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]](http://latex.artofproblemsolving.com/d/0/9/d09caec6074d6abf81a6e3a7755b2eecc103bc41.png)
Solution 1
Let the length of the small square be , intuitively, the length of the big square is
. It can be seen that the width of the rectangle is
. Thus, the length of the rectangle is
times as large as the width. The answer is
.
Solution 2
We can say the area of one small square is , so
of the area of the larger square is
so the area of the large square is
, so each side is
so the length of the rectangle is
and the width of the rectangle is
so
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 | |
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