# Difference between revisions of "2009 AMC 12B Problems/Problem 9"

## 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 $\boxed 6$.

## See also

 2009 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 8 Followed byProblem 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.

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