In this page area of quadrilateral we are going to see the formula to be used to find the area of the quadrilateral.To find the area we are given four vertices.We need to consider the given vertices as

(x1,y1), (x2,y2), (x3,y3),(x4,y4)

The formula to be used to find the area is

Required area = 12 {x1y2+x2y3+x3y4+x4y1)-(x2y1+x3y2+x4y3+x1y4)} We need to apply the the appropriate values in the formula to get the area.Now let us see the example problems to understand this topic more clear

Example 1:

Find the area formed by the points (-3,-9) (-1,6) (3,9) and (5,-8).

Now we have

x1 = -3 , y1 = -9 , x2 = -1 , y2 = 6 , x3 = 3 , y3 = 9 , x4 = 5 , y4 = -8

= 1/2 {(-3(6)+(-1)9+3(-8)+5(-9))-((-1)(-9)+3(6)+5(9)+(-3)(-8))}

= 1/2 {(-18-9-24-45) - (9+18+45+24)}

= 1/2 {-96 -96}

= 1/2 {-192}

= -192/2

= -96

Area cannot be negative. So the answer is 96 Square Units.

Example 2:

Find the area formed by the points (5,8) (6,3) (3,1) and (2,6)

Now we have

x1 = 5 , y1 = 8 , x2 = 6 , y2 = 3 , x3 = 3 , y3 = 1 , x4 = 2 , y4 = 6

= 1/2 {(5(3)+6(1)+3(6)+2(8))-(6(8)+3(3)+2(1)+5(6)}

= 1/2 {(15+6+18+16) - (48+9+2+30)}

= 1/2 {55 -89}

= 1/2 {-34}

= -34/2

= -17

Area cannot be negative. So the answer is 17 Square units.

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Area of quadrilateral to Analytical Geometry

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