Difference between revisions of "2005 AMC 8 Problems/Problem 7"
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Find the length of the diagonal of the rectangle to find the length of the direct line to the starting time using Pythagorean Theorem. | Find the length of the diagonal of the rectangle to find the length of the direct line to the starting time using Pythagorean Theorem. | ||
| − | <cmath>\sqrt{(\frac12+\frac12)^2+(\frac34)^2} = \sqrt{1+\frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac54 = \boxed{\textbf{(B)}\ 1 \tfrac14}</cmath> | + | <cmath>\sqrt{\left(\frac12+\frac12\right)^2+\left(\frac34\right)^2} = \sqrt{1+\frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac54 = \boxed{\textbf{(B)}\ 1 \tfrac14}</cmath> |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2005|num-b=6|num-a=8}} | {{AMC8 box|year=2005|num-b=6|num-a=8}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 18:24, 15 December 2020
Problem
Bill walks 
mile south, then 
mile east, and finally mile south. How many miles is he, in a direct line, from his starting point?
Solution
Draw a picture.
![[asy] unitsize(3cm); draw((0,0)--(0,-.5)--(.75,-.5)--(.75,-1),linewidth(1pt)); label("$\frac12$",(0,0)--(0,-.5),W); label("$\frac12$",(.75,-.5)--(.75,-1),E); label("$\frac34$",(.75,-.5)--(0,-.5),N); draw((0,0)--(.75,0)--(.75,-1)--(0,-1)--cycle,gray); [/asy]](http://latex.artofproblemsolving.com/6/1/1/61183e6c6cffe3acc5a9f72a236d911bb70b3b9a.png)
Find the length of the diagonal of the rectangle to find the length of the direct line to the starting time using Pythagorean Theorem.
![\[\sqrt{\left(\frac12+\frac12\right)^2+\left(\frac34\right)^2} = \sqrt{1+\frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac54 = \boxed{\textbf{(B)}\ 1 \tfrac14}\]](http://latex.artofproblemsolving.com/5/e/6/5e67012dbc82179be02fa20e3559741dd6ee65e7.png)
See Also
| 2005 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
