https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Aarushyadav&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T12:01:28ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_6&diff=1102842003 AMC 10B Problems/Problem 62019-10-14T00:01:16Z<p>Aarushyadav: /* Problem */</p>
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<div>{{duplicate|[[2003 AMC 12B Problems|2003 AMC 12B #5]] and [[2003 AMC 10B Problems|2003 AMC 10B #6]]}}<br />
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==Problem==<br />
<br />
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?<br />
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<math>\textbf{(A) } 20 \qquad\textbf{(B) } 20.5 \qquad\textbf{(C) } 21 \qquad\textbf{(D) } 21.5 \qquad\textbf{(E) } 22 </math><br />
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==Solution 1==<br />
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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.<br />
<br />
<cmath>\begin{align*}(4x)^2+(3x)^2&=27^2\\<br />
16x^2+9x^2&=729\\<br />
25x^2&=729\\<br />
x^2&=\frac{729}{25}\\<br />
x&=\frac{27}{5}\\<br />
x&=5.4\end{align*}</cmath><br />
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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>.<br />
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== Solution 2 ==<br />
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>.<br />
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==See Also==<br />
<br />
{{AMC12 box|year=2003|ab=B|num-b=4|num-a=6}}<br />
{{AMC10 box|year=2003|ab=B|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Aarushyadavhttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_6&diff=1102832003 AMC 10B Problems/Problem 62019-10-14T00:01:01Z<p>Aarushyadav: /* Problem */</p>
<hr />
<div>{{duplicate|[[2003 AMC 12B Problems|2003 AMC 12B #5]] and [[2003 AMC 10B Problems|2003 AMC 10B #6]]}}<br />
<br />
==Problem==<br />
<br />
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 orizontal length of a "<math>27</math>-inch" television screen is closest, in inches, to which of the following?<br />
<br />
<math>\textbf{(A) } 20 \qquad\textbf{(B) } 20.5 \qquad\textbf{(C) } 21 \qquad\textbf{(D) } 21.5 \qquad\textbf{(E) } 22 </math><br />
<br />
==Solution 1==<br />
<br />
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.<br />
<br />
<cmath>\begin{align*}(4x)^2+(3x)^2&=27^2\\<br />
16x^2+9x^2&=729\\<br />
25x^2&=729\\<br />
x^2&=\frac{729}{25}\\<br />
x&=\frac{27}{5}\\<br />
x&=5.4\end{align*}</cmath><br />
<br />
The horizontal length is <math>5.4\times4=21.6</math>, which is closest to <math>\boxed{\textbf{(D) \ } 21.5}</math>.<br />
<br />
== Solution 2 ==<br />
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>.<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2003|ab=B|num-b=4|num-a=6}}<br />
{{AMC10 box|year=2003|ab=B|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Aarushyadav