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Find the area of quadrilateral ABCD whose vertices are A(-3,-1) B(-2,-4) C(4,-1) D(3,4) ?????

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## Answers

A(-3,1), B(-2,-4), C(4,-1), D(3,4) are the vertices of quadrilateral ABCD

Let AC be the diagonal of quad ABCD

therefore, ar of quad ABCD= ar triangle ABC+ ar triangle ADC

ar of triangle ABC= 1/2[x1(y2-y3) -x2(y3-y1) -x3(y1-y2)]

= 1/2[ (-3) (-4+1) + (-2) (-1+1) +4(-1+4) ]

=1/2[9+0+12]

=21/2 sq units

ar of triangle ADC= 1/2[ (-3) (4+1) + 3(-1+1) + 4(-1-4)]

=1/2[-15+0-20]

= -35/2 sq units (since area cant be negative so area of triangle ADC=35/2 sq units)

Hence ar of quadrilateral ABCD= 21/2 + ( 35/2)

=56/2= 28 sq units

### This Is a Certified Answer

Here, we observe here that two points are on a horizontal line. A and C are on the line y = -1 or y + 1 = 0

So, perpendicular from B on to AC = |-4 + 1| /1 = 3 units

Perpendicular from D on to AC = | 4 + 1 | / 1 = 5 units

AC = 4 - (-3) = 7 units

Total area of ABCD = 1/2 * 7 * (3 + 5) = 28 square units