The vertices of quadrilateral ABCD are
A(-4, -2), B(-3, -5), C (3, -2) and D (2, 3).
Find the area of ABCD.
Answers
Answer:
28 square unit
Step-by-step explanation:
Join AC of the quadrilateral ABCD such that it gets divided into two triangles that are ΔACD and ΔABC.
Now, area of the triangle is given as:
A=\frac{1}{2}(x_{1}(y_{2}-y_{3})+(x_{2}(y_{3}-y_{2})+(x_{3}(y_{1}-y_{2}))A=
2
1
(x
1
(y
2
−y
3
)+(x
2
(y
3
−y
2
)+(x
3
(y
1
−y
2
))
From ΔABC, substituting the values in the formula of the area, we have
A=\frac{1}{2}((-4)(-5-(-2))+(-3)(-2-(-2))+(3)(-2-(-2)))A=
2
1
((−4)(−5−(−2))+(−3)(−2−(−2))+(3)(−2−(−2)))
A=\frac{12+0+9}{2}=\frac{21}{2}sq unitsA=
2
12+0+9
=
2
21
squnits
From ΔACD, substituting the values in the formula of the area, we have
A=\frac{1}{2}((-4)(-2-(-3))+3(3-(-2))+2(-2-(-2)))A=
2
1
((−4)(−2−(−3))+3(3−(−2))+2(−2−(−2)))
A=\frac{20+15+0}{2}=\frac{35}{2}sq unitsA=
2
20+15+0
=
2
35
squnits
Now, area ABCD=arΔACD +arΔABC
=\frac{21}{2}+\frac{35}{2}=28 sq units
2
21
+
2
35
=28squnits
Thus, the area of ABCD is 28 square units.