Difference between revisions of "2015 AMC 10A Problems/Problem 11"
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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== | ||
Revision as of 18:53, 16 April 2015
- The following problem is from both the 2015 AMC 12A #8 and 2015 AMC 10A #11, so both problems redirect to this page.
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 
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 | |
