Difference between revisions of "2016 AMC 8 Problems/Problem 2"
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===Solution 3(a check)=== | ===Solution 3(a check)=== | ||
We can find the area of the entire rectangle, DCBA to be <math>8 \cdot 6=48</math> and find DCM area to be <math>\frac{6 \cdot 4}{2} = 12</math> and BCA to be <math>\frac{6 \cdot 8}{2}=24</math> <math>48-12-24=</math> <math>\boxed{\textbf{(A) } 12}</math>. | We can find the area of the entire rectangle, DCBA to be <math>8 \cdot 6=48</math> and find DCM area to be <math>\frac{6 \cdot 4}{2} = 12</math> and BCA to be <math>\frac{6 \cdot 8}{2}=24</math> <math>48-12-24=</math> <math>\boxed{\textbf{(A) } 12}</math>. | ||
| + | |||
| + | ==Solution 4== | ||
| + | A triangle is half of a rectangle. So Since M is the midpoint of DA, we can see that triangle DAC is half of the whole rectangle. And it is also true that triangle MAC is half of triangle DAC, so triangle MAC would just be 1/4 of the whole rectangle. Since the rectangle's area is 48, 1/4 of 48 would be 12. Which gives us the answer <math>\boxed{\textbf{(A) } 12}</math>. | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 22:32, 8 November 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, DCBA to be 
and find DCM area to be 
and BCA to be 
.
Solution 4
A triangle is half of a rectangle. So Since M is the midpoint of DA, we can see that triangle DAC is half of the whole rectangle. And it is also true that triangle MAC is half of triangle DAC, so triangle MAC would just be 1/4 of the whole rectangle. Since the rectangle's area is 48, 1/4 of 48 would be 12. Which gives us the answer .
Video Solution
https://www.youtube.com/watch?v=LqnQQcUVJmA (has questions 1-5)
| 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. 
