Question

Find the area of the quadrilateral whose vertices are at (-9,0), (-8,6),(-1,-2) and (-6, -3)

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Answer 1

Step-by-step explanation:

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Answer 2

The area of the quadrilateral for the given vertices are 34 square units.

Step 1: Given data          

Vertices of quadrilateral are given as (-9,0), (-8,6), (-1,-2), and (-6,-3).

Step 2:

To find

The area of the quadrilateral for the given vertices.            

Step 3:

Formula used for the area of quadrilateral.                            

        Area of quadrilateral$=\frac{1}{2}\left|\begin{array}{ll}x_{1}-x_{3} & x_{2}-x_{4} \\y_{1}-y_{3} & y_{2}-y_{4}\end{array}\right|$          

Name the vertices

$x_{1} = -9 , y_{1} = 0 \\x_{2} = -8, y_{2} = 6\\x_{3} = -1, y_{3} = 2\\x_{4} = -6, y_{4} = -3$                          

Substitute those values in the above formula,

We get, Area of quadrilateral $=\frac{1}{2}\left|\begin{array}{ll}-9-(-1) & -8-(-6) \\0-(-2)& 6-(-3)\end{array}\right|$

Area of quadrilateral $=\frac{1}{2}\left|\begin{array}{ll}-9+1 & -8+6 \\0+2& 6+3\end{array}\right|$

Area of quadrilateral $=\frac{1}{2}\left|\begin{array}{ll}-8 & -2 \\+2& +9\end{array}\right|$

While cross multiplying, we get

Area of quadrilateral $=\frac{1}{2} \left|\begin{array}{ll}(-8\times9) -(-2\times2)\end{array}\right|$

Area of quadrilateral $=\frac{1}{2} \left|\begin{array}{ll}-72 -(-4)\end{array}\right|$

Area of quadrilateral $=\frac{1}{2} \left|\begin{array}{ll}-72 +4\end{array}\right|$

Area of quadrilateral $=\frac{1}{2} \left|\begin{array}{ll}-68\end{array}\right|$

Because of mod, neglecting the negative values, we get

Area of quadrilateral $=\frac{1}{2} \times 68$

Area of quadrilateral = 34 square units.

Hence, the area of quadrilateral is 34 square units.          

To learn more...

1) If a(-2,1),b(9,0),c(4,b) and d(1,a) are the vertices of Parallogram ABC find the value of a and b hence find the length of side

brainly.in/question/3142431

2) Find the area of the quadrilateral field abcd whose sides ab=40m ,bc=28m,cd=15m,ad=9m tringlea=90

brainly.in/question/1988542