Difference between revisions of "2016 AMC 8 Problems/Problem 2"

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==Solution==
 
==Solution==
{{solution}}
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[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,0));
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label("<math>A</math>", (0,0), SW);
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label("<math>B</math>", (6, 0), SE);
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label("<math>C</math>", (6,8), NE);
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label("<math>D</math>", (0, 8), NW);
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label("<math>M</math>", (0, 4), W);
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label("<math>4</math>", (0, 2), W);
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label("<math>6</math>", (3, 0), S);[/asy]
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The area of <math>\triangle AMC = \frac{1}{2} \cdot 6 \cdot 4 = \boxed{\text{(A) }12}</math>.
  
 
{{AMC8 box|year=2016|num-b=1|num-a=3}}
 
{{AMC8 box|year=2016|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 09:54, 23 November 2016

In rectangle $ABCD$, $AB=6$ and $AD=8$. Point $M$ is the midpoint of $\overline{AD}$. What is the area of $\triangle AMC$?

$\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24$

Solution

[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,0)); label("$A$", (0,0), SW); label("$B$", (6, 0), SE); label("$C$", (6,8), NE); label("$D$", (0, 8), NW); label("$M$", (0, 4), W); label("$4$", (0, 2), W); label("$6$", (3, 0), S);[/asy] The area of $\triangle AMC = \frac{1}{2} \cdot 6 \cdot 4 = \boxed{\text{(A) }12}$.

2016 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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
All AJHSME/AMC 8 Problems and Solutions

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