Question

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.​

Answer 1

Step-by-step explanation:

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.

Answer 2

Step-by-step explanation:

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