2005 AMC 8 Problems/Problem 7

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

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]

Find the length of the diagonal of the rectangle to find the length of the direct line to the starting time using Pythagorean Theorem.

\[\sqrt{(\frac12+\frac12)^2+(\frac34)^2} = \sqrt{1+\frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac54 = \boxed{\textbf{(B)}\ 1 \tfrac14}\]

See Also

2005 AMC 8 (ProblemsAnswer KeyResources)
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

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