Question

find the area of a regular hexagon ABCDE which each side 13 cm height is 23 CM as shown in the figure.​

Answer 1

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

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.