Difference between revisions of "2024 AMC 10A Problems/Problem 16"
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==Solution 1== | ==Solution 1== | ||
− | Using the rectangle with | + | Using the rectangle with area <math>1</math>, let its short side be <math>x</math> and the long side be <math>y</math>. Observe that for every rectangle, since ratios of the side length of the rectangles are directly proportional to the ratios of the square roots of the areas (For example, each side of the rectangle with area <math>9</math> is <math>\sqrt{9}=3</math> times that of the rectangle with area <math>1</math>), as they are all similar to each other. |
The side opposite <math>AB</math> on the large rectangle is hence written as <math>6x + 4x + 2y\sqrt{2} + 3y\sqrt{2} = 10x+5y\sqrt{2}</math>. However, <math>AB</math> can be written as <math>4y\sqrt{2}+5x+7x = 4y\sqrt{2}+12x</math>. Since the two lengths are equal, we can write <math>10x+5y\sqrt{2} = 4y\sqrt{2}+12x</math>, or <math>y\sqrt{2} = 2x</math>. Therefore, we can write <math>y=x\sqrt{2}</math>. | The side opposite <math>AB</math> on the large rectangle is hence written as <math>6x + 4x + 2y\sqrt{2} + 3y\sqrt{2} = 10x+5y\sqrt{2}</math>. However, <math>AB</math> can be written as <math>4y\sqrt{2}+5x+7x = 4y\sqrt{2}+12x</math>. Since the two lengths are equal, we can write <math>10x+5y\sqrt{2} = 4y\sqrt{2}+12x</math>, or <math>y\sqrt{2} = 2x</math>. Therefore, we can write <math>y=x\sqrt{2}</math>. |
Revision as of 21:55, 9 November 2024
Contents
[hide]Problem
All of the rectangles in the figure below, which is drawn to scale, are similar to the enclosing rectangle. Each number represents the area of the rectangle. What is length ?
Solution 1
Using the rectangle with area , let its short side be and the long side be . Observe that for every rectangle, since ratios of the side length of the rectangles are directly proportional to the ratios of the square roots of the areas (For example, each side of the rectangle with area is times that of the rectangle with area ), as they are all similar to each other.
The side opposite on the large rectangle is hence written as . However, can be written as . Since the two lengths are equal, we can write , or . Therefore, we can write .
Since , we have , which we can evaluate as . From this, we can plug back in to to find . Substituting into , we have which can be evaluated to .
~i_am_suk_at_math_2
Remark
We know that the area is an integer, so after finding , AB must contains a fourth root. The only such option is .
Solution 2
Let the rectangle's height be the length The entire rectangle has an area of We will be using this fact for ratios.
Note that the short side of the rectangle with area 32 will have a height of We use because it is apparent that the height of the rectangle with area is the shorter side, corresponding with
Similarly, the long side of the rectangle with area 36 has a height of
Noting that the total height of the big rectangle has height we have the equation
Since the area is equal to 200, we have:
~mathboy282
Video Solution by Pi Academy
https://youtu.be/fW7OGWee31c?si=oq7toGPh2QaksLHE
Video Solution 1 by Power Solve == https://youtu.be/8abGnAJZ3AM
See also
2024 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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.