Art of Problem Solving
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2016 AMC 8 Problems/Problem 2: Difference between revisions

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==Solution==
==Solution==
<asy>draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8));
<asy>draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0));
label("$A$", (0,0), SW);
label("$A$", (0,0), SW);
label("$B$", (6, 0), SE);
label("$B$", (6, 0), SE);

Revision as of 09:06, 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)--(6,8)--(0,0)); 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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