Difference between revisions of "2012 AMC 10B Problems/Problem 12"
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If point B is due east of point A and point C is due north of point B, <math>\angle CBA</math> is a right angle. And if <math>\angle BAC = 45^\circ</math>, <math>\triangle CBA</math> is a 45-45-90 triangle. Thus, the lengths of sides <math>CB</math>, <math>BA</math>, and <math>AC</math> are in the ratio <math>1:1:\sqrt 2</math>, and <math>CB</math> is <math>10 \sqrt 2 \div \sqrt 2 = 10</math>. | If point B is due east of point A and point C is due north of point B, <math>\angle CBA</math> is a right angle. And if <math>\angle BAC = 45^\circ</math>, <math>\triangle CBA</math> is a 45-45-90 triangle. Thus, the lengths of sides <math>CB</math>, <math>BA</math>, and <math>AC</math> are in the ratio <math>1:1:\sqrt 2</math>, and <math>CB</math> is <math>10 \sqrt 2 \div \sqrt 2 = 10</math>. | ||
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+ | <math>\triangle DBA</math> is clearly a right triangle with <math>C</math> on the side <math>DB</math>. <math>DC</math> is 20, so <math>DB = DC + CB = 20 + 10 = 30</math>. | ||
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+ | By the Pythagorean Theorem, <math>DA = \sqrt {DB^2 + BA^2} = \sqrt {30^2 + 10 ^2} = \sqrt {1000}</math>. | ||
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+ | <math>31^2 = 961</math>, and <math>32^2 = 1024</math>. Thus, <math>\sqrt {1000}</math> must be between <math>31</math> and <math>32</math>. The answer is <math>\boxed {B}</math>. |
Revision as of 19:52, 25 February 2012
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
.