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

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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 \boxed{\textbf{(A) } 12}$.  
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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 <math>\boxed{\textbf{(A) } 12}</math>.  
 
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Revision as of 18:09, 4 March 2020

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 $\boxed{\textbf{(A) } 12}$.

2016 AMC 8 (Problems • Answer Key • Resources)
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Problem 1 		Followed by
Problem 3
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All AJHSME/AMC 8 Problems and Solutions

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