2002 AMC 8 Problems/Problem 20
Contents
Problem
The area of triangle 
is 8 square inches. Points 
and 
are midpoints of congruent segments 
and 
. Altitude 
bisects 
. The area (in square inches) of the shaded region is
![[asy] /* AMC8 2002 #20 Problem */ draw((0,0)--(10,0)--(5,4)--cycle); draw((2.5,2)--(7.5,2)); draw((5,4)--(5,0)); fill((0,0)--(2.5,2)--(5,2)--(5,0)--cycle, mediumgrey); label(scale(0.8)*"$X$", (5,4), N); label(scale(0.8)*"$Y$", (0,0), W); label(scale(0.8)*"$Z$", (10,0), E); label(scale(0.8)*"$A$", (2.5,2.2), W); label(scale(0.8)*"$B$", (7.5,2.2), E); label(scale(0.8)*"$C$", (5,0), S); fill((0,-.8)--(1,-.8)--(1,-.95)--cycle, white);[/asy]](http://latex.artofproblemsolving.com/3/a/8/3a879a53da609c8f6948eef78732e18f14a90fe4.png)
Solution 1
We know the area of triangle is 
square inches. The area of a triangle can also be represented as 
or in this problem 
. By solving, we have ![\[\frac{XC\cdot YZ}{2} = 8,\]](http://latex.artofproblemsolving.com/b/4/e/b4e08e6d320f1c6830cef20de6bbf4399b5722e2.png)
![\[XC\cdot YZ = 16.\]](http://latex.artofproblemsolving.com/8/4/6/84667c66408f8ed434a0d7fefd303a5b81f24cbb.png)
With SAS congruence, triangles 
and 
are congruent. Hence, triangle 
. (Let's say point 
is the intersection between line segments 
and 
.) We can find the area of the trapezoid 
by subtracting the area of triangle 
from 
.
We find the area of triangle by the formula- 
. 
is 
of 
from solution 1. The area of is ![\[\frac{\frac{XC}{2}\cdot \frac{YZ}{4}}{2} = \frac{16}{16} = 1\]](http://latex.artofproblemsolving.com/1/f/3/1f3effd38d41efdd1f62bcc3171bfcec110527db.png)
.
Therefore, the area of the shaded area- trapezoid has area 
.
- sarah07
Video Solution
https://www.youtube.com/watch?v=zwy5U5IQi88 ~David
See Also
| 2002 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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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