Answer
132 $in^{2}$
Work Step by Step
Lets give the name ABCD for the given quadrilateral
lets divide the quadrilateral ABCD into two right triangles ABC and ACD
Area of quadrilateral = area of 1 triangle + area of 2 triangle
In $\triangle$ ABC By Pythagoras theorem
$AB^{2}$ = $AC^{2}$ + $BC^{2}$
$20^{2}$ = $AC^{2}$ + $16^{2}$
400 = $AC^{2}$ + 256
$AC^{2}$ = 400 - 256 = 144
AC = 12 in
Area of triangle whose base has length b and altitude has length h is given by A = $\frac{1}{2}$ bh
From given diagram base b = 16 in
h = 12 in
Area = $\frac{1}{2}$ * 16 * 12 = 96 $in^{2}$
Similarly in right triangle ADC base b = 6 in
h = 12 in
Area = $\frac{1}{2}$ * 6 * 12 = 36 $in^{2}$
The area of the given diagram = Area of ABC + Area of ADC
= 96 + 36 = 132 $in^{2}$
