+296 Rectangle ABCD is cut by two lines as shown. Find the area of AEXH.
https://latex.artofproblemsolving.com/6/6/c/66cb67543703cafc48a84fd49998af9ddbc7d9c1.png
+128474 Not the easiest way to solve this, I suspect
Note that the distance from X to AB = (1/2)AD = (1/2) (9) = 4.5 units
The line containing HF has a slope of 5/8
Extend BA in the direction of A and FH in the direction of H to meet at I
So..... the length of IA can be found as
5/8 = 2/ IA
5/16 = 1/ IA
IA = 16/5
So the area of triangle IAH = (1/2)(Base)(Height) = (1/2)(IA)(HA) = (1/2)(16/5) (2) = 16/5 = 3.2 units
And the area of triangle XEI = (1/2) (Base)(Height) = (1/2)(5 + 16/5) (4.5) = 2.25 ( 41/5) = 18.45 units
So [AEXH] = [ XEI ] - [ IAH] = [ 18.45 ] - [3.2 ] = 15.25 units^2
Here's a pic :
