Given:
ABCD is a quadrilateral
AC = 12, DE = 4 and FB = 5
To find:
The area of the polygon.
Solution:
AC bisects the quadrilateral into two triangles.
Area of triangle:
[tex]$A=\frac{1}{2}\times\text{base}\times\text{height}[/tex]
Area of triangle DAC:
[tex]$A=\frac{1}{2}\times AC \times DE[/tex]
[tex]$A=\frac{1}{2}\times 12 \times 4[/tex]
[tex]A=24[/tex]
Area of triangle DAC = 24 square units.
Area of triangle BAC:
[tex]$A=\frac{1}{2}\times AC \times FB[/tex]
[tex]$A=\frac{1}{2}\times 12 \times 5[/tex]
[tex]A=30[/tex]
Area of triangle BAC = 30 square units.
Area of polygon = Area of triangle DAC + Area of triangle BAC
= 24 square units + 30 square units
= 54 square units
The area of the polygon is 54 square units.
