Area Of Quadrilateral
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
The formula to be used to find the area is
Required area = 1⁄2 {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
= 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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- Orthocentre of a triangle
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- Locus
- Perpendicular distance
- Angle between two straight lines
- Equation of a circle
- With center and radius
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- Length of the tangent to a circle
- Equation of the tangent to a circle
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