Answers

2015-01-02T14:46:26+05:30

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

2 4 2
THANK YOU
perhaps a simpler method exists
2015-01-02T15:27:03+05:30

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Please see diagram.  In such cases, it is simpler to first draw the diagram after plotting the points.   In general, area of a quadrilateral can be found by splitting it into two triangles by a diagonal and finding each area by using the formula.

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 

1 5 1
please click on thank you link and select best answer
Comment has been deleted