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Difference between revisions of "2005 AMC 8 Problems/Problem 7"

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==Solution==
 
==Solution==
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Draw a picture.
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We have a right-angle triangle with sides <math>3/4</math>and 1. We can divide the simple pythagorean theorem identity 3,4,5 by 4 to get 3/4, 1, 5/4. This shows us the answer is D.
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<asy>
 
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unitsize(3cm);
 
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draw((0,0)--(0,-.5)--(.75,-.5)--(.75,-1),linewidth(1pt));
 
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label("$\frac12$",(0,0)--(0,-.5),W);
 
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label("$\frac12$",(.75,-.5)--(.75,-1),E);
 
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label("$\frac34$",(.75,-.5)--(0,-.5),N);
 
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draw((0,0)--(.75,0)--(.75,-1)--(0,-1)--cycle,gray);
 
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</asy>
 
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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.
 
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<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}}
 
{{MAA Notice}}

Revision as of 16:19, 16 December 2024

Problem

Bill walks $\tfrac12$ mile south, then $\tfrac34$ mile east, and finally $\tfrac12$ mile south. How many miles is he, in a direct line, from his starting point?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 1\tfrac14\qquad\textbf{(C)}\ 1\tfrac12\qquad\textbf{(D)}\ 1\tfrac34\qquad\textbf{(E)}\ 2$

Solution

We have a right-angle triangle with sides $3/4$and 1. We can divide the simple pythagorean theorem identity 3,4,5 by 4 to get 3/4, 1, 5/4. This shows us the answer is D.

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. AMC logo.png

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