2015 AMC 10A Problems/Problem 11

Revision as of 19:22, 4 February 2015 by Suli (talk | contribs) (Solution)

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

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