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(ii)     Polygon ABCDE is divided into parts as shown below (Fig 11.18). Find its area if $AD = 8 cm$ , $AH = 6 cm$ , $AG = 4 cm$ , $AF = 3 cm$          and perpendiculars $BF = 2 cm$ , $CH = 3 cm$ ,   $EG = 2.5 cm$ . Area of Polygon ABCDE = area of $\Delta AFB+...$

Area of $\Delta AFB=\frac{1}{2}\times AF\times BF=\frac{1}{2}\times 3\times 2=....$

Area of trapezium $FBCH=FH\times \frac{(BF+CH)}{2}$

                                                 $=3\times \frac{(2+3)}{2}[FH=AH-AF]$

Area of $\Delta CHD=\frac{1}{2}\times HD\times CH=.....,$   Area of $\Delta ADE=\frac{1}{2}\times AD\times GE=...$

           So, the area of polygon ABCDE = ....

Answers (1)

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Area of Polygon ABCDE = area of $\traingle$ $\bigtriangleup AFB$ + area of trapezium BCHF + area of $\traingle$ $\bigtriangleup CDH$ +area of $\traingle$ $\bigtriangleup AED$ $=(\frac{1}{2}\times AF\times BF)+\left ( FH\times \left ( BF+CH \right )\div 2 \right )+(\frac{1}{2}\times HD\times CH)+(\frac{1}{2}\times AD\times EG)$

$=(\frac{1}{2}\times 3\times2)+\left ( 3\times \left ( 2+3 \right )\div 2 \right )+(\frac{1}{2}\times 2\times 3)+(\frac{1}{2}\times 8\times2.5)$

$= 3+7.5+3+10$

$=23.5 cm^{2}$

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seema garhwal

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