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)

Answers (1)

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 =  \frac{1}{2}|(x_3 - x_1) X (x_3-x_2)|

Here "X" represents a cross product.

\frac{1}{2}|(9, -3, 0) X (7, -7, -2)|

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

⇒ 1/2 |6, 18, -42|

⇒   \frac{1}{2} |(6)^2 + (18)^2 + (-42)^2 |

⇒ 1062 sq units

Consider ΔACD

Area of ΔACD =  \frac{1}{2}|(x_3 - x_1) X (x_3-x_2)|

Here "X" represents cross product.

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

\frac{1}{2}|(7, -7, -2) X (-2, -4, -2)|

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

⇒ 1/2 |6, 18, -42|

⇒   \frac{1}{2} |(6)^2 + (18)^2 + (-42)^2 |

⇒ 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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