Difference between revisions of "2010 AMC 10A Problems/Problem 2"
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==Solution 2== | ==Solution 2== | ||
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> | 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> | ||
+ | |||
+ | ==Solution 3== | ||
+ | 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>. | ||
+ | |||
+ | ~ youtube.com/indianmathguy | ||
==Video Solution== | ==Video Solution== |
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]](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 large 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
Solution 3
Let the side length of one of the squares equal . Then, the width of the rectangle will be
, and since the width of the rectangle is the same as the length of the entire shape, the length of the rectangle is
. The ratio between the two is therefore
, so our answer is
.
~ 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.