Difference between revisions of "1995 AHSME Problems/Problem 10"

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== Solution ==
 
== Solution ==
{{image}}
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<center><asy>
 
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defaultpen(fontsize(8));
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draw((0,0)--(6,6)--(-6,6)--(0,0));
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draw((0,-1)--(0,8));
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draw((-7,0)--(7,0));
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label("$(6,6)$",(6,6), (1,1));label("$(-6,6)$",(-6,6),(-1,1));
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label("$y=x$",(3,3),(1,-1));label("$y=-x$",(-3,3),(-1,-0.5));
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label("$6$",(-3,6),(0,1));label("$6$",(3,6),(0,1));
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label("$6$",(0,3),(-1,0));
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</asy></center>
 
The height of the triangle is <math>y = 6</math>, and the width of the triangle is <math>|x_1| + |x_2| = 2y = 12</math>. Thus the area of the triangle is <math>\frac 12 \cdot 6 \cdot 12 = 36\ \mathrm{(E)}</math>.
 
The height of the triangle is <math>y = 6</math>, and the width of the triangle is <math>|x_1| + |x_2| = 2y = 12</math>. Thus the area of the triangle is <math>\frac 12 \cdot 6 \cdot 12 = 36\ \mathrm{(E)}</math>.
  
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[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 12:59, 5 July 2013

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
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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