Question

Find the area of the pentagon shown in the figure if AD=10 cm,AG=8 cm,AH=6 cm,AF=5 cm and BF=5 cm,CG=7 cm and EH=3 cm.

Answer 1

In the figure,here are three triangles and one trapezium.

AD=10 cm, BF=5 cm

AG=8 cm, CG=7 cm

AH=6 cm,EH=3 cm

AF=5 cm

$Nowareaof△ADE=\frac{1}{2}\times AD\times EH=\frac{1}{2}\times 10\times 3c{m}^2=15c{m}^{2}Areaof△ABF=\frac{1}{2}\times AF\times BF=\frac{1}{2}\times 5\times 5=\frac{25}{2}c{m}^2=12.5c{m}^{2}Areaof△CDG=\frac{1}{2}\times GD\times CG=\frac{1}{2}[AD-AG]\times CG$

$=\frac{1}{2}(10-8)\times 7c{m}^2=\frac{1}{2}\times 2\times 7=7c{m}^{2}AreaoftrapeziumBFGC=\frac{1}{2}(BF+CG)\times FG=\frac{1}{2}(5+7)\times (AG-AF)=\frac{1}{2}\times 12\times (8-5)c{m}^2=6\times 3=18c{m}^2\therefore AreaofthefigureABCDE=15+12.5+7+18=52.5c{m}^2$