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:
[tex]h= \sqrt{ x^{2} - (x/2)^{2}}= \sqrt{ x^{2}- x^{2}/4}= \sqrt{3x^2/4}= \frac{ \sqrt{3} }{2}x [/tex]
thus,
[tex]Area_{triangle}= \frac{1}{2}\cdot base \cdot height\\\\18= \frac{1}{2}\cdot x \cdot \frac{ \sqrt{3} }{2}x\\\\ \frac{18 \cdot 4}{ \sqrt{3}}=x^2 [/tex]
thus:
[tex]x^2 =\frac{18 \cdot 4}{ \sqrt{3}}= \frac{18 \cdot 4 \cdot\ \sqrt{3} }{3}=24 \sqrt{3} [/tex] (cm^2)
but [tex] x^{2} [/tex] is exactly the base area, since the base is a square of sidelength = x cm.
So, the total surface area = base area + lateral area = [tex]24 \sqrt{3}+72[/tex] cm^2
Answer: [tex]24 \sqrt{3}+72[/tex] cm^2
