Question

find the area of a quadrilateral of vertices (3, 1),(- 3, 5),(- 1, 4),(0, 3)​

Answer 1

Step-by-step explanation:

Let the vertices of quadrilateral are A(-4, -2), B(-3, -5), C(3, -2) and D(2, 3). = (10.5 + 17.5) = 28 sq. units. Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts.

Answer 2

Step-by-step explanation:

Area of a quadrilateralABCD=Area of △ABD+Area of △BCD

Area of a triangle=

2

1

∣

∣

∣

∣

∣

∣

∣

∣

x

1

x

2

x

3

y

1

y

2

y

3

1

1

1

∣

∣

∣

∣

∣

∣

∣

∣

Area of a △ABD=

2

1

∣

∣

∣

∣

∣

∣

∣

∣

−3

−5

1

4

−6

2

1

1

1

∣

∣

∣

∣

∣

∣

∣

∣

=

2

1

[−3(−6−2)−4(−5−1)+1(−10+6)]

=

2

1

[24+24−4]=

2

44

=22

Area of a △BCD=

2

1

∣

∣

∣

∣

∣

∣

∣

∣

$\infty \pi \sec( \gamma \beta )$

−5

4

1

−6

−1

2

1

1

1

∣

∣

∣

∣

∣

∣

∣

∣

=

2

1

[−5(−1−2)+6(4−1)+1(8+1)]

=

2

1

[15+18+9]=

2

42

=21

∴ Area of a quadrilateralABCD=Area of △ABD+Area of △BCD=22+21=43sq.units.