Difference between revisions of "2003 AMC 10B Problems/Problem 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 | + | 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? |
<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> |
Revision as of 19:01, 13 October 2019
- The following problem is from both the 2003 AMC 12B #5 and 2003 AMC 10B #6, so both problems redirect to this page.
Contents
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 orizontal 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.