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 $\frac {|3 + 0 - 7|}{\sqrt {1^2 + 1^2}} = 2\sqrt 2$ Therefore its area is $\frac 12 \cdot 3\sqrt 2 \cdot 2\sqrt 2 = \boxed{6}$ . The answer is $\mathrm{(A)}$ .

2009 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2009 AMC 12B Problems/Problem 9

Contents

Problem

Solution

Solution 1

Solution 2

See also