Difference between revisions of "2015 AMC 10A Problems/Problem 11"
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==Solution== | ==Solution== | ||
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> | 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> | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/Ep-IQmOK-nU | ||
| + | |||
| + | ~savannahsolver | ||
== See also == | == See also == | ||
{{AMC10 box|year=2015|ab=A|num-b=10|num-a=12}} | {{AMC10 box|year=2015|ab=A|num-b=10|num-a=12}} | ||
{{AMC12 box|year=2015|ab=A|num-b=7|num-a=9}} | {{AMC12 box|year=2015|ab=A|num-b=7|num-a=9}} | ||
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 02:52, 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. 
