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

Answer:

$(24\sqrt{3}+72)\text{ square unit}$

Step-by-step explanation:

Since, the area of a regular hexagon is,

$A=\frac{3\sqrt{3}}{2}a^2$

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,

$A_B=\frac{3\sqrt{3}}{2}(4)^2$

$=\frac{48\sqrt{3}}{2}$

$=24\sqrt{3}\text{ square unit}$

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,

$A_L=6\times \frac{1}{2}\times 4\times 6$

$=\frac{144}{2}$

$=72\text{ square unit}$

Hence, the total area of the pyramid is,

$T.A.=A_B+A_L$

$=(24\sqrt{3}+72)\text{ square unit}$

Answer 2

Answer:

$24\sqrt3+72$ Square units

Step-by-step 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=$\frac{3\sqrt3}{2}a^2$

Where a= Side length

Now, area of hexagonal  base=$\frac{3\sqrt3}{2}(4)^2=24\sqrt3$ square units

Area of triangular face=$\frac{1}{2}\times base\times height=\frac{1}{2}\times 6\times 4=12$ square units

In pyramid , there are 6 triangular faces.

Therefore, lateral area of pyramid=$6\times 12=72$ square units

Substitute the values in the given formula then, we get

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