Math, asked by bajajpriyanka, 16 days ago

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

Answered by kanishkrajput560
0

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.

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