Knockout NEET July 2020
An exhaustive E-learning program for the complete preparation of NEET..
₹ 15999/-
₹ 6999/-
Buy Now
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