# Find the Area of the Following Polygon, If Al = 10 Cm, Am = 20 Cm, An = 50 Cm, Ao = 60 Cm And Ad = 90 Cm. - Mathematics

Sum

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)
$=(\frac{1}{2}\times AM\times MF)+[\frac{1}{2} \times (MF+OE)\times(OM)]+(\frac{1}{2}\times OD\times OE)+(\frac{1}{2}\times DN\times NC) +[ \frac{1}{2} \times (LB+NC)\times(NL)]+(\frac{1}{2} \times AL\times LB)$
$=(\frac{1}{2}\times20\times20)+[\frac{1}{2} \times (20+60)\times(40)]+(\frac{1}{2} \times 30\times60)+(\frac{1}{2}\times40\times40) +[ \frac{1}{2} \times (30+40)\times(40)]+(\frac{1}{2} \times 10 \times 30)$
$=200+1600+900+800+1400+150$
${=5050 cm}^2$

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#### APPEARS IN

RD Sharma Class 8 Maths
Chapter 20 Mensuration - I (Area of a Trapezium and a Polygon)
Exercise 20.3 | Q 4 | Page 29