Answer:
[tex](24\sqrt{3}+72)\text{ square unit}[/tex]
Step-by-step explanation:
Since, the area of a regular hexagon is,
[tex]A=\frac{3\sqrt{3}}{2}a^2[/tex]
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,
[tex]A_B=\frac{3\sqrt{3}}{2}(4)^2[/tex]
[tex]=\frac{48\sqrt{3}}{2}[/tex]
[tex]=24\sqrt{3}\text{ square unit}[/tex]
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,
[tex]A_L=6\times \frac{1}{2}\times 4\times 6[/tex]
[tex]=\frac{144}{2}[/tex]
[tex]=72\text{ square unit}[/tex]
Hence, the total area of the pyramid is,
[tex]T.A.=A_B+A_L[/tex]
[tex]=(24\sqrt{3}+72)\text{ square unit}[/tex]
Answer:
[tex]24\sqrt3+72[/tex] 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=[tex]A_B+A_L[/tex]
Where [tex]A_B[/tex]=Base area
[tex]A_L[/tex]=Lateral area
Area of hexagonal base=[tex]\frac{3\sqrt3}{2}a^2[/tex]
Where a= Side length
Now, area of hexagonal base=[tex]\frac{3\sqrt3}{2}(4)^2=24\sqrt3[/tex] square units
Area of triangular face=[tex]\frac{1}{2}\times base\times height=\frac{1}{2}\times 6\times 4=12[/tex] square units
In pyramid , there are 6 triangular faces.
Therefore, lateral area of pyramid=[tex]6\times 12=72[/tex] square units
Substitute the values in the given formula then, we get
Total lateral area of given pyramid=[tex]24\sqrt3+72[/tex] Square units
