Answer :
for get the length of apothem write sin60 = a/4 so sqrt3 /2 = a/4 so a = 4sqrt3 /2
a = 2sqrt3
so area of base = 6(4*2sqrt3)/2 = 6*4sqrt3 = 24sqrt3 unit squared
the lateral area = 4*4*6/2 = 4*2*6 = 48 unit squared
total area = 24sqrt3 +48 = 24(sqrt3 +2) unit squared
hope this will help you
a = 2sqrt3
so area of base = 6(4*2sqrt3)/2 = 6*4sqrt3 = 24sqrt3 unit squared
the lateral area = 4*4*6/2 = 4*2*6 = 48 unit squared
total area = 24sqrt3 +48 = 24(sqrt3 +2) unit squared
hope this will help you
Answer:
113.57 unit²
Step-by-step explanation:
Surface area of a pyramid with regular hexagonal base
= Area of slant sides + area of Hexagonal base
Area of one slant side = [tex]\frac{1}{2}[/tex] (side of base) × slant height
= [tex]\frac{1}{2}[/tex] × 6 × 4
= 3 × 4
= 12 unit²
Since hexagonal pyramid has 6 slant sides.
So area of sic slant sides = 6 × 12 = 72 unit²
Now for the area of hexagonal base we will take triangle ABC.
∠BAC = 60° [angle formed at center = [tex]\frac{360}{\text{number of sides}}[/tex] ]
and ∠CAD = 30°
Now tan 30° = [tex]\frac{DC}{AD}[/tex] = [tex]\frac{2}{AD}[/tex]
[tex]\frac{1}{\sqrt{3}}=\frac{2}{AD}[/tex]
AD = [tex]2\sqrt{3}[/tex]
Now hexagonal base area = 6 × [[tex]\frac{1}{2}[/tex] (BC)(AD)]
6 × [[tex]\frac{4}{2}[/tex] × [tex]2\sqrt{3}[/tex]] = [tex]24\sqrt{3}[/tex]
Therefore area of the pyramid = 72 + 41.57 = 113.57 unit²