# 2016 AMC 8 Problems/Problem 2

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

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

$[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0)); label("A", (0,0), SW); label("B", (6, 0), SE); label("C", (6,8), NE); label("D", (0, 8), NW); label("M", (0, 4), W); label("4", (0, 2), W); label("6", (3, 0), S);[/asy]$

## Contents

### Solution 1

Use the triangle area formula for triangles: $A = \frac{bh}{2},$ where $A$ is the area, $b$ is the base, and $h$ is the height. This equation gives us $A = \frac{4 \cdot 6}{2} = \frac{24}{2} =\boxed{\textbf{(A) } 12}$.

### Solution 2

A triangle with the same height and base as a rectangle is half of the rectangle's area. This means that a triangle with half of the base of the rectangle and also the same height means its area is one quarter of the rectangle's area. Therefore, we get $\frac{48}{4} =\boxed{\textbf{(A) } 12}$.

### Solution 3(a check)

We can find the area of the entire rectangle, DCBA to be $8 \cdot 6=48$ and find DCM area to be $\frac{6 \cdot 4}{2} = 12$ and BCA to be $\frac{6 \cdot 8}{2}=24$ $48-12-24=$ $\boxed{\textbf{(A) } 12}$.