Asked by Jethro
                A lateral edge of the hexagonal pyramid is 2.0m long and its base perimeter is 6m determine the volume of the pyramid and its lateral area
            
            
        Answers
                    Answered by
            Reiny
            
    The base perimeter is 6 m, so each side of the hexagon is 1 m
Let's concentrate on one of the triangles of the pyramid, it will be isosceles with sides
2 m, 2m and 1 m. Draw a perpendicular and you will have a right-angled triangle, let
the height be h
then h^2 + (1/2)^2 = 2^2
h^2 = 15/4
h = √15/2
So the area of one of those triangles = (1/2)(1)(√15/2) = √15/4
So 6 of them would be 3√15/2 m^2 <---- lateral surface area
No look at the base, it is also made up of 6 isosceles triangle, sketch one of them
the area of one of them = (1/2)(1)(1)sin60° = (1/2)(√3/2) = √3/4
and with 6 of those, we get a base area of 3√3/2 m^2
We need the height of the pyramid, call it k
Back to the equilateral triangle of the base, all sides have 1, so with basic
geometry we see that the "height" in one of these is √3/2
k^2 + (√3/2)^2 = (√15/2)^2
k^2 + 3/4 = 15/4
k^2 = 3
k = √3, <---- nice
volume of pyramid = (1/3)(base area)(height)
= (1/3)(3√3/2)(√3) = 3/2 m^3
better check my arithmetic on that, should have written it out first.
    
Let's concentrate on one of the triangles of the pyramid, it will be isosceles with sides
2 m, 2m and 1 m. Draw a perpendicular and you will have a right-angled triangle, let
the height be h
then h^2 + (1/2)^2 = 2^2
h^2 = 15/4
h = √15/2
So the area of one of those triangles = (1/2)(1)(√15/2) = √15/4
So 6 of them would be 3√15/2 m^2 <---- lateral surface area
No look at the base, it is also made up of 6 isosceles triangle, sketch one of them
the area of one of them = (1/2)(1)(1)sin60° = (1/2)(√3/2) = √3/4
and with 6 of those, we get a base area of 3√3/2 m^2
We need the height of the pyramid, call it k
Back to the equilateral triangle of the base, all sides have 1, so with basic
geometry we see that the "height" in one of these is √3/2
k^2 + (√3/2)^2 = (√15/2)^2
k^2 + 3/4 = 15/4
k^2 = 3
k = √3, <---- nice
volume of pyramid = (1/3)(base area)(height)
= (1/3)(3√3/2)(√3) = 3/2 m^3
better check my arithmetic on that, should have written it out first.
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.