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

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− | Let the rectangle have length <math>4x</math> and width <math>3x</math>. Then by 3-4-5 triangles (or the Pythagorean Theorem), we have <math>d = 5x</math>, and so <math> | + | Let the rectangle have length <math>4x</math> and width <math>3x</math>. Then by 3-4-5 triangles (or the Pythagorean Theorem), we have <math>d = 5x</math>, and so <math>d = \dfrac{x}{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> |

## Revision as of 18:45, 4 February 2015

## 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 ?

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

## Solution

Let the rectangle have length and width . Then by 3-4-5 triangles (or the Pythagorean Theorem), we have , and so . Hence, the area of the rectangle is , so the answer is