2000 AMC 8 Problems/Problem 25
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
![[asy] pair A,B,C,D; A = (0,8); B = (9,8); C = (9,0); D = (0,0); draw(A--B--C--D--A--(9,4)--(4.5,0)--cycle); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW);[/asy]](http://latex.artofproblemsolving.com/a/e/2/ae2ad9d6b6a31a8972be6fd88ebf1263dcdbf75f.png)
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 
.
![$[\triangle ADN] = \frac{1}{2}\cdot 6\cdot 6 = 18$](http://latex.artofproblemsolving.com/4/2/b/42b2c422f2233d6eb1b70dd5edbe833e717ece0b.png)
![$[\triangle MNC] = \frac{1}{2}\cdot 3\cdot 6 = 9$](http://latex.artofproblemsolving.com/4/d/8/4d8b19458ef0897e7ae7739d74e38a10903ae4a6.png)
![$[\triangle ABM] = \frac{1}{2}\cdot 12\cdot 3 = 18$](http://latex.artofproblemsolving.com/5/8/4/5844ee44242ea54a7bfe3a072058f64ae1f75f40.png)
![$[\triangle AMN] = [\square ABCD] - [\triangle ADN] - [\triangle MNC] - [\triangle ABM]$](http://latex.artofproblemsolving.com/e/f/2/ef262f2dfb9e242afc07854ed32f26e7425b195f.png)
![$[\triangle AMN] = 72 - 18 - 9 - 18$](http://latex.artofproblemsolving.com/9/1/9/919a5d523f49978744cf5558a88386a2797dd349.png)
![$[\triangle AMN] = 27$](http://latex.artofproblemsolving.com/2/5/9/259887dc6a12bde654d497515f1b767be2c9c42e.png)
, 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
M4st3r0fm4th
| 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. 
