Question 1194213
<br>
{{{drawing(400,400,0,20,0,20
,line(2,11,8,11),line(8,11,8,5),line(8,5,16,5),line(16,5,16,11),line(16,11,18,11),line(18,11,18,2),line(18,2,2,2),line(2,2,2,11)
,locate(3,12,"(1)"),locate(5,12,6),locate(6.5,9,"(2)"),locate(7,8,x),
locate(10,6,"(3)"),locate(12,6,y),locate(14.5,9,"(4)"),locate(15,8,x)
,locate(16,12,"(5)"),locate(17.5,12,2)
,locate(18.5,5,"(6)"),locate(19,4,9)
,locate(10,2,"(7)"),locate(12,2,16)
,locate(.5,5,"(8)")
,green(line(14,5,14,2)),locate(14.5,4,3)
)}}}
The figure is a closed figure; the eighth side ends where the first side starts.<br>
That means the "rights" and "lefts" must be equal, and the "ups" and "downs" must be equal.<br>
rights: side 1 (6), side 3 (unknown), and side 5 (2)
lefts: side 7 (16)<br>
6+y+2 = 16
y=8
So side 3 has length 8.<br>
ups and downs....<br>
side 2 (down) and side 4 (up) are the same length, so those cancel each other out.  The only other down is side 6 (9), so side 8 (up) is also 9.<br>
The vertical distance between sides 3 and 7 is 3; since side 6 is 9, the length of sides 2 and 4 is 9-3 = 6.<br>
So the figure is a rectangle with sides 16 and 9, with a rectangle with side lengths 6 and 8 removed.<br>
The area of the figure is then (16*9)-(6*8) = 144-48 = 96.<br>
ANSWER: 96 square units<br>