(4,4)--(0,4)--cycle); draw((0,3)--(0,4)--(1,4)--(1,3)--cycle); draw((1,3)--(1,4)--(2,4)--(2,3)--cycle); draw((2,3)--(2,4)--(3,4)--(3,3)--cycle); draw((3,3)--(3,4)--(4,4)--(4,3)--cycle); [/asy]

$\mathrm{(A)}\ \dfrac{5}{4} \qquad \mathrm{(B)}\ \dfrac{4}{3} \qquad \mathrm{(C)}\ \dfrac{3}{2} \qquad \mathrm{(D)}\ 2 \qquad \mathrm{(E)}\ 3$

Solution 1

Let the length of the small square be $x$, intuitively, the length of the big square is $4x$. It can be seen that the width of the rectangle is $3x$. Thus, the length of the rectangle is $4x/3x = 4/3$ times large as the width. The answer is $\boxed{B}$.

Solution 2

We can say the smaller squares area is $x^2$, so $\dfrac{1}{4} of the area of the larger square is$4x^2$so the large squares are is$16x^2$, so each side is$4x$so length is$4x$and the width is$4x-x=3x$so$\dfrac{4x}{3x}=\dfrac{4}{3}$

See also

2010 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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All AMC 10 Problems and Solutions

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