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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

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

Solution

defaultpen(fontsize(8));draw((0,0)--(6,6)--(-6,6)--(0,0));draw((0,-1)--(0,8));draw((-7,0)--(7,0));label("$(6,6)$",(...

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)
Preceded by Problem 9	Followed by Problem 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

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Category: Introductory Geometry Problems

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