Difference between revisions of "2002 AMC 8 Problems/Problem 20"
| Line 23: | Line 23: | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2002|num-b=19|num-a=21}} | {{AMC8 box|year=2002|num-b=19|num-a=21}} | ||
| + | {{MAA Notice}} | ||
Revision as of 23:46, 4 July 2013
Problem
The area of triangle 
is 8 square inches. Points 
and 
are midpoints of congruent segments 
and 
. Altitude 
bisects 
. What is the area (in square inches) of the shaded region?
![[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
The shaded region is a right trapezoid. Assume WLOG that 
. Then because the area of 
is equal to 8, the height of the triangle 
. Because the line 
is a midsegment, the top base of the triangle is 
. Also, divides 
in two, so the height of the trapezoid is 
. The bottom base is 
. The area of the shaded region is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
