Difference between revisions of "2009 AMC 12B Problems/Problem 9"
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label( "$\frac 4{\sqrt 2}$", C--(7,0), NE ); | label( "$\frac 4{\sqrt 2}$", C--(7,0), NE ); | ||
</asy> | </asy> | ||
+ | |||
+ | === Solution 3 === | ||
+ | By Shoelace, our area is: | ||
+ | <cmath>\frac {1}{2} \cdot |(3\cdot 3 + 0 \cdot y+ 0 \cdot x)-(0*0+3(x+y))|.</cmath> | ||
+ | We know <math>x+y=7</math> so we get: | ||
+ | <cmath>\frac {1}{2} \cdot |9-21|=\boxed 6</cmath> | ||
== See also == | == See also == |
Revision as of 18:51, 4 February 2017
Problem
Triangle has vertices , , and , where is on the line . What is the area of ?
Solution
Solution 1
Because the line is parallel to , the area of is independent of the location of on that line. Therefore it may be assumed that is . In that case the triangle has base and altitude , so its area is .
Solution 2
The base of the triangle is . Its altitude is the distance between the point and the parallel line , which is . Therefore its area is . The answer is .
Solution 3
By Shoelace, our area is: We know so we get:
See also
2009 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.