Difference between revisions of "2010 AMC 10A Problems/Problem 2"

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==Solution 1==
 
==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>.
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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 as large as the width. The answer is <math>\boxed{B}</math>.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 08:33, 4 July 2021

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 as 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 $\dfrac{4x}{3x}=\dfrac{4}{3}$

Video Solution

https://youtu.be/C1VCk_9A2KE?t=80

~IceMatrix

See also

2010 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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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