Question

find the area of quadrilateral whose points are (-3,-8),(6,-6),(4,2),(-8,2)​

Answer 1

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Answer 2

Answer:

Hello.......

Let the ABCD is quadrilateral and the diagonal is AC ,which divides the quadrilateral into two triangles....

And the points A(-3,-8),B (6,-6) ,c (4,2),D(-8,2)

Area of quadrilateral =

Area of ∆ ABC + Area of ∆ ACD

Area of triangle =

1/2 | x1(y2-y3) + x2(y3-y1) + x3(y1-y2) |

Area of ∆ ABC =

= 1/2 | -3(-6-2) + 6{2-(-8)} + 4{-8-(-6)} |

= 1/2 | -3 (-8) + 6(10)+4(-2) |

= 1/2 | 24+60-8 |

= 1/2 | 76 |

= 38 sq . units...

Area of ∆ ACD

= 1/2 | -3(2-2)+4(2-(-8)+ (-8)(-8-2) |

= 1/2 | -3(0)+4(10)+(-8)(-10) |

= 1/2 | 0+40+80 |

= 1/2 |120 |

= 60 sq.units...

:. Area of quadrilateral

= 38 +60 = 98 sq .units....

Hope it helps.....

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