Find the area of the parallelogram with vertices at A(5,9,1) B(5,3,8) C(13,19,-9) D(13,13,-2)  

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You may split the parallelogram in two triangles and you may evaluate the areas of the triangles using the following formula called Heron's formula, such that:

`A = sqrt(p(p - l_1)(p - l_2)(p - l_2))`

l_1,l_2,l_3 represent the lengths of sides of triangle

`p = (l_1+l_2+l_3)/2 ` (half perimeter)

You need to evaluate the area of triangle `Delta ABC` , such that:

`A_(Delta ABC) = sqrt(p(p-AB)(p-AC)(p-BC))`

`p = (AB+BC+AC)/2`

`AB = sqrt((5-5)^2 + (3-9)^2 + (8-1)^2)`

`AB = sqrt(36 + 49) => AB = sqrt 45 = 3sqrt5`

`AC = sqrt((13-5)^2 + (19-9)^2 + (-9-1)^2)`

`AC = sqrt(64 + 100 + 100) => AC = sqrt 264 = 2sqrt 66`

`BC = sqrt((13-5)^2 + (19-3)^2 + (-9-8)^2)`

`BC = sqrt(64 + 256 + 289) => BC = sqrt609`

`p = (AB+BC+AC)/2 = (3sqrt5 + 2sqrt 66 + sqrt609)/2`

`p-AB = (-AB+BC+AC)/2 = (-3sqrt5 + 2sqrt 66 + sqrt609)/2`

`p-BC = (AB-BC+AC)/2 = (3sqrt5 - 2sqrt 66 + sqrt609)/2`

`p-AC = (AB+BC-AC)/2 = (3sqrt5 + 2sqrt 66 - sqrt609)/2`

`p*(p-AB) = ((2sqrt 66 + sqrt609)^2 - 45)/4`

`(p-BC)(p-AC) = (45 - (2sqrt 66 - sqrt609)^2)/4`

`(45 - (2sqrt 66 - sqrt609)^2)/4*((2sqrt 66 + sqrt609)^2 - 45)/4 = (45(2sqrt 66 + sqrt609)^2 - 45^2 - (2sqrt 66 - sqrt609)^2(2sqrt 66 + sqrt609)^2 + 45(2sqrt 66 - sqrt609)^2)/16`

You need to evaluate the area of triangle `Delta ADC` using the same Heron's formula.

Hence, evaluating the area of parallelogram ABCD yields `sqrt(p(p-AB)(p-AC)(p-BC)) + sqrt(p(p-AD)(p-AC)(p-DC)).`

3sqrt5

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