The volume of the pyramid=47040[tex]m^3[/tex]
Answer:
Solution Given:
The total surface area of the pyramid = [tex]9072\:m^2[/tex]
let the height be AB be h.
length of side[l]=56 m
The volume of the pyramid=?
let slant height be AC=a.
and BC=b
we have,
The perimeter of base[P]=4l=4*56=224m
BC=28m
Now;
The total surface area of the pyramid = [tex]9072\:m^2[/tex]
lateral= surface area +base area=[tex]9072\:m^2[/tex]
[tex]\frac{1}{2}*P*a+l^2=9072\:m^2[/tex]
[tex]\frac{1}{2}*224*a+56^2=9072\:m^2[/tex]
[tex]112a=9072-3136[/tex]
[tex]a=\frac{5936}{112}=53m[/tex]
now
By using Pythagoras law;
[tex]AC^2=AB^2+BC^2[/tex]
[tex]a^2=h^2+b^2[/tex]
[tex]53^2=h^2+28^2[/tex]
[tex]h^2=2809-784[/tex]
[tex]h=\sqrt{2025}[/tex]
h=45m
now,
The volume of the pyramid=[tex]\frac{1}{3}*l^2*h[/tex]
[tex]\frac{1}{3}*56^2*45=47040m^3[/tex]
Step-by-step explanation:
