Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2009 AMC 12B Problems/Problem 9"
(→‎Solution 4)
 
(4 intermediate revisions by 2 users not shown)
Line 39: Line 39:
 
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>
 +
 +
=== Solution 4 ===
 +
WLOG, let the coordinates of <math>C</math> be <math>(3,4)</math> , or any coordinate, for that matter. Applying the shoelace formula, we get the area as <math>\boxed 6</math>.
 +
 +
~coolmath2017
  
 
== See also ==
 
== See also ==

Latest revision as of 20:19, 27 April 2020

Problem

Triangle $ABC$ has vertices $A = (3,0)$, $B = (0,3)$, and $C$, where $C$ is on the line $x + y = 7$. What is the area of $\triangle ABC$?


$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 12\qquad \mathrm{(E)}\ 14$

Solution

Solution 1

Because the line $x + y = 7$ is parallel to $\overline {AB}$, the area of $\triangle ABC$ is independent of the location of $C$ on that line. Therefore it may be assumed that $C$ is $(7,0)$. In that case the triangle has base $AC = 4$ and altitude $3$, so its area is $\frac 12 \cdot 4 \cdot 3 = \boxed {6}$.

Solution 2

The base of the triangle is $AB = \sqrt{3^2 + 3^2} = 3\sqrt 2$. Its altitude is the distance between the point $A$ and the parallel line $x + y = 7$, which is $\frac 4{\sqrt 2} = 2\sqrt 2$. Therefore its area is $\frac 12 \cdot 3\sqrt 2 \cdot 2\sqrt 2 = \boxed{6}$. The answer is $\mathrm{(A)}$.

[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]

Solution 3

By Shoelace, our area is: \[\frac {1}{2} \cdot |(3\cdot 3 + 0 \cdot y+ 0 \cdot x)-(0*0+3(x+y))|.\] We know $x+y=7$ so we get: \[\frac {1}{2} \cdot |9-21|=\boxed 6\]

Solution 4

WLOG, let the coordinates of $C$ be $(3,4)$ , or any coordinate, for that matter. Applying the shoelace formula, we get the area as $\boxed 6$.

~coolmath2017

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_9&oldid=121766"