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

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Line 12: Line 12:
 
label("$4$", (0, 2), W);
 
label("$4$", (0, 2), W);
 
label("$6$", (3, 0), S);</asy>
 
label("$6$", (3, 0), S);</asy>
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The area of <math>\triangle AMC = \frac{1}{2} \cdot 6 \cdot 4 = \boxed{\text{(A) }12}</math>.
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===Solution 1===
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We simply use the base times height formula for triangles (<math>\frac{bh}{2}</math>), giving us <math>\frac{4 \cdot 6}{2} = \frac{24}{2} = 12</math>.
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===Solution 2===
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 +
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>.
  
 
{{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 13:45, 23 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 ($\frac{bh}{2}$), giving us $\frac{4 \cdot 6}{2} = \frac{24}{2} = 12$.

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 $\frac{48}{4} = 12$.

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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