Question

Find the area of hexagon ABCDEF in which BL IAD, CM LAD,
EN LAD and FP 1 AD such that AP = 6 cm, PL = 2 cm.
LN = 8 cm, NM = 2 cm, MD = 3 cm, FP = S cm. EN = 12 cm.
BL = 8 cm and CM = 6 cm.​

Answer 1

Step-by-step explanation:

The given details are,

BL⊥AD, CM⊥AD, EN⊥AD and FP⊥AD 

$$AP=6 cm, PL= 2 cm, LN=8 cm, NM=2 cm, MD=3 cm, FP=8 cm, EN=12 cm. BL= 8 cm , 

CM=6cm.

AL=AP+PL=6+2=8cm

PN=PL+LN=2+8=10cm

LM=LN+NM=8+2=10cm

ND=NM+MD=2+3=5cm

By Using the formula,

Area (hex. ABCDEF) =area(△APF)+area(△DEN)+area(△ABL)+area(△CMD)+area(Trap.PNEF)+area(Trap.LMCB)

Area of triangle =1/2×base×height

Area of trapezium =1/2×(sum of parallel sides)×height

∴lets calculate,

Area(△APF)=1/2(AP)×(FP)=1/2×6×8=24cm2 

Area(△DEN)=1/2(ND)×(EN)=1/2×5×12=30cm2

Area(△ABL)=1/2(AL)×(BL)=1/2×