Find the area of a regular hexagon ABCDEF in which each side measures 13 cm and whose height is 23 cm, as shown in the given figure.

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


AB2 = AL2 + LB2


132 = x2 + LB2


132 = 52 + LB2


169 = 25 + LB2


LB2 = 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 cm2.


Area(ABCDEFGH) = 2 × area (Trap. ABCD) = 2 × 216 = 432 cm2


Area(ABCDEFGH) = 432 cm2.


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