Question

Find the area of the given hexagon ABCDEF in which each one of BG, CI, EJ and FH is perpendicular to AD and it is being given that AG=6cm, AH=10cm, AI=18cm, AJ=21cm, AD=27cm, BG=5cm, CI=6cm, EJ=4cm and FH=6cm.​

Answer 1

Step-by-step explanation:

AJ=6cm;AK=10cm;AL=18cm;AM=21cm;AD=27cm;BJ=5cm;CL=6cm;EM=4cm

and FK=6cm.

LD=AD−AL=27−18=9cm

MD=AD−AM=27−21=6cm

JL=AD−AJ−LD=27−6−9=12cm

KM=AD−AK−MD=26−10−6=10cm

Area of triangles ABC,CLD,DME,AKF=(AJ×BJ)/2+(CL×LD)/2+(AK×KF)/2+(MD×ME)/2

Area of trapezium BJLC and trapezium KMEF=JL(BJ+CL)/2+KM(ME+KF)/2

Total area =(AJ×BJ)/2+(CL×LD)/2+(AK×KF)/2+(MD×ME)/2+JL(BJ+CL)/2+KM(ME+KF)/2

A=(6×5)/2+(6×9)/2