Find the area of the following polygon,

The given polygon is:

Given:

$\mathrm{AL}=10 \mathrm{~cm}, \mathrm{AM}=20 \mathrm{~cm}, \mathrm{AN}=50 \mathrm{~cm}$

$\mathrm{AO}=60 \mathrm{~cm}, \mathrm{AD}=90 \mathrm{~cm}$

Hence, we have the following:

$\mathrm{MO}=\mathrm{AO}-\mathrm{AM}=60-20=40 \mathrm{~cm}$

$\mathrm{OD}=\mathrm{AD}-\mathrm{AO}=90-60=30 \mathrm{~cm}$

$\mathrm{ND}=\mathrm{AD}-\mathrm{AN}=90-50=40 \mathrm{~cm}$

$\mathrm{LN}=\mathrm{AN}-\mathrm{AL}=50-10=40 \mathrm{~cm}$

From given figure:

Area of Polygon $=($ Area of triangle $\mathrm{AMF})+($ Area of trapezium MOEF $)+($ Area of triangle $\mathrm{EOD})+($ Area of triangle $\mathrm{DNC})+$ (Area of trapezium NLBC ) $+$ (Area of triangle ALB)

$=\left(\frac{1}{2} \times \mathrm{AM} \times \mathrm{MF}\right)+\left[\frac{1}{2} \times(\mathrm{MF}+\mathrm{OE}) \times(\mathrm{OM})\right]+\left(\frac{1}{2} \times \mathrm{OD} \times \mathrm{OE}\right)+\left(\frac{1}{2} \times \mathrm{DN} \times \mathrm{NC}\right)+\left[\frac{1}{2} \times(\mathrm{LB}+\mathrm{NC}) \times(\mathrm{NL})\right]+\left(\frac{1}{2} \times \mathrm{AL} \times \mathrm{LB}\right)$

$=\left(\frac{1}{2} \times 20 \times 20\right)+\left[\frac{1}{2} \times(20+60) \times(40)\right]+\left(\frac{1}{2} \times 30 \times 60\right)+\left(\frac{1}{2} \times 40 \times 40\right)+\left[\frac{1}{2} \times(30+40) \times(40)\right]+\left(\frac{1}{2} \times 10 \times 30\right)$

$=200+1600+900+800+1400+150$

$=5050 \mathrm{~cm}^{2}$