Question

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

Answer 1

`24\sqrt3+72` Square units

Explanation:

We are given that a pyramid which has a regular hexagonal base.

Side length of hexagonal base=Base of triangular face=4 units

Height of triangle=6 units

We have to find total area of the pyramid.

The total area of pyramid=
`A_B+A_L`

Where
`A_B`=Base area

`A_L`=Lateral area

Area of hexagonal base=
`(3\sqrt3)/(2)a^2`

Where a= Side length

Now, area of hexagonal base=
`(3\sqrt3)/(2)(4)^2=24\sqrt3` square units

Area of triangular face=
`(1)/(2)* base* height=(1)/(2)* 6* 4=12` square units

In pyramid , there are 6 triangular faces.

Therefore, lateral area of pyramid=
`6* 12=72` square units

Substitute the values in the given formula then, we get

Total lateral area of given pyramid=
`24\sqrt3+72` Square units

Answer 2

`(24√(3)+72)\text{ square unit}`

Explanation:

Since, the area of a regular hexagon is,

`A=(3√(3))/(2)a^2`

Where, a is the side of the hexagon,

Here, the base of the pyramid is a regular hexagon having side length,

a = 4 unit,

Thus, the base area of the pyramid is,

`A_B=(3√(3))/(2)(4)^2`

`=(48√(3))/(2)`

`=24√(3)\text{ square unit}`

Now, the lateral face of the pyramid is a triangle having base = 4 unit and height = 6 unit,

Also, a hexagonal pyramid has 6 triangular faces,

So, the total lateral area of the pyramid is,

`A_L=6* (1)/(2)* 4* 6`

`=(144)/(2)`

`=72\text{ square unit}`

Hence, the total area of the pyramid is,

`T.A.=A_B+A_L`

`=(24√(3)+72)\text{ square unit}`