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

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== Problem 2 ==
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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?
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<center><asy>
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unitsize(8mm);
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defaultpen(linewidth(.8pt));
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draw((0,0)--(4,0)--(4,4)--(0,4)--cycle);
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draw((0,3)--(0,4)--(1,4)--(1,3)--cycle);
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draw((1,3)--(1,4)--(2,4)--(2,3)--cycle);
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draw((2,3)--(2,4)--(3,4)--(3,3)--cycle);
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draw((3,3)--(3,4)--(4,4)--(4,3)--cycle);
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</asy></center>
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<math>
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\mathrm{(A)}\ \dfrac{5}{4}
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\qquad
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\mathrm{(B)}\ \dfrac{4}{3}
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\qquad
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\mathrm{(C)}\ \dfrac{3}{2}
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\qquad
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\mathrm{(D)}\ 2
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\qquad
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\mathrm{(E)}\ 3
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</math>
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==Solution 1==
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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>.
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==Solution 2==
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We can say the area of one small square is <math>x^2</math>, so <math>\dfrac{1}{4}</math> of the area of the large square is <math>4x^2</math> so the area of the large square is <math>16x^2</math>, so each side is <math>4x</math> so the length of the rectangle is <math>4x</math> and the width of the rectangle is <math>4x-x=3x</math> so <math>\dfrac{4x}{3x}=\dfrac{4}{3}</math>
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==Solution 3==
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Let the side length of one of the squares equal <math>1</math>. Then, the width of the rectangle will be <math>4</math>, and since the width of the rectangle is the same as the length of the entire shape, the length of the rectangle is <math>4 - 1 = 3</math>. The ratio between the two is therefore <math>\frac{4}{3}</math> , so our answer is <math>\boxed{B}</math>.
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~ youtube.com/indianmathguy
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==Video Solution==
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https://youtu.be/C1VCk_9A2KE?t=80
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~IceMatrix
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== See also ==
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{{AMC10 box|year=2010|ab=A|num-b=1|num-a=3}}
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{{MAA Notice}}

Latest revision as of 16:33, 5 February 2024

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 area of one small square is $x^2$, so $\dfrac{1}{4}$ of the area of the large square is $4x^2$ so the area of the large square is $16x^2$, so each side is $4x$ so the length of the rectangle is $4x$ and the width of the rectangle is $4x-x=3x$ so $\dfrac{4x}{3x}=\dfrac{4}{3}$

Solution 3

Let the side length of one of the squares equal $1$. Then, the width of the rectangle will be $4$, and since the width of the rectangle is the same as the length of the entire shape, the length of the rectangle is $4 - 1 = 3$. The ratio between the two is therefore $\frac{4}{3}$ , so our answer is $\boxed{B}$.

~ youtube.com/indianmathguy

Video Solution

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

~IceMatrix

See also

2010 AMC 10A (Problems • Answer Key • Resources)
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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