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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) ?

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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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, area of quad ABCD= area triangle ABC+ area triangle ADC

area 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

Similarly,

area 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

Posted by

Suraj Bhandari

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