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

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===Solution 1===
 
===Solution 1===
  
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We simply use the base times height formula for triangles <math>A = \frac{bh}{2},</math> where <math>A</math> is the area, <math>b</math> is the base, and <math>h</math> is the height. This equation gives us <math>A = \frac{4 \cdot 6}{2} = \frac{24}{2} =\textbf{(A) } 12</math>.
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We simply use the base times height formula for triangles <math>A = \frac{bh}{2},</math> where <math>A</math> is the area, <math>b</math> is the base, and <math>h</math> is the height. This equation gives us <math>A = \frac{4 \cdot 6}{2} = \frac{24}{2} =\boxed{\textbf{(A) } 12}</math>.
  
 
===Solution 2===
 
===Solution 2===
  
−
Usually, 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 <math>\frac{48}{4} = 12</math>.
+
Usually, 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 <math>\frac{48}{4} =\boxed{\textbf{(A) } 12}</math>.
  
 
{{AMC8 box|year=2016|num-b=1|num-a=3}}
 
{{AMC8 box|year=2016|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:49, 27 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]

Solution 1

We simply use the base times height 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

Usually, 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 $\frac{48}{4} =\boxed{\textbf{(A) } 12}$.

2016 AMC 8 (Problems • Answer Key • Resources)
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All AJHSME/AMC 8 Problems and Solutions

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