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 $\boxed 6$ $ .

2009 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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All AMC 12 Problems and Solutions

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