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Question

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

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Solution

By joining A and C, we get two triangles ABC and ACD

Let A (3, -1), B (9, -5) and C (14, 0) and D (9, 19)

(x1=3,y1=1),(x2=9,y2=5),(x3=14,y3=0),(x4=9,y4=19)

Then

Area of triangle ABC

=12(x1(y2y3)+x2(y3y1)+x3(y1y2))


=12×(3(50)+(9)(0+1)+(14)(1+5))


=12×(3(5)+9(1)+14(4))


=12×(15+9+56)

=12×(50)

= 25 sq.units

Area of triangle ACD

12×(x1(y2y3)+x2(y3y1)+x3(y1y2))

=12×(3(019)+(14)(19+(1))+(9)(10)))


=12×(3(19)+14(20)+9(1))


=12×(57+2809)

=12×(214)

= 107 sq.units

So, the area of the quadrilateral is 25 + 107 = 132 sq. units


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