(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)

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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