Difference between revisions of "2015 AMC 10A Problems/Problem 11"
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+ | {{duplicate|[[2015 AMC 12A Problems|2015 AMC 12A #8]] and [[2015 AMC 10A Problems|2015 AMC 10A #11]]}} | ||
==Problem 11== | ==Problem 11== | ||
The ratio of the length to the width of a rectangle is <math>4</math> : <math>3</math>. If the rectangle has diagonal of length <math>d</math>, then the area may be expressed as <math>kd^2</math> for some constant <math>k</math>. What is <math>k</math>? | The ratio of the length to the width of a rectangle is <math>4</math> : <math>3</math>. If the rectangle has diagonal of length <math>d</math>, then the area may be expressed as <math>kd^2</math> for some constant <math>k</math>. What is <math>k</math>? | ||
− | <math>\textbf{(A)}\ \frac{2}{7}\qquad\textbf{(B)}\ \frac{3}{7}\qquad\textbf{(C)}\ \frac{12}{25}\qquad\textbf{(D) | + | <math>\textbf{(A)}\ \frac{2}{7}\qquad\textbf{(B)}\ \frac{3}{7}\qquad\textbf{(C)}\ \frac{12}{25}\qquad\textbf{(D)}\ \frac{16}{25}\qquad\textbf{(E)}\ \frac{3}{4}</math> |
==Solution== | ==Solution== | ||
− | Let the rectangle have length <math>4x</math> and width <math>3x</math>. Then by 3-4-5 triangles (or the Pythagorean Theorem), we have <math>d = 5x</math>, and so <math> | + | Let the rectangle have length <math>4x</math> and width <math>3x</math>. Then by <math>3-4-5</math> triangles (or the Pythagorean Theorem), we have <math>d = 5x</math>, and so <math>x = \dfrac{d}{5}</math>. Hence, the area of the rectangle is <math>3x \cdot 4x = 12x^2 = \dfrac{12d^2}{25}</math>, so the answer is <math>\boxed{\textbf{(C) }\frac{12}{25}}</math> |
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+ | ==Video Solution== | ||
+ | https://youtu.be/Ep-IQmOK-nU | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | == See also == | ||
+ | {{AMC10 box|year=2015|ab=A|num-b=10|num-a=12}} | ||
+ | {{AMC12 box|year=2015|ab=A|num-b=7|num-a=9}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 03:52, 13 January 2021
- The following problem is from both the 2015 AMC 12A #8 and 2015 AMC 10A #11, so both problems redirect to this page.
Contents
Problem 11
The ratio of the length to the width of a rectangle is : . If the rectangle has diagonal of length , then the area may be expressed as for some constant . What is ?
Solution
Let the rectangle have length and width . Then by triangles (or the Pythagorean Theorem), we have , and so . Hence, the area of the rectangle is , so the answer is
Video Solution
~savannahsolver
See also
2015 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 |
2015 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.