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== | ||
− | 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? | + | 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> | ||
− | ==Solution== | + | ==Solution 1== |
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− | <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. |
− | < | + | <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> | + | The horizontal length is <math>5.4\times4=21.6</math>, which is closest to <math>\boxed{\textbf{(D) \ } 21.5}</math>. |
− | + | == Solution 2 == | |
+ | 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== | ||
+ | {{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}} |
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.
Contents
[hide]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 . The horizontal length of a "-inch" television screen is closest, in inches, to which of the following?
Solution 1
If you divide the television screen into two right triangles, the legs are in the ratio of , and we can let one leg be and the other be . Then we can use the Pythagorean Theorem.
The horizontal length is , which is closest to .
Solution 2
One can realize that the diagonal, vertical, and horizontal lengths all make up a triangle. Therefore, the horizontal length, being the in the ratio, is simply times the hypotenuse. .
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.