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  • 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.

(ii) Polygon ABCDE is divided into parts as shown below (Fig 11.18). Find its area if $A D=8 \mathrm{~cm}, A H=6 \mathrm{~cm}, A G=4 \mathrm{~cm}, A F=3 \mathrm{~cm}$ and perpendiculars $B F=2 \mathrm{~cm}, C H=3 \mathrm{~cm}, E G=2.5 \mathrm{~cm}$. Area of Polygon ABCDE = area of $\triangle A F B+\ldots$

Area of $\Delta A F B=\frac{1}{2} \times A F \times B F=\frac{1}{2} \times 3 \times 2=\ldots$

Area of trapezium $F B C H=F H \times \frac{(B F+C H)}{2}$ 

$=3 \times \frac{(2+3)}{2}[F H=A H-A F]$

Area of $\Delta C H D=\frac{1}{2} \times H D \times C H=\ldots$
Area of $\triangle A D E=\frac{1}{2} \times A D \times G E=\ldots$

So, the area of polygon $ABCDE = $

Answers (1)

best_answer

Area of Polygon ABCDE = area of \bigtriangleup AFB + area of trapezium BCHF + area of \bigtriangleup CDH+area of \bigtriangleup AED   =(\frac{1}{2}\times AF\times BF)+ ( FH\times \frac{( BF+CH)}{2})+(\frac{1}{2}\times HD\times CH)+(\frac{1}{2}\times AD\times EG) =(\frac{1}{2}\times 3\times2)+( 3\times \frac{( 2+3 )}{2})+(\frac{1}{2}\times 2\times 3)+(\frac{1}{2}\times 8\times2.5)

= 3+7.5+3+10

=23.5 cm^{2}

 

Posted by

seema garhwal

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