Find the area of the parallelogram whose vertices are A (-5,2,5) B (-3,6,7) C (4,-1,5) D (2,-5,3)

Given quadrilateral has vertices A (-5,2,5) B (-3,6,7)  C (4,-1,5) D (2,-5,3)

Consider ΔABC

Area of ΔABC =

Here "X" represents a cross product.

⇒ 1/2 | (-3×-2)-(0×-7), (0×7) - (9×-2), (9×-7) - (-3×7)|

⇒ 1/2 |6, 18, -42|

⇒

⇒ 1062 sq units

Consider ΔACD

Area of ΔACD =

Here "X" represents cross product.

A (-5,2,5) C (4,-1,5) D (2,-5,3)

⇒ 1/2 | (-7×-2)-(-2×-4), (-2×-2) - (7×-2), (7×-4) - (-7×-2)|

⇒ 1/2 |6, 18, -42|

⇒

⇒ 1062 sq units

So, the area of the quadrilateral ABCD = Area of ΔABC + Area of ΔACD
= 1062 + 1062
= 2124 sq units

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