Find the area of the quadrilateral $ABDC$

Question

If $A=(2,1), B(8,1), C(4,3), D(6,6)$ then find the area of the quadrilateral $ABDC$.

My Attempt:

Area of quadrilateral= area of triangle ABD + area of triangle ADC.

Area of triangle ABD= $\frac12|2(1-6)+8(6-1)+6(1-1)|=15$

Area of triangle ADC=$\frac12|2(6-3)+6(3-1)+4(1-6)|=1$

So, area of quadrilateral=16. But the answer given is 14. Maybe they didn't take the mod sign?

Also, just to be safe, let me also see the area of quadrilateral ABCD which will be area of triangle ABC + area of triangle ACD

Area of triangle ABC= $\frac12|2(1-3)+8(3-1)+4(1-1)|=6$

So, area of quadrilateral ABCD=7.

Is everything correct here?

Answer 1

It's a known "triangle": the point $C$ is located within the triangle $ABD$, so if you want to calculate the surface of the quadrilateral, you need to subtract $ADC$ from $ABD$ instead of adding it :-)

This can clearly be seen from the following image:

Answer 2

The “shoelace algorithm”

Place your points into two columns repeating the first point:

$\begin{array}{} 2&1\\ 8&1\\ 4&3\\ 6&6\\ 2&1\end{array}$

Multiply diagonally adding the results when multiplying from Northwest to Southeast and subtracting when multiplying from Southwest to Northeast.

$2-8+24-4+24-18+6-12 = 14$

If the points are arranged counter-clockwise you should get a positive result. If you get a negative result, take the absolute value.