Difference between revisions of "2016 AMC 8 Problems/Problem 2"
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+ | == Problem == | ||
+ | |||
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>\triangle AMC</math>? | 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>\triangle AMC</math>? | ||
− | <math>\ | + | <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> | ||
+ | |||
+ | <math>\text{(A) }12\qquad\text{(B) }15\qquad\text{(C) }18\qquad\text{(D) }20\qquad \text{(E) }24</math> | ||
+ | |||
+ | |||
+ | |||
+ | ==Solutions== | ||
+ | |||
+ | ===Solution 1=== | ||
+ | |||
+ | Use the triangle area 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} =\boxed{\textbf{(A) } 12}</math>. | ||
+ | |||
+ | ===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 <math>\frac{48}{4} =\boxed{\textbf{(A) } 12}</math>. | ||
+ | |||
+ | ===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>. | ||
+ | |||
+ | ===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== | ||
+ | https://www.youtube.com/watch?v=LqnQQcUVJmA (has questions 1-5) | ||
− | |||
− | |||
+ | ==See Also== | ||
{{AMC8 box|year=2016|num-b=1|num-a=3}} | {{AMC8 box|year=2016|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 17:30, 21 April 2021
Contents
Problem
In rectangle , and . Point is the midpoint of . What is the area of ?
Solutions
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)
See Also
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.