the points of quadrilateral are A(-5,7),B(-4,-5),C(-1,-6),D(4,5) find the area of quadrilateral
Answers
In the quadrilateral ABCD , BD IS JOINED.
In ΔABD , X1 = (-5) , X2 = ( - 4 ) , X3 = 4
Y1 = 7 , Y2 = ( - 5 ) , Y3 = 5
∴ ar(Δ ABD) = 1/2 [X1(Y2 -Y3) + X2(Y3 - Y1) + X3(Y1 - Y2)] SQ. UNITS
= 1/2 [(-5) (- 5 -5 ) + (-4) ( 5 - 7 ) + 4 ( 7 + 5 )] SQ. UNITS
=1/2 [ ( - 5) (-10) + (-4) (-2) + ( 4 × 12 )] sq. units
= 1/2 [ 50 + 8 + 48 ] sq. units
=[ 1/2 × 106 ] sq. units
= 53 sq. units
NOW , IN Δ BCD , X1 = ( -1 ) , X2 = ( -4 ) , X3 = 4
Y1 = ( - 6 ) , Y2 = ( - 5 ) , Y3 = 5
∴ar( Δ BCD ) = 1/2 [X1 (Y2 - Y3 ) + X2 ( Y3 - Y1 ) + X3 ( Y1 - Y2 ) SQ. UNITS
= 1/2 [(-1) (- 5 - 5 ) + ( - 4 ) ( 5 + 6 ) + 4 ( -6 + 5 )] SQ. UNITS
= 1/2 [ ( - 1 ) ( - 10 ) + ( - 4 ) 11 + 4 (- 1 )] SQ. UNITS
= 1/2 [10 - 44 - 4 ] SQ. UNITS
= ( - 38 / 2 ) SQ. UNITS
= (- 19 ) SQ. UNITS
= 19 SQ. UNITS ( ∵ area cannot be negative ∴ are is taken as positive )
∴ ar ( quadrilateral ABCD ) = ar ( Δ ABD ) + ar ( Δ BCD ) sq. units
= ( 53 + 19 ) sq. units
= 72 sq. units
Hence , area of quadrilateral ABCD = 72 sq. units