Answer:
1.25
Step-by-step explanation:
I think it would be the second question!
A pyramid is regular if its base is a regular polygon, that is a polygon with equal sides and angle measures.
(and the lateral edges of the pyramid are also equal to each other)
Thus a regular rectangular pyramid is a regular pyramid with a square base, of side length say
x.
The lateral faces are equilateral triangles of side length
x.
The lateral surface area is 72 cm^2, thus the area of one face is 72/4=36/2=18 cm^2.
now we need to find
x. Consider the picture attached, showing one lateral face of the pyramid.
by the Pythagorean theorem:

thus,

thus:

(cm^2)
but

is exactly the base area, since the base is a square of sidelength =
x cm.
So, the total surface area = base area + lateral area =

cm^2
Answer:

cm^2
Answer: do it have a answer choice if it do then its B,C and e
Step-by-step explanation: