2010 AMC 10A Problems/Problem 2

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}$

2010 AMC 10A (Problems • Answer Key • Resources)
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2010 AMC 10A Problems/Problem 2

Contents

Problem 2

Solution 1

Solution 2

See also