question_answer

Polygon ABCDE is divided into parts as shown below: 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.

Answer:

Given, polygon ABCDE is divided into four parts, so it is clear from given figure that

Area of polygon ABCDE = Area of \[\Delta AFB\]= Area of trapezium FBCH + Area of \[\Delta CHD\]+ Area of \[\Delta ADE\]                                                                            ....(i)

Also, AD = 8 cm, AH = 6 cm, AG = 4 cm, AF = 3 cm, BF = 2 cm, CH = 3 cm and EG = 2.5 cm

Now, are of

\[\Delta AFB=\frac{1}{2}\times AF\times BF=\frac{1}{2}\times 3\times 2=3c{{m}^{2}}\]

Area of trapezium FBCH

\[=\text{ }\frac{1}{2}\text{ }\times \text{FH}\times \left( BF+CH \right)=\frac{1}{2}\times \text{3}\times (2+3)\]

[\[\because FH=AH-AF\text{ }=\text{ }6-3\text{ }=\text{ }cm\]]

\[=\frac{1}{2}\times 3\times 5=\frac{1}{2}=7.5\,c{{m}^{2}}\]

Area of

\[\Delta CHD\text{ }=\frac{1}{2}\times HD\times CH\text{ }=\frac{1}{2}\times \left( AD-AH \right)\times CH\]

\[\left[ \because HD=AD-AH \right]\]

\[=\frac{1}{2}\times (8-6)\times 3=\frac{1}{2}\times 2\times 3=3c{{m}^{2}}\]

Now,     area of\[\Delta ADE=\frac{1}{2}\text{ }\times AD\times GE\]

\[=\frac{1}{2}\times 8\times 2.5\]

\[=4\times 2.5=10c{{m}^{2}}\]

On putting all these values is Eq. (i), we get

Area of polygon ABCDE \[=\left( 3+7.5+3+10 \right)\]= \[23.5\text{ }c{{m}^{2}}\]

Hence, area of polygon ABCDE is 23.5\[c{{m}^{2}}.\]