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Find the Area of Quadrilateral Abcd Whose Vertices Are A(-5, 7), B(-4, -5) C(-1,-6) and D(4,5) - Mathematics

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Find the area of quadrilateral ABCD whose vertices are A(-5, 7), B(-4, -5) C(-1,-6) and D(4,5)

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Solution

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

`" let"  A (x_1,y_1)=A(-5,7) , B(x_2,y_2) = B(-4,-5) , C (x_3,y_3) = c (-1,-6) and D(x_4,y_4) = D(4,5)`

Then 

`"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[-5(-5+6)-4(-6-7)-1(7+5)]`

`=1/2[-5+52-12]=35/2` sq. units

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

`=1/2 [-5(-6-5)-1(5-7)+4(7+6)]`

`=1/2[55+2+52]=109/2 `sq. units

So, the area of the quadrilateral ABCD is `35/2+109/2=72 ` sq .units.

Concept: Coordinate Geometry
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APPEARS IN

RS Aggarwal Secondary School Class 10 Maths
Chapter 16 Coordinate Geomentry
Q 5

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