Question 870943
As written, there are no units to the {{{10sqrt(2)}}} and {{{12}}} length of the diagonal and height,
but I will assume the measurements are all in inches.
Looking at the base, from the bottom, it looks like this:
{{{drawing(300,300,-5.5,5.5,-5.5,5.5,
rectangle(-5,-5,5,5),rectangle(5,-5,4.8,-4.8),
green(line(-5,-5,5,5)),
locate(0,0.2,green(10sqrt(2))),
locate(-0.2,-5.02,x),locate(5.1,0,x),
circle(-5,0,0.1),locate(-4.8,0.2,M),
circle(5,0,0.1),locate(4.6,0.2,N),
circle(0,0,0.1),locate(-0.5,0.4,C)
)}}} C is the center of the base; M and N are the midpoints of opposite sides of the base.
The side of the square base of the pyramid measures {{{x}}} inches.
The diagonal length, in inches, is {{{10sqrt(2)}}} .
According to the Pythagorean theorem,
{{{x^2+x^2=(10sqrt(2))^2}}}
{{{2x^2=10^2(sqrt(2))^2}}}
{{{2x^2=10^2*2}}}
{{{x^2=10^2}}}
{{{x=10}}}
Cutting the pyramid into halves through the vertex and the middle of opposite sides,
the section would look look this:
{{{drawing(300,375,-6,6,-1.5,13.5,
green(line(0,0,0,12)),
green(rectangle(0,0,-0.3,0.3)),
triangle(-5,0,5,0,0,12),
locate(-2.2,0.55,5),locate(2.3,0.55,5),
locate(0.1,6.2,green(12)),
locate(-2.5,6.1,y),locate(-0.1,-0.02,C),
locate(-5.1,-0.02,M),locate(4.9,-0.02,N),
locate(-0.1,12.55,V)
)}}} V is the vertex of the pyramid; C is the center of the base; M and N are the midpoints of opposite sides of the base.
The slant height, measured from the midpoint of a side of the square base to the vertex, is {{{y}}} inches.
According to the Pythagorean theorem,
{{{y^2=5^2+12^2}}}
{{{y^2=25+144}}}
{{{y^2=169}}}
{{{y^2=13^2}}}
{{{y=13}}}
The lateral surface of the pyramid is made of {{{4}}} triangular sides.
Each of those triangles has a base measuring {{{10}}} inches (the side of the square base of the pyramid),
and a height of {{{13}}} inches (the slant height of the pyramid.
The area of a triangle is calculated as {{{(1/2)*base*height}}} .
The total lateral surface area of the pyramid, in square inches, is
{{{4*(1/2)*10*13=highlight(260)}}}