The total area of pyramid is 113.569 square units
The equation to find the total area of the pyramid is Total area = base area + lateral area.
Solution:
Given, A pyramid has a regular hexagonal base with side lengths of 4 and a slant height of 6.
The total area of pyramid is given by:
---- eqn 1
Where,
![A_(b) \text { is the base area and } A_(l) \text { is the lateral area }](https://img.qammunity.org/2020/formulas/mathematics/middle-school/znn9k43oxxgql8hxxor7okfbaj8z34nqyd.png)
The area of base is given as:
![A_(b)=(3 √(3))/(2) l^(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/cawozl2rz3vpljb1zqmipn6qbg02h7ovdz.png)
Where "l" is the side of hexagon.
Substituting we get,
![\begin{array}{l}{A_(b)=(3 √(3))/(2)(4)^(2)} \\\\ {=(3 √(3))/(2) * 16=3 √(3) * 8} \\\\ {A_(b)=24 √(3)}\end{array}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/aie2hlp3o2ql09yekih87e83vcm047lomh.png)
The lateral area is given as:
![A_(l)=3 b h](https://img.qammunity.org/2020/formulas/mathematics/middle-school/kso58z7b4n7pc69pb73g3em2bnmi4ezhvx.png)
Where,
b: base of the triangle
h: height of the triangle
Substituting we get,
![\begin{array}{l}{A_(l)=3 *(4) *(6)} \\\\ {A_(l)=72}\end{array}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/3vv875h2w39lj0t5hi27yqex2edwpfoy35.png)
Plugging in the values we found in eqn 1 we get,
![A=24 √(3)+72](https://img.qammunity.org/2020/formulas/mathematics/middle-school/yaaja5r8zzpcqz21x91kclnn7gb9jrzwjv.png)
A = 113.569 square units
Summarizing the results:
The total area of pyramid is 113.569 square units approximately
The equation used to find total area of pyramid is Total area = base area + lateral area.