Difference between revisions of "2003 AMC 10B Problems/Problem 6"

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x&=5.4\end{align*}</cmath>
 
x&=5.4\end{align*}</cmath>
  
The horizontal length is <math>5.4\times4=21.6</math>, which is closest to <math>\boxed{\mathrm{(D) \ } 21.5}</math>.
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The horizontal length is <math>5.4\times4=21.6</math>, which is closest to <math>\boxed{\textbf{(D) \ } 21.5}</math>.
  
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2003|ab=B|num-b=5|num-a=7}}
 
{{AMC10 box|year=2003|ab=B|num-b=5|num-a=7}}

Revision as of 17:53, 26 November 2011

Problem

Many television screens are rectangles that are measured by the length of their diagonals. The ratio of the horizontal length to the height in a standard television screen is $4 : 3$. The horizontal length of a "$27$-inch" television screen is closest, in inches, to which of the following?

$\textbf{(A) } 20 \qquad\textbf{(B) } 20.5 \qquad\textbf{(C) } 21 \qquad\textbf{(D) } 21.5 \qquad\textbf{(E) } 22$

Solution

If you divide the television screen into two right triangles, the legs are in the ratio of $4 : 3$, and we can let one leg be $4x$ and the other be $3x$. Then we can use the Pythagorean Theorem.

\begin{align*}(4x)^2+(3x)^2&=27^2\\ 16x^2+9x^2&=729\\ 25x^2&=729\\ x^2&=\frac{729}{25}\\ x&=\frac{27}{5}\\ x&=5.4\end{align*}

The horizontal length is $5.4\times4=21.6$, which is closest to $\boxed{\textbf{(D) \ } 21.5}$.

See Also

2003 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AMC 10 Problems and Solutions
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