Difference between revisions of "2016 AMC 8 Problems/Problem 2"
(=) |
m (→Solution 1) |
||
| Line 15: | Line 15: | ||
===Solution 1=== | ===Solution 1=== | ||
| − | We simply use the base times height formula for triangles | + | 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>. |
===Solution 2=== | ===Solution 2=== | ||
Revision as of 17:54, 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)--(6,8)--(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
We simply use the base times height formula for triangles 
where 
is the area, 
is the base, and 
is the height. This equation gives us 
.
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 
.
| 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. 
