Difference between revisions of "2009 AMC 12B Problems/Problem 9"
VelaDabant (talk | contribs) (New page: == Problem == Triangle <math>ABC</math> has vertices <math>A = (3,0)</math>, <math>B = (0,3)</math>, and <math>C</math>, where <math>C</math> is on the line <math>x + y = 7</math>. What i...) |
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=== Solution 2 === | === Solution 2 === | ||
| − | The base of the triangle is <math>AB = \sqrt{3^2 + 3^2} = 3\sqrt 2</math>. Its altitude is the distance between the point <math>A</math> and the parallel line <math>x + y = 7</math>, which is | + | The base of the triangle is <math>AB = \sqrt{3^2 + 3^2} = 3\sqrt 2</math>. Its altitude is the distance between the point <math>A</math> and the parallel line <math>x + y = 7</math>, which is <math>\frac 4{\sqrt 2} = 2\sqrt 2</math>. Therefore its area is <math>\frac 12 \cdot 3\sqrt 2 \cdot 2\sqrt 2 = \boxed{6}</math>. The answer is <math>\mathrm{(A)}</math>. |
| − | < | + | |
| − | Therefore its area is <math>\frac 12 \cdot 3\sqrt 2 \cdot 2\sqrt 2 = \boxed{6}</math>. The answer is <math>\mathrm{(A)}</math>. | + | <asy> |
| + | unitsize(0.75cm); | ||
| + | defaultpen(0.8); | ||
| + | pair A=(3,0), B=(0,3); | ||
| + | draw ( (-1,0) -- (9,0), dashed ); | ||
| + | draw ( (0,-1) -- (0,9), dashed ); | ||
| + | dot(A); dot(B); draw(A--B); | ||
| + | draw ( (-1,8) -- (8,-1) ); | ||
| + | label( "$A$", A, S ); | ||
| + | label( "$B$", B, W ); | ||
| + | label( "$3$", A--(0,0), S ); | ||
| + | label( "$3$", B--(0,0), W ); | ||
| + | label( "$x+y=7$", (8,-1), SE ); | ||
| + | pair C = intersectionpoint(A--(10,7),(7,0)--(0,7)); | ||
| + | draw( A--C, dashed ); | ||
| + | draw(rightanglemark(A,C,(7,0))); | ||
| + | draw(rightanglemark(C,A,B)); | ||
| + | label( "$4$", A--(7,0), S ); | ||
| + | label( "$3\sqrt 2$", 0.67*B+0.33*A, NE ); | ||
| + | label( "$\frac 4{\sqrt 2}$", A--C, NW ); | ||
| + | label( "$\frac 4{\sqrt 2}$", C--(7,0), NE ); | ||
| + | </asy> | ||
== See also == | == See also == | ||
{{AMC12 box|year=2009|ab=B|num-b=8|num-a=10}} | {{AMC12 box|year=2009|ab=B|num-b=8|num-a=10}} | ||
Revision as of 16:03, 2 March 2009
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 
.
![[asy] unitsize(0.75cm); defaultpen(0.8); pair A=(3,0), B=(0,3); draw ( (-1,0) -- (9,0), dashed ); draw ( (0,-1) -- (0,9), dashed ); dot(A); dot(B); draw(A--B); draw ( (-1,8) -- (8,-1) ); label( "$A$", A, S ); label( "$B$", B, W ); label( "$3$", A--(0,0), S ); label( "$3$", B--(0,0), W ); label( "$x+y=7$", (8,-1), SE ); pair C = intersectionpoint(A--(10,7),(7,0)--(0,7)); draw( A--C, dashed ); draw(rightanglemark(A,C,(7,0))); draw(rightanglemark(C,A,B)); label( "$4$", A--(7,0), S ); label( "$3\sqrt 2$", 0.67*B+0.33*A, NE ); label( "$\frac 4{\sqrt 2}$", A--C, NW ); label( "$\frac 4{\sqrt 2}$", C--(7,0), NE ); [/asy]](http://latex.artofproblemsolving.com/f/a/b/fab711cac413774cb38c162f63d0bf76374d2421.png)
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 | |
