Question

Find the area of the quadrilateral whose vertices are at (-9,-2),(-8,-4),(2,2),(1,-3)

Answer 1

Answer:

The area of the quadrilateral is 35 sq. units.

Step-by-step explanation:

Given:

Let the vertices be A( -9, -2), B( -8, -4), C( 2, 2), D( 1, -3).

Construction:

Joining AC,

There are 2 triangles formed ABC & BCD.

Hence,

Area of quadrilateral ABCD = Area of Δ ABC + Area of Δ BCD.

Finding area of Δ ABC :

Area of triangle ABC = $\frac{1}{2}$ [$x_{1}(y_{2} - y_{3} ) + x_{2} (y_{3} - y_{1} ) + x_{3} (y_{1} - y_{2} )$]

here,

x1=-9, x2=-8, x3=2

y1=-2, y2=-4, y3=2

Putting the values,

Area of the triangles = 13 sq. units.

Finding area of Δ BCD :

Area of triangle BCD = $\frac{1}{2}$ [$x_{1}(y_{2} - y_{3} ) + x_{2} (y_{3} - y_{1} ) + x_{3} (y_{1} - y_{2} )$]

Here, x1=-8, x2=1, x3=2

y1=-4, y2=-3, y3=2

Putting the values,

Area of triangle BCD = 22 sq. units.

∴Area of quadrilateral ABCD = Area of Δ ABC + Area of Δ BCD.

                                               = (13+22)sq. units

                                              =35 sq. units.