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

Revision as of 09:16, 20 May 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 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://www.youtube.com/watch?v=dQw4w9WgXcQ

~IceMatrix

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 35: / Line 35: @@
 ==Video Solution==
-https://youtu.be/C1VCk_9A2KE?t=80
+https://www.youtube.com/watch?v=dQw4w9WgXcQ
 ~IceMatrix

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