Find the area of the following polygon, if AL = 10 cm, AM = 20 cm, AN = 50 cm, AO = 60 cm and AD = 90 cm.
Solution
The given polygon is:
Given:
AL=10 cm, AM=20 cm, AN=50 cm
\[AO=60 cm, AD=90 cm\]
Hence, we have the following:
\[MO=AO-AM=60-20=40 cm\]
\[OD=AD-AO=90-60=30 cm\]
\[ND=AD-AN=90-50=40 cm\]
\[LN=AN-AL=50-10=40 cm\]
From given figure:
Area of Polygon=(Area of triangle AMF)+(Area of trapezium MOEF)+(Area of triangle EOD)+(Area of triangle DNC)+ (Area of trapezium NLBC )+(Area of triangle ALB)
\[=(\frac{1}{2}\times AM\times MF)+[\frac{1}{2} \times (MF+OE)\times(OM)]+(\frac{1}{2}\times OD\times OE)+(\frac{1}{2}\times DN\times NC) +[ \frac{1}{2} \times (LB+NC)\times(NL)]+(\frac{1}{2} \times AL\times LB)\]
\[=(\frac{1}{2}\times20\times20)+[\frac{1}{2} \times (20+60)\times(40)]+(\frac{1}{2} \times 30\times60)+(\frac{1}{2}\times40\times40) +[ \frac{1}{2} \times (30+40)\times(40)]+(\frac{1}{2} \times 10 \times 30)\]
\[=200+1600+900+800+1400+150\]
\[ {=5050 cm}^2\]
