Find the area of quadrilateral joining the points (3, 2) (5, -2) (2, -5) and (0, 1) in order
Answers
Answer:
Let us plot the points roughly and take the vertices in counter clock-wise direction.
Let the vertices be
A(−4,−2),B(−3,−5),C(3,−2) and D(2,3)
Area of the quadrilateral ABCD
=
2
1
{(x
1
y
2
+x
2
y
3
+x
3
y
1
+x
4
y
1
)−(x
2
y
1
+x
3
y
2
+x
1
y
3
+x
1
y
4
)}
=
2
1
[(20+6+9−4)−(6−15−4−12)]
=
2
1
(31+25)=28 sq.units.
Step-by-step explanation:
Answer:
Let A(–4,–2), B(–3,–5), C(3,–2) and D(2,3) be the vertices of the quadrilateral ABCD.
Area of a quadrilateral ABCD= Area of △ABC+ Area of △ACD
By using a formula for the area of a triangle =
2
1
∣x
1
(y
2
−y
3
)+x
2
(y
3
−y
2
)+x
3
(y
1
−y
2
)∣
Area of △ABC
=
2
1
[−4(−5+2)+−3(−2+2)+3(−2+5)]
=
2
1
[12+9]
=
2
21
sq.units
Area of △ACD=
2
1
[−4(3+2)+−2(−2+2)+3(−2−3)]
=
2
1
[−20−15]
=
2
35
sq.units
∴Area of quadilateral=
2
21
+
2
35
=
2
56
=28 sq.unit
Step-by-step explanation: