Difference between revisions of "2016 AMC 8 Problems/Problem 2"
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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>. | 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>. | ||
| + | ===Solution 3(a check)=== | ||
| + | We can find the area of the entire rectangle, | ||
{{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 18:06, 4 March 2020
In rectangle 
, 
and 
. Point 
is the midpoint of 
. What is the area of 
?
![[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]](http://latex.artofproblemsolving.com/b/5/d/b5d7e3f5ecfd57e90e56a341f52fc4b9b93b2457.png)
Solution 1
Use the triangle area formula for triangles: 
where 
is the area, 
is the base, and 
is the height. This equation gives us 
.
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 
.
Solution 3(a check)
We can find the area of the entire rectangle,
| 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. 
