Area of quadrilateral formed by the vertices $(- 16)$ , $(- 3, - 9)$ , $(5, - 8)$ and $(3, 9)$ is units.

96

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Solution

The correct option is A $96$
Area of polygon
$= \frac{1}{2} | (x_{1} y_{2} - x_{2} y_{1}) + (x_{3} y_{4} - y_{3} x_{4}) + . . . + (x_{n} y_{1} - x_{1} y_{n}) |$
Given the points, $A (- 1, 6)$
$B (- 3, - 9)$
$C (5, - 8)$
$D (3, 9)$
$= \frac{1}{2} | (9 + 18) + (+ 45 + 24) + (+ 24 + 45) + (+ 9 + 18) | ∴ a r e a = \frac{1}{2} | 192 | = 96 u n i t s$

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Area of quadrilateral formed by the vertices (−16), (−3,−9), (5,−8) and (3,9) is units.

The correct option is A 96Area of polygon=12|(x1y2−x2y1)+(x3y4−y3x4)+...+(xny1−x1yn)|Given the points,A(−1,6)B(−3,−9)C(5,−8)D(3,9)=12|(9+18)+(+45+24)+(+24+45)+(+9+18)|∴area=12|192|=96units

Area of quadrilateral formed by the vertices $(- 16)$ , $(- 3, - 9)$ , $(5, - 8)$ and $(3, 9)$ is units.