I think I have the answer, need confirmation please...

[Gr8fu13]

Find the surface area of a right regular hexagonal pyramid with sides 2cm and slant height 7cm. Round to nearest hundred in final answer.

I used the pythagorean Theorem to find the apothem:
2^2 + b^2 = 7^2
4 + b^2 = 49
b^2 = 45
b = 3 square root of 5

Then I found tha lateral sides by doing:
1/2 x 2 x 2 = 2cm^2

There are 6 sides so the lateral surface area of the pyramid is:
6(2) = 12cm^2

The area of the base is:
1/2(3 square root of 5)(12) = 18 square root of 5 cm^2

Total surface area is base, plus area of lateral surface of pyramid, so I had this:
12 + 18 square root of 5 = 52.25cm^2 for the surface area

I just noticed while typing this, the question says to find the surface area of a right regular hexagonal pyramid; it says nothing about the TOTAL surface area. Now I don't know what to do

 

[Gr8fu13]

If the base is 2 and the height is 7 then the way to find the area would be to use the formula:
a=1/2bh

So, 1/2 (2*7)=7

Is this correct?

 

[soroban]

Hello, Gr8fu13!

Find the surface area of a right regular hexagonal pyramid with sides 2 cm
and slant height 7 cm. .Round to nearest hundredth in final answer.

Note that the slant height is 7 cm.
The side panels are isosceles triangle with base 2 and equal sides 7.

            *
           /:\
          / : \
       7 /  :  \ 7
        /   :h  \
       /    :    \
      * - - + - - *
         1     1

The height \(\displaystyle h\) is given by: .\(\displaystyle h^2 + 1^2 \:=\:7^2\)

Hence: .\(\displaystyle h^2 \:=\:48 \quad\Rightarrow\quad h \:=\:\sqrt{48} \:=\:4\sqrt{3}\)

The area of a triangle is: .\(\displaystyle A \:=\:\frac{1}{2}bh \:=\:\frac{1}{2}(2)(4\sqrt{3}) \:=\:4\sqrt{3}\)

The area of six triangles is: .\(\displaystyle 24\sqrt{3}\)


The base is regular hexagon, composed of six equilateral triangles with side 2.
The area of an equilateral triangle of side \(\displaystyle s\) is: .\(\displaystyle \dfrac{\sqrt{3}}{4}s^2\)
The area of one equilateral triangle is: .\(\displaystyle \dfrac{\sqrt{3}}{4}(2^2) \:=\:\sqrt{3}\)

The area of the base (six equilateral triangles) is: .\(\displaystyle 6\sqrt{3}\)


The total surface area is: .\(\displaystyle 24\sqrt{3} + 6\sqrt{3} \:=\:30\sqrt{3}\)

. . which is approximately \(\displaystyle 51.96\text{ cm}^2.\)

 

[Gr8fu13]

Thanks for clarification, somehow I came up with something close