Difference between revisions of "2000 AMC 8 Problems/Problem 25"
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<math>[\triangle AMN] = 27</math>, and the answer is <math>\boxed{B}</math> | <math>[\triangle AMN] = 27</math>, and the answer is <math>\boxed{B}</math> | ||
− | ==Solution 2== | + | ==Solution 3== |
+ | Let's assume, for simplicity, that the sides of the rectangle are <math>9</math> and <math>8.</math> The area of the 3 triangles would then be <math>8\cdot\frac{9}{2}\cdot\frac{1}{2} = 18,</math> | ||
+ | <math>4\cdot\frac{9}{2}\cdot\frac{1}{2} = 9,</math> | ||
+ | <math>4\cdot 9\cdot\frac{1}{2} = 18.</math> | ||
+ | Adding these up, we get <math>45</math>, and subtracting that from <math>72</math>, we get <math>27</math>, so the answer is <math>\boxed{B}</math> | ||
− | + | ~ilee0820 | |
− | + | == Video Solution == | |
+ | https://youtu.be/yoIO9q_GTig. Soo, DRMS, NM | ||
− | + | https://www.youtube.com/watch?v=XxQwfirFn4M ~David | |
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==See Also== | ==See Also== |
Latest revision as of 10:10, 12 June 2024
Problem
The area of rectangle is
units squared. If point
and the midpoints of
and
are joined to form a triangle, the area of that triangle is
Solution 1
To quickly solve this multiple choice problem, make the (not necessarily valid, but very convenient) assumption that can have any dimension. Give the rectangle dimensions of
and
, which is the easiest way to avoid fractions. Labelling the right midpoint as
, and the bottom midpoint as
, we know that
, and
.
, and the answer is
Solution 3
Let's assume, for simplicity, that the sides of the rectangle are and
The area of the 3 triangles would then be
Adding these up, we get
, and subtracting that from
, we get
, so the answer is
~ilee0820
Video Solution
https://youtu.be/yoIO9q_GTig. Soo, DRMS, NM
https://www.youtube.com/watch?v=XxQwfirFn4M ~David
See Also
2000 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Question | |
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.