2015 AMC 10A Problems/Problem 11

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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 $k$?

$\textbf{(A)}\ \frac{2}{7}\qquad\textbf{(B)}\ \frac{3}{7}\qquad\textbf{(C)}\ \frac{12}{25}\qquad\textbf{(D)}}\ \frac{16}{25}\qquad\textbf{(E)}\ \frac{3}{4}$ (Error compiling LaTeX. ! Extra }, or forgotten $.)

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 $d = \dfrac{x}{5}$. Hence, the area of the rectangle is $3x \cdot 4x = 12x^2 = \dfrac{12d^2}{25}$, so the answer is $\boxed{\textbf{(C) }\frac{12}{25}}$

See also

2015 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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
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