Difference between revisions of "2016 AMC 8 Problems/Problem 2"
(Created page with "In rectangle <math>ABCD</math>, <math>AB=6</math> and <math>AD=8</math>. Point <math>M</math> is the midpoint of <math>\overline{AD}</math>. What is the area of <math>\triangl...") |
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==Solution== | ==Solution== | ||
| − | {{ | + | [asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,0)); |
| + | label("<math>A</math>", (0,0), SW); | ||
| + | label("<math>B</math>", (6, 0), SE); | ||
| + | label("<math>C</math>", (6,8), NE); | ||
| + | label("<math>D</math>", (0, 8), NW); | ||
| + | label("<math>M</math>", (0, 4), W); | ||
| + | label("<math>4</math>", (0, 2), W); | ||
| + | label("<math>6</math>", (3, 0), S);[/asy] | ||
| + | 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 09:54, 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("
", (0,0), SW);
label("
", (6, 0), SE);
label("
", (6,8), NE);
label("
", (0, 8), NW);
label("", (0, 4), W);
label("
", (0, 2), W);
label("
", (3, 0), S);[/asy]
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. 
