find the area of the pentagon whose vertices have co -ordinate (4,3),(-5,6),(0,-7),­(3,-6)&(-7,-2).

Answer 1

I would re-list them in counterclockwise order:
(4,3),(-5,6),(-7,-2),(0,-7),­(3,-6)

Now list them in a column, in order, then repeating the first one listed
(4,3)
(-5,6)
(-7,-2)
(0,-7)
­(3,-6)
(4,3)
area = (1/2)| sum of downproducts - sum of upproduct|
= (1/2)|(24+10 + 49+0+9) - (-15-42+0-21-24)|
= 97

check my arithmetic

see
http://www.mathopenref.com/coordpolygonarea.html