Question 966755
<TABLE>
  <TR>
  <TD>

A sketch of your pyramid is schematically shown in the &nbsp;<B>Figure 1a</B>&nbsp; (not in the scale, though). 

The base of the given pyramid is a square, &nbsp;of course. 


Draw the straight interval from the base center to the middle of the base side. 

Its length is &nbsp;16/2 = 8 cm. 


Next, &nbsp;draw the slant height from the vertex of the pyramid to that point &nbsp;(<B>Figure 1b</B>).

Find the length of the slant height from the right triangle. &nbsp;It is 


{{{h[slant]}}} = {{{sqrt(6^2 + 8^2)}}} = {{{sqrt(36 + 64)}}} = {{{sqrt(100)}}} = {{{10}}} {{{cm}}}. 


Notice that the slant height has the same length for all lateral faces of the pyramid.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

 </TD>
  <TD>
{{{drawing( 220, 200,  -2.5, 3.5, -0.5, 3.8, 

            line (  0.0, 0.0,  3.0,  0.0),
            line (  0.0, 0.0, -2.0,  0.8),
      green(line ( -2.0, 0.8,  1.0,  0.8)),
      green(line (  1.0, 0.8,  3.0,  0.0)),

            line (  0.0, 0.0,  0.5,  3.4),
            line (  3.0, 0.0,  0.5,  3.4),
      green(line (  1.0, 0.8,  0.5,  3.4)),
            line ( -2.0, 0.8,  0.5,  3.4),

        red(line ( 0.5,  3.4,  0.5,  0.4)),

          locate ( 1.0,  0.0, 16),
          locate (-1.6,  0.5, 16),
      red(locate ( 0.54, 1.7, 6))
)}}}


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<B>Figure 1a</B>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;


 </TD>
  <TD>
{{{drawing( 220, 200,  -2.5, 3.5, -0.5, 3.8, 

            line (  0.0, 0.0,  3.0,  0.0),
            line (  0.0, 0.0, -2.0,  0.8),
      green(line ( -2.0, 0.8,  1.0,  0.8)),
      green(line (  1.0, 0.8,  3.0,  0.0)),

            line (  0.0, 0.0,  0.5,  3.4),
            line (  3.0, 0.0,  0.5,  3.4),
      green(line (  1.0, 0.8,  0.5,  3.4)),
            line ( -2.0, 0.8,  0.5,  3.4),

        red(line ( 0.5,  3.4,  0.5,  0.4)),

          locate ( 1.0,  0.0, 16),
          locate (-1.6,  0.5, 16),
      red(locate ( 0.54, 1.7, 6)),

        red(line ( 0.5,  3.4,  1.5,  0.0)),
        red(line ( 0.5,  0.4,  1.5,  0.0)),

      red(locate ( 1.2, 1.4, h)),
      red(locate ( 1.4, 1.25, slant))
)}}}


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<B>Figure 1b</B>


 </TD>
 </TR>
</TABLE>

Now, &nbsp;the area of each triangular lateral face is &nbsp;{{{1/2}}}.{{{16*10}}} = {{{80}}} {{{cm^2}}}. 

Hence, &nbsp;the lateral area of the pyramid is four times this value, &nbsp;i.e. &nbsp;{{{4*80}}} = {{{320}}} {{{cm^2}}}. 

To get the total surface area of the pyramid, &nbsp;you need to add the area of the base square, &nbsp;which is &nbsp;{{{16^2}}} = {{{256}}} {{{cm^2}}}. 


<B>Answer</B>. &nbsp;The lateral surface area of the pyramid is &nbsp;{{{320}}} {{{cm^2}}}. 

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The total surface area of the pyramid is &nbsp;{{{320}}} + {{{256}}} = {{{576}}} {{{cm^2}}}.