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> | ||
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==Solution 2== | ==Solution 2== | ||
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<math>[\triangle AMN] = \frac{3}{8}lh = \frac{3}{8}\cdot 72 = 27</math>, and the answer is <math>\boxed{B}</math> | <math>[\triangle AMN] = \frac{3}{8}lh = \frac{3}{8}\cdot 72 = 27</math>, and the answer is <math>\boxed{B}</math> | ||
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==Solution 3== | ==Solution 3== | ||
− | Let's assume | + | Let's assume WLOG 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\frac{9}{2}\cdot\frac{1}{2} = 9,</math> | ||
<math>4\cdot 9\cdot\frac{1}{2} = 18.</math> | <math>4\cdot 9\cdot\frac{1}{2} = 18.</math> | ||
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==See Also== | ==See Also== | ||
+ | M4st3r0fm4th | ||
+ | |||
+ | SlimeKnight | ||
− | {{AMC8 box|year=2000|num-b=24|after=Last<br | + | {{AMC8 box|year=2000|num-b=24|after=Last<br>Question}} |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 17:26, 27 October 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 2
The above answer is fast, but satisfying, and assumes that the area of is independent of the dimensions of the rectangle. All in all, it's a very good answer though. However this is an alternative if you don't get the above answer. Label and
Labelling and as the right and lower midpoints respectively, and redoing all the work above, we get:
, and the answer is
Solution 3
Let's assume WLOG 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
M4st3r0fm4th
SlimeKnight
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 |
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