Question

If A(-5 7) B(-4 -5) C(-1 -6) and D(4 5) are the vertices of a parallelogram taken in order then find the area.​

Answer 1

Let the vertices of the quadrilateral be A(−5,7),B(−4,−5),C(−1,−6),D(4,5).

Joining AC there are two triangles ABC,ADC

Therefore area of quadrilateral ABCD=Area of ΔABC+Area of ΔADC

Area of ΔABC=

2

1

[x

1

(y

2

−y

3

)+x

2

(y

3

−y

1

)+x

3

(y

1

−y

2

)]

Here x

1

=−5,y

1

=7,x

2

=−4,y

2

=−5,x

3

=−1,y

3

=−6

Area of ΔABC=

2

1

[−5(−5+6)−4(−6−7)−1(7+5)]=

2

1

(35) square units.

Area of ΔADC=

2

1

[x

1

(y

2

−y

3

)+x

2

(y

3

−y

1

)+x

3

(y

1

−y

2

)]

Here x

1

=−5,y

1

=7,x

2

=4,y

2

=5,x

3

=−1,y

3

=−6

Area of ΔADC=

2

1

[−5(5+6)+4(−6−7)−1(7−5)]=

2

1

(−109)=

2

1

(109) square units. (since area cannot be negative)

Therefore area of quadrilateral ABCD=Area of ΔABC+Area of ΔADC

=

2

1

(35)+

2

1

109

=

2

1

(144)

=72

Therefore area of quadrilateral ABCD=72 square units