Difference between revisions of "2016 AMC 8 Problems/Problem 2"
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==Solution== | ==Solution== | ||
| − | + | <asy>draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,0)); | |
| − | label(" | + | label("$A$", (0,0), SW); |
| − | label(" | + | label("$B$", (6, 0), SE); |
| − | label(" | + | label("$C$", (6,8), NE); |
| − | label(" | + | label("$D$", (0, 8), NW); |
| − | label(" | + | label("$M$", (0, 4), W); |
| − | label(" | + | label("$4$", (0, 2), W); |
| − | label(" | + | label("$6$", (3, 0), S);</asy> |
The area of <math>\triangle AMC = \frac{1}{2} \cdot 6 \cdot 4 = \boxed{\text{(A) }12}</math>. | The area of <math>\triangle AMC = \frac{1}{2} \cdot 6 \cdot 4 = \boxed{\text{(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 08:55, 23 November 2016
In rectangle 
, 
and 
. Point 
is the midpoint of 
. What is the area of 
?
Solution
![[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,0)); 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]](http://latex.artofproblemsolving.com/6/d/7/6d7fda92fee0516c931f5098d45bd1eccf8f492f.png)
The area of 
.
| 2016 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
