# Difference between revisions of "2015 AMC 10A Problems/Problem 11"

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 $4$ : $3$. If the rectangle has diagonal of length $d$, then the area may be expressed as $kd^2$ for some constant $k$. What is $k$?

$\textbf{(A)}\ \frac{2}{7}\qquad\textbf{(B)}\ \frac{3}{7}\qquad\textbf{(C)}\ \frac{12}{25}\qquad\textbf{(D)}\ \frac{16}{25}\qquad\textbf{(E)}\ \frac{3}{4}$

## Solution

Let the rectangle have length $4x$ and width $3x$. Then by $3-4-5$ triangles (or the Pythagorean Theorem), we have $d = 5x$, and so $x = \dfrac{d}{5}$. Hence, the area of the rectangle is $3x \cdot 4x = 12x^2 = \dfrac{12d^2}{25}$, so the answer is $\boxed{\textbf{(C) }\frac{12}{25}}$

~savannahsolver

## See also

 2015 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 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 byProblem 7 Followed byProblem 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.

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