1995 AHSME Problems/Problem 10

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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 (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 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 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

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