Answer:
[tex]51m^2[/tex]
Step-by-step explanation:
The lateral area of the square pyramid is given by;
[tex]A_L=a\sqrt{a^2+4h^2}[/tex]
where [tex]a=4m,h=6m[/tex].
We substitute the values to obtain;
[tex]A_L=4\sqrt{4^2+4(6^2)}[/tex]
[tex]A_L=4\sqrt{160}[/tex]
[tex]A_L=16\sqrt{10}[/tex]
[tex]A_L=50.5964[/tex]
[tex]A_L\approx 51m^2[/tex]
