Find the area of the quadrangle with vetices (4,1), (-6,5),(-2,-6) and (3,-4)

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The quadrangle with vertices (4,1), (-6,5),(-2,-6) and (3,-4) can be divided into two triangles. One with vertices (4,1), (-6,5) and (-2,-6) and the other with vertices (4,1), (-2,-6) and (3, -4).

For the triangle with vertices (4,1), (-6,5) and (-2,-6), the length of the sides is :

`sqrt((4 + 6)^2 + (1 - 5)^2) = sqrt(100 + 16) = sqrt 116`

`sqrt((4 +2)^2 + (1 + 6)^2) = sqrt (36 + 49) = sqrt 85`

`sqrt((-6 + 2)^2 + (5 + 6)^2) = sqrt(16 + 121) = sqrt 137`

The area of the triangle using Heron's formula is:

`sqrt(((sqrt 116 + sqrt 85 + sqrt 137)/2)*((sqrt 116 - sqrt 85 + sqrt 137)/2))` `*sqrt(((sqrt 116 + sqrt 85 - sqrt 137)/2))*((-sqrt 116 + sqrt 85 + sqrt 137)/2)`

= 47

For the triangle with vertices (4,1), (-2,-6) and (3,-4), the length of the sides is :

`sqrt((4 + 2)^2 + (1 +6)^2) = sqrt(36 + 49) = sqrt 85`

`sqrt((4 - 3)^2 + (1 + 4)^2) = sqrt (1 + 25) = sqrt 26`

`sqrt((-2 - 3)^2 + (-6 + 4)^2) = sqrt(25 + 4) = sqrt 29`

The area of the triangle using Heron's formula is:

`sqrt(((sqrt 85 + sqrt 26 + sqrt 29)/2)*((sqrt 85 - sqrt 26 + sqrt 29)/2))`

`* sqrt(((sqrt 85 + sqrt 26 - sqrt 29)/2)*((-sqrt 85 + sqrt 26 + sqrt 29)/2))`

= 11.5

Adding the area of the two triangles gives 47 + 11.5 = 58.5

The area of the quadrangle with vertices (4,1), (-6,5),(-2,-6) and (3,-4) is 58.5

 

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