GEOMETRY-PYRAMID

You can use the Pythagoras' theorem multiple times here.
You have four triangles of the area (4*6)/2, and one square of the area (6*6) and hence,

area = ((4*6)/2)*4+6*6
= 84.

p=24cm
So base leg=24cm/4=6cm
area base =6cm*6cm=36cm^2

cross sectional area is 3*14^.5 & this goes from corner to corner of the base (yes it is a isosceles triangle).

The triangle area is 1/2bh=3*14^.5 cm^2
the base (c) is determined from pythagoreus (a^2+b^2=c^2) where a=b=6cm
c=2^.5*6 cm

1/2ch=1/2*2^.5*6h=3*14^.5 or h=7^.5 cm
the height of the pyramis is 7^.5cm

now to determine the height of the four triangular faces of the triangle (also from pythagoreus)

h'^2=(b/2)^2+h^2=3^2+(7^.5)^2=9+7=16 cm2

h'=4cm

a pyramid side area is 1/2bh'=1/2*6cm*4cm=12cm^2=12cm^2

and there are four sides 4*12cm^2=48cm^2

add this to the base

SA pyramid=48cm^2+36cm^2=84cm^2

Rap

Heres some rough stuff I did on a piece of paper. Got the same answer as raprap so it's probably right

Thank you raprap and Red888, You hav been very helpful

I can't see the image.

Edit: Ah, there it is.