Q. 84.2( 32 Votes )

# Find the area of

Answer :

Given: a regular hexagon ABCDEF

AB = BC = CD = DE = EF = FA = 13 cm

AD = 23 cm

Here AL = MD

Therefore Let AL = MD = x

Here AD = AL + LM + MD

23 = 13 + 2x

2x = 23 – 13 = 10

x = 5

Now,

In ABL using Pythagoras theorem

AB^{2} = AL^{2} + LB^{2}

13^{2} = x^{2} + LB^{2}

13^{2} = 5^{2} + LB^{2}

169 = 25 + LB^{2}

LB^{2} = 169 – 25 = 144

LB = 12

Here area (Trap. ABCD) = area (Trap. AFED)

Therefore,

Area (Hex. ABCDEF) = 2 × area (Trap. ABCD)

Area of trapezium = × (sum of parallel sides) × height

Area (Trap. ABCD) = × (BC + AD) × LB = × (13 + 23) × 12 = 216 cm^{2}.

Area(ABCDEFGH) = 2 × area (Trap. ABCD) = 2 × 216 = 432 cm^{2}

Area(ABCDEFGH) = 432 cm^{2}.

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