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Find the Area of Quadrilateral Abcd Whose Vertices Are A(-3, -1), B(-2,-4) C(4,-1) and D(3,4) - Mathematics

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Find the area of quadrilateral ABCD whose vertices are A(-3, -1), B(-2,-4) C(4,-1) and D(3,4)

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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(-2,-4) , C(x_3,y_3) = C(4,-1) and  Then  D (x_4,y_4)= D(3,4)`

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

`=1/2 [-3(-4+1)-2(-1+1)+4(-1+4)]`

`=1/2 [9-0+12]=21/2 ` 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(-1-4)+4(4+1)+3(-1+1)]`

`=1/2 [15+20+0]=35/2` sq. units

So, the area of the quadrilateral ABCD is `21/2+35/2=28 `.sq units sq units

Concept: Coordinate Geometry
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APPEARS IN

RS Aggarwal Secondary School Class 10 Maths
Chapter 16 Coordinate Geomentry
Q 4

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