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

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 $\frac 12 \cdot 6 \cdot 12 = 36\ \mathrm{(E)}$ .

1995 AHSME (Problems)
Preceded by Problem 9	Followed by Problem 11
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1995 AHSME Problems/Problem 10

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Problem

Solution

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