Find the area of a quadrilateral ABCD, the coordinates of whose varities are A (-3, 2), B (5, 4), C (7, -6) and (-5, -4).

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Solution

Area of triangle is given by $\frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$

Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ADC

$= \frac{1}{2} | - 3 (4 + 6) + 5 (- 6 - 2) + 7 (2 - 4) | + \frac{1}{2} | - 3 (- 4 + 6) - 5 (- 6 - 2) + 7 (2 + 4) |$

$= 42 + 38 = 80 s q . u n i t s$

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Similar questions

Find the area of a quadrilateral $A B C D$ , the coordinates of whose vertices are $A (- 3, 2), B (5, 4), C (7, - 6)$ and $D (- 5, - 4)$ .

A (- 3, 2), B (5, 4), C (7, - 6)

D (- 5, - 4)

A (- 3, 2), B (5, 4), C (7, - 6)

D (- 5, - 4)

A (5, 4),

B (- 3, 2),

C (- 5, - 4)

D (7, - 6)

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