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Difference between revisions of "2003 AMC 10B Problems/Problem 6"
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{{duplicate|[[2003 AMC 12B Problems|2003 AMC 12B #5]] and [[2003 AMC 10B Problems|2003 AMC 10B #6]]}}
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==Problem==
 
==Problem==
  
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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 <math>4 : 3</math>. The horizontal length of a "<math>27</math>-inch" television screen is closest, in inches, to which of the following?
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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 <math>4:3</math>. The horizontal length of a "<math>27</math>-inch" television screen is closest, in inches, to which of the following?
  
 
<math>\textbf{(A) } 20 \qquad\textbf{(B) } 20.5 \qquad\textbf{(C) } 21 \qquad\textbf{(D) } 21.5 \qquad\textbf{(E) } 22 </math>
 
<math>\textbf{(A) } 20 \qquad\textbf{(B) } 20.5 \qquad\textbf{(C) } 21 \qquad\textbf{(D) } 21.5 \qquad\textbf{(E) } 22 </math>
  
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==Solution==
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==Solution 1==
  
 
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.
 
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.
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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>.
 
The horizontal length is <math>5.4\times4=21.6</math>, which is closest to <math>\boxed{\textbf{(D) \ } 21.5}</math>.
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== Solution 2 ==
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One can realize that the diagonal, vertical, and horizontal lengths all make up a <math>3,4,5</math> triangle. Therefore, the horizontal length, being the <math>4</math> in the <math>4 : 3</math> ratio, is simply <math>\frac{4}{5}</math> times the hypotenuse. <math>\frac{4}{5}\cdot27=21.6 \approx \boxed{\textbf{(D) } 21.5}</math>.
  
 
==See Also==
 
==See Also==
  
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{{AMC12 box|year=2003|ab=B|num-b=4|num-a=6}}
 
{{AMC10 box|year=2003|ab=B|num-b=5|num-a=7}}
 
{{AMC10 box|year=2003|ab=B|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:53, 20 October 2020

The following problem is from both the 2003 AMC 12B #5 and 2003 AMC 10B #6, so both problems redirect to this page.

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 1

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 $\boxed{\textbf{(D) \ } 21.5}$.

Solution 2

One can realize that the diagonal, vertical, and horizontal lengths all make up a $3,4,5$ triangle. Therefore, the horizontal length, being the $4$ in the $4 : 3$ ratio, is simply $\frac{4}{5}$ times the hypotenuse. $\frac{4}{5}\cdot27=21.6 \approx \boxed{\textbf{(D) } 21.5}$.

See Also

2003 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
2003 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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