Question

The lateral edges of a regular hexagonal are all 20cm long and all base edges are 16cm long.What is volume of the pyramid.

Answer 1

The volume of the pyramid is $V=\frac{1}{3}\times$ base $\times$ height

Since the base is a hexagon we use formula for area of hexagon

Base Area $=\frac{3}{2}\sqrt{3}{s}^2$ here is the length of side

Base Area $=\frac{3}{2}\sqrt{3}\times {16}^2=384\sqrt{3}$

The height of the pyramid can be formed by Pthyagorean theorem

$h=\sqrt{{20}^2-{16}^2}=\sqrt{400-256}=\sqrt{144}=12$

Volume is $=\frac{1}{3}\left(384\sqrt{3}\right)\left(12\right)=1536\sqrt{3}=2660{cm}^3$

Now we have refined the surface area and add all area of forces. The area of six lateral forces & $6$ times the area of one.

Area of triangle $=\frac{1}{2}\times$ base $\times$ height

we have an isosceles triangle with sides cocmad base $10$ cm

Height $=\sqrt{{20}^2-5^2}=\sqrt{375}$

Area of six lateral forces $=6\times \frac{1}{2}\times 16\times \sqrt{375}=48\sqrt{375}$

Surface of the pyramid $=384\sqrt{3}+48\sqrt{375}=1595{cm}^2$