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]$

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*6=48 and find DCM area to be 6*4/2=12 and BCA to be 6*8/2=24 48-12-24=12

2016 AMC 8 (Problems • Answer Key • Resources)
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2016 AMC 8 Problems/Problem 2

Solution 1

Solution 2

Solution 3(a check)