Question 655154
since you have square pyramid (and assume a = 4 in, b = 5 in) which is a container that may {{{not}}} have a top, means you will not need the area of the base {{{a^2=4^2=16}}}

so, you will calculate lateral area:

{{{L = 4 * (1/2)a*s = 2a*s }}}

{{{L = 4 * (1/2)*4in*5in}}}

{{{L =  2*4*5in^2}}}

 {{{L = highlight(40in^2)}}}

if you include base area of a square pyramid (square):

       {{{ B = a^2}}}

{{{ B = (4in)^2}}}

{{{ B = 16in^2}}}


you will have:

    Total Surface Area of a square pyramid:

        {{{A = L + B =40in^2+16in^2}}}

 {{{ highlight(A =56in^2)}}}