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Find the area of quadrilateral PQRS whose vertices are P(–5, –3), Q(–4, –6), R(2, –3) and S(1, 2).

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Question

Find the area of quadrilateral PQRS whose vertices are P(–5, –3), Q(–4, –6), R(2, –3) and S(1, 2).

Sum
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Solution

By joining P and R, we get two triangles PQR and PRS.

`"let"  P (x_1, y_1) = P (-5,-3), Q(x_2,y_2) = Q(-4,-6), R (x_3,y_3) = R(2,-3) and  . Then  S(x_4,y_4) = S(1,2)`

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

`=1/2 [-5(-6+3)-4(-3+3)+2(-3+6)}`

`=1/2 [15-0+6]=21/2 sq. units`

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

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

`=1/2[25+10+0]=35/2 sq. units`

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

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Chapter 6: Coordinate Geometry - EXERCISE 6C [Page 341]

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R.S. Aggarwal Mathematics [English] Class 10
Chapter 6 Coordinate Geometry
EXERCISE 6C | Q 3. | Page 341
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