Find the area of the quadrilateral whose vertices are
A(-4,-2), B(0,5), C(9,3) and D(7,-4).

I thought it was a trapezoid after plotting and connecting each point and thus my answer was area = 40 but I was wrong.

2. [quote=magentarita;282648]Find the area of the quadrilateral whose vertices are
A(-4,-2), B(0,5), C(9,3) and D(7,-4).

I thought it was a trapezoid after plotting and connecting each point and thus my answer was area = 40 but I was wrong.[/quote

ignore this

3. Originally Posted by magentarita
Find the area of the quadrilateral whose vertices are
A(-4,-2), B(0,5), C(9,3) and D(7,-4).

I thought it was a trapezoid after plotting and connecting each point and thus my answer was area = 40 but I was wrong.
I've attached a sketch of the quadrilateral. By first inspection it looks like an irregular quadrilateral.

Construct a rectangular "container" such that the quadrilateral is inscribed into this container. Calculate the area of the rectangle and subtract the areas of the 4 right triangles:

$a_{rectangle} = 9 \cdot 13 = 117$
$a_1=\frac12 \cdot 2 \cdot 11 = 11$
$a_2=\frac12 \cdot 2 \cdot 7 = 7$
$a_3=\frac12 \cdot 2 \cdot 9 = 9$
$a_4=\frac12 \cdot 4 \cdot 7 = 14$

Therefore $\boxed{a_{ABCD} = 76}$

4. I see...

Originally Posted by earboth
I've attached a sketch of the quadrilateral. By first inspection it looks like an irregular quadrilateral.

Construct a rectangular "container" such that the quadrilateral is inscribed into this container. Calculate the area of the rectangle and subtract the areas of the 4 right triangles:

$a_{rectangle} = 9 \cdot 13 = 117$
$a_1=\frac12 \cdot 2 \cdot 11 = 11$
$a_2=\frac12 \cdot 2 \cdot 7 = 7$
$a_3=\frac12 \cdot 2 \cdot 9 = 9$
$a_4=\frac12 \cdot 4 \cdot 7 = 14$

Therefore $\boxed{a_{ABCD} = 76}$