1995 AHSME Problems/Problem 10

Problem

The area of the triangle bounded by the lines $y = x, y = - x$ and $y = 6$ is

$\mathrm{(A) \ 12 } \qquad \mathrm{(B) \ 12\sqrt{2} } \qquad \mathrm{(C) \ 24 } \qquad \mathrm{(D) \ 24\sqrt{2} } \qquad \mathrm{(E) \ 36 }$

Solution

[asy] defaultpen(fontsize(8)); draw((0,0)--(6,6)--(-6,6)--(0,0)); draw((0,-1)--(0,8)); draw((-7,0)--(7,0)); label("$(6,6)$",(6,6), (1,1));label("$(-6,6)$",(-6,6),(-1,1)); label("$y=x$",(3,3),(1,-1));label("$y=-x$",(-3,3),(-1,-0.5)); label("$6$",(-3,6),(0,1));label("$6$",(3,6),(0,1)); label("$6$",(0,3),(-1,0)); [/asy]

The height of the triangle is $y = 6$, and the width of the triangle is $|x_1| + |x_2| = 2y = 12$. Thus the area of the triangle is $\frac 12 \cdot 6 \cdot 12 = 36\ \mathrm{(E)}$.

See also

1995 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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