Question

find the area of quadrilateral ABCD whose vertices are (3, -1) B(9, -5) C(14, 0) and D(9, 19) ​

Answer 1

Step-by-step explanation:

let sides of quadrilateral is A(3,-1) ,B(9,-5) ,C(14,0) and D(9,19)

divide quadrilateral in two parts by joining point B and D.

by formula of area of triangle

In ΔABD

1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]

1/2[3(-5-19)+9(19+1)+9(-1+5)]

1/2[3(-24)+9(20)+9(4)]

1/2[144]

72

In ΔBCD

1/2[9(0-19)+14(19+5)+9(-5-0)]

1/2[9(-19)+14(24)+9(-5)]

1/2[120]

60

area of quadrilateral =area of ΔABD + area of ΔBCD

=72+60

=132sq. unit