Difference between revisions of "2010 AMC 10A Problems/Problem 2"
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| − | 4/3 | + | == 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? | ||
| + | |||
| + | <center><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></center> | ||
| + | |||
| + | <math> | ||
| + | \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 | ||
| + | </math> | ||
| + | |||
| + | ==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>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | We can say the smaller squares area is <math>x^2</math>, so <math>\dfrac{1}{4}</math> of the area of the larger square is <math>4x^2</math> so the large squares are is <math>16x^2</math>, so each side is <math>4x</math> so length is <math>4x</math> and the width is <math>4x-x=3x</math> so <math>\dfrac{4x}{3x}=\dfrac{4}{3}</math> | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/C1VCk_9A2KE?t=80 | ||
| + | |||
| + | ~IceMatrix | ||
| + | |||
| + | == See also == | ||
| + | {{AMC10 box|year=2010|ab=A|num-b=1|num-a=3}} | ||
| + | {{MAA Notice}} | ||
Revision as of 06:16, 26 May 2020
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 large as the width. The answer is 
.
Solution 2
We can say the smaller squares area is 
, so 
of the area of the larger square is 
so the large squares are is 
, so each side is so length is and the width 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 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
