Difference between revisions of "2012 AMC 10B Problems/Problem 12"
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<math>\textbf{(A)}\ 30\ \text{and}\ 31 \qquad\textbf{(B)}\ 31\ \text{and}\ 32 \qquad\textbf{(C)}\ 32\ \text{and}\ 33 \qquad\textbf{(D)}\ 33\ \text{and}\ 34 \qquad\textbf{(E)}\ 34\ \text{and}\ 35</math> | <math>\textbf{(A)}\ 30\ \text{and}\ 31 \qquad\textbf{(B)}\ 31\ \text{and}\ 32 \qquad\textbf{(C)}\ 32\ \text{and}\ 33 \qquad\textbf{(D)}\ 33\ \text{and}\ 34 \qquad\textbf{(E)}\ 34\ \text{and}\ 35</math> | ||
+ | [[Category: Introductory Geometry Problems]] | ||
== Solution == | == Solution == |
Revision as of 11:46, 13 August 2014
Problem
Point B is due east of point A. Point C is due north of point B. The distance between points A and C is , and
. Point D is 20 meters due north of point C. The distance AD is between which two integers?
Solution
If point B is due east of point A and point C is due north of point B, is a right angle. And if
,
is a 45-45-90 triangle. Thus, the lengths of sides
,
, and
are in the ratio
, and
is
.
is clearly a right triangle with
on the side
.
is 20, so
.
By the Pythagorean Theorem, .
, and
. Thus,
must be between
and
. The answer is
.
See Also
2012 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.