Difference between revisions of "2010 AMC 10A Problems/Problem 2"
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==Solution 2== | ==Solution 2== | ||
− | We can say the | + | 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 larger 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> |
==Video Solution== | ==Video Solution== |
Revision as of 08:36, 4 July 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?
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 | |
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 |
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