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> | ||
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+ | ==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>B</math>. | ||
+ | |||
+ | ~ abed_nadir | ||
==Video Solution== | ==Video Solution== |
Revision as of 12:11, 11 November 2023
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 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 .
~ abed_nadir
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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