Find the perimeter of quadrilateral a(-5, 4), b(0, 3), c(4,-1), d(4,-5)

Explanation:
Let us divide the quadrilateral formed by
A
(
6
,
5
)
,
B
(
2
,
−
4
)
,
C
(
−
5
,
2
)
and
D
(
−
3
,
6
)
in two parts,
Δ
A
B
C
and
Δ
A
C
D
.
Now let us find the lengths of their sides.
A
B
=
√
(
2
−
6
)
2
+
(
−
4
−
5
)
2
=
√
16
+
81
=
√
97
=
9.8489
B
C
=
√
(
−
5
−
2
)
2
+
(
2
+
4
)
2
=
√
49
+
36
=
√
85
=
9.2195
A
C
=
√
(
−
5
−
6
)
2
+
(
2
−
5
)
2
=
√
121
+
9
=
√
130
=
11.4018
Now as
s
1
=
1
2
(
9.8489
+
9.2195
+
11.4018
)
=
1
2
×
30.4702
=
15.2351
And area of
Δ
A
B
C
=
√
15.2351
(
15.2351
−
9.8489
)
(
15.2351
−
9.2195
)
(
15.2351
−
11.4018
)
=
√
15.2351
×
5.3862
×
6.0156
×
3.8333
=
√
1892.2545
=
43.5
A
D
=
√
(
−
3
−
6
)
2
+
(
6
−
5
)
2
=
√
81
+
1
=
√
82
=
9.0554
C
D
=
√
(
−
5
+
3
)
2
+
(
2
−
6
)
2
=
√
4
+
16
=
√
20
=
4.4721
s
2
=
1
2
(
9.0554
+
4.4721
+
11.4018
)
=
1
2
×
24.9293
=
12.4647
And area of
Δ
A
C
D
=
√
12.4647
(
12.4647
−
9.0554
)
(
12.4647
−
4.4721
)
(
12.4647
−
11.4018
)
=
√
12.4647
×
3.4093
×
7.9926
×
1.0636
=
√
361.2547
=
19.01
Hence area of quadrilateral
A
B
C
D
is
43.5
+
19.0
=
62.5
units.

Equatorial triangle and obtuce angle