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Find the Area of a Quadrilateral Abcd Whose Vertices Area A(3, -1), B(9, -5) C(14, 0) and D(9, 19). - Mathematics

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Find the area of a quadrilateral ABCD whose vertices area 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(x_1,y_1)=A(3,-1),B(x_2,y_2)=B(9,-5),C(x_3,y_3)=C(14,0) and D(x_4 , y_4 ) = D(9,19)`

Then,

`"Area of"  ΔABC = 1/2 [ x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1 - y_2)]`

`=1/2 [ 3(-5-0) +9(0+1)+14(-1+5)]`

`=1/2 [-15+9+56] = 25` sq.units

`Area of  ΔACD = 1/2 [ x_1 (y_3-y_4)+x_3(y_4-y_1)+x_4(y_1-y_3)]`

`=1/2 [ 3(0-19)+14(19+1)+9(-1-0)]`

`=1/2 [ -57 +280-9] = 107 sq. units`

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

 

Concept: Coordinate Geometry
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APPEARS IN

RS Aggarwal Secondary School Class 10 Maths
Chapter 16 Coordinate Geomentry
Q 2

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