All edges of a square pyramid are of length 12 cm.
 $ \left( a \right) $ What is the area of one lateral face of it?
 $ \left( b \right) $ What is the surface area of this pyramid?
 $ \left( c \right) $ Prove that if the length of the edges is doubled then the new surface area of the pyramid is 4 times the previous one.

Hint: In this particular question first draw the pictorial representation of the given problem it will give us a clear picture of what we have to find out then use the concept that if all the sides of a triangle is equal so it is an equilateral triangle, so use these concepts to reach the solution of the question.

Complete step-by-step answer:

Given data:
All edges of a square pyramid are of length 12 cm as shown in the above figure.
It consists of 1 square and four equilateral triangles which are given as,
Square = ABCD
Equilateral triangles = ABE, BCE, CDE, DAE.
 $ \left( a \right) $ What is the area of one lateral face of it?
As we see from the figure that the base of the pyramid is square and it has four equilateral triangles as all the edges of the pyramid are equal so if all the sides of a triangle is equal so it is an equilateral triangle.
Now as we know that the area of an equilateral triangle = $ \dfrac{{\sqrt 3 }}{4}{\left( {{\text{side}}} \right)^2} $ sq. units.
So the surface area of one lateral face of the pyramid = $ \dfrac{{\sqrt 3 }}{4}{\left( {{\text{12}}} \right)^2} = 36\sqrt 3 $ sq. cm.
 $ \left( b \right) $ What is the surface area of this pyramid?
The surface area of the pyramid = area of the base (square) + surface area of four equilateral triangles.
So, the surface area of the pyramid = $ {\left( {{\text{side}}} \right)^2} + 4\left[ {\dfrac{{\sqrt 3 }}{4}{{\left( {{\text{side}}} \right)}^2}} \right] $
Now substitute the values we have,
So, the surface area of the pyramid = $ {\left( {{\text{12}}} \right)^2} + 4\left[ {\dfrac{{\sqrt 3 }}{4}{{\left( {12} \right)}^2}} \right] = 144 + 144\sqrt 3 = 144\left( {1 + \sqrt 3 } \right) $ Sq. cm.
 $ \Rightarrow {\left( {S.A} \right)_1} = 144\left( {1 + \sqrt 3 } \right) $ Sq. cm.
 $ \left( c \right) $ Prove that if the length of the edges is doubled then the new surface area of the pyramid is 4 times the previous one.
Let the length of the edge of the new pyramid be a’.
So the length of the edge of the new pyramid = 2 times the length of the previous edge of the pyramid.
Therefore, a’ = 2(12) = 24 cm.
So the surface area of the new pyramid = $ {\left( {{\text{new side}}} \right)^2} + 4\left[ {\dfrac{{\sqrt 3 }}{4}{{\left( {{\text{new side}}} \right)}^2}} \right] $
 $ \Rightarrow {\left( {S.A} \right)_2} = {\left( {{\text{24}}} \right)^2} + 4\left[ {\dfrac{{\sqrt 3 }}{4}{{\left( {{\text{24}}} \right)}^2}} \right] $
 $ \Rightarrow {\left( {S.A} \right)_2} = 576 + 576\sqrt 3 = 576\left( {1 + \sqrt 3 } \right) $ Sq. cm.
 $ \Rightarrow {\left( {S.A} \right)_2} = \left( 4 \right)144\left( {1 + \sqrt 3 } \right) $
Therefore, $ {\left( {S.A} \right)_2} = 4{\left( {S.A} \right)_1} $
Hence proved.

Note: Whenever we face such types of question the key concept we have to remember is that always recall the formula of the area of the equilateral triangle which is stated above and the total surface area of the square pyramid is the sum of the area of the base which is in square form and the area of the 4 equilateral triangles.