Difference between revisions of "2015 AMC 10A Problems/Problem 11"
(→Solution) |
m (→Solution) |
||
| Line 6: | Line 6: | ||
==Solution== | ==Solution== | ||
| − | 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>x = \dfrac{x}{5}</math>. Hence, the area of the rectangle is <math>3x \cdot 4x = 12x^2 = \dfrac{12d^2}{25}</math>, so <math>k = \dfrac{12}{25} | + | 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>x = \dfrac{x}{5}</math>. Hence, the area of the rectangle is <math>3x \cdot 4x = 12x^2 = \dfrac{12d^2}{25}</math>, so <math>k = \dfrac{12}{25}</math> <math>\textbf{(C)}</math>. |
Revision as of 18:22, 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. Unknown error_msg)
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 
.
