Question

A pyramid has a regular hexagonal base with the side lengths of 4 and a slant height of 6. Find the total area of the pyramid..

Answer 1

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 

Answer 2

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 = $\frac{1}{2}$ (side of base) × slant height

                                    = $\frac{1}{2}$ × 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 = $\frac{360}{\text{number of sides}}$ ]

and ∠CAD = 30°

Now tan 30° = $\frac{DC}{AD}$ = $\frac{2}{AD}$

$\frac{1}{\sqrt{3}}=\frac{2}{AD}$

AD = $2\sqrt{3}$

Now hexagonal base area = 6 × [$\frac{1}{2}$ (BC)(AD)]

6 × [$\frac{4}{2}$ × $2\sqrt{3}$] =  $24\sqrt{3}$

Therefore area of the pyramid = 72 + 41.57 = 113.57 unit²