Difference between revisions of "2003 AMC 10B Problems/Problem 6"
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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? | + | 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> |
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.