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

(Created page with "==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 ...")
 
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==Solution==
 
==Solution==
  
Because the legs are in the ratio of <math>4 : 3</math>, let one leg be <math>4x</math> and the other be <math>3x</math>.
+
If you divide the television screen into two right triangles, the legs are in the ratio of <math>4 : 3</math>, and we can let one leg be <math>4x</math> and the other be <math>3x</math>. Then we can use the Pythagorean Theorem.
  
Then we use the Pythagorean Theorem:
+
<cmath>\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*}</cmath>
  
<math>(4x)^2+(3x)^2=27^2</math>
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The horizontal length is <math>5.4\times4=21.6</math>, which is closest to <math>\boxed{\mathrm{(D) \ } 21.5}</math>.
 
 
<math>16x^2+9x^2=729</math>
 
 
 
<math>25x^2=729</math>
 
 
 
<math>x^2=\frac{729}{25}</math>
 
 
 
<math>x=\frac{27}{5}</math>
 
 
 
<math>x=5.4</math>
 
 
 
The horizontal length is <math>5.4\cdot4=21.6</math>, which is closest to <math>\boxed{21.5 \text{ (D)}}</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 02:15, 10 June 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 $\boxed{\mathrm{(D) \ } 21.5}$.

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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