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2016 AMC 8 Problems/Problem 2

Revision as of 22:32, 8 November 2020 by Suakim (talk | contribs) (→‎Solution 3(a check))

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 $8 \cdot 6=48$ and find DCM area to be $\frac{6 \cdot 4}{2} = 12$ and BCA to be $\frac{6 \cdot 8}{2}=24$ $48-12-24=$ $\boxed{\textbf{(A) } 12}$.

Solution 4

A triangle is half of a rectangle. So Since M is the midpoint of DA, we can see that triangle DAC is half of the whole rectangle. And it is also true that triangle MAC is half of triangle DAC, so triangle MAC would just be 1/4 of the whole rectangle. Since the rectangle's area is 48, 1/4 of 48 would be 12. Which gives us the answer $\boxed{\textbf{(A) } 12}$.

Video Solution

https://www.youtube.com/watch?v=LqnQQcUVJmA (has questions 1-5)

2016 AMC 8 (Problems • Answer Key • Resources)
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