Question

E
F
4. Find the area of hexagon ABCDEF in which BL 1 AD, CM IAD,
EN I AD and FP AD such that AP = 6 cm, PL = 2 cm,
LN = 8 cm. NM = 2 cm, MD = 3 cm, FP = 8 cm, EN = 12 cm.
BL = 8 cm and CM = 6 cm.
AS
M
N
B
O​

Answer 1

Answer:

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=24cm

2

Area(△DEN)=1/2(ND)×(EN)=1/2×5×12=30cm

2

Area(△ABL)=1/2(AL)×(BL)=1/2×8×8=32cm

2

Area(△CMD)=1/2(MD)×(CM)=1/2×3×6=9cm

2

Area(Trap.DMNE)=1/2×(FP+EN)×ON=1/2×(18+12)×1=100cm

2

Area(Trap.DMNE)=1/2×(BL+CM)×LM=1/2×(BL+CM)×LM=1/2×(8+6)×10=70cm

2

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

=24+30+32+9+100+70

=265cm

2