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Method 1:
A = 2 1 ∣ ∣ ∣ ∣ x 1 y 1 x 2 y 2 x 3 y 3 x 4 y 4 x 1 x 1 ∣ ∣ ∣ ∣ = 2 1 ∣ ∣ ∣ ∣ 5 3 0 5 − 8 0 4 − 2 5 3 ∣ ∣ ∣ ∣ = 2 1 [ 2 5 + 0 + 1 6 + 1 2 − ( 0 − 4 0 + 1 0 − 1 0 ) ] = 2 1 ( 5 3 + 5 0 ) = 2 1 ( 1 0 3 ) = 5 1 . 5
Method 2:
Enclosed it with a rectangle then subtract the areas of the four triangles.
A
=
1
3
(
7
)
−
2
1
[
8
(
5
)
+
5
(
2
)
+
1
2
(
2
)
+
5
(
1
)
]
=
9
1
−
2
1
(
7
9
)
=
9
1
−
3
9
.
5
=
5
1
.
5