For a better understanding of the solution/expression provided here please find the diagram of the side view of the pyramid in the attached file.
BC is the base. [tex] BC=8 [/tex].
AP is the height. [tex] AP=10 [/tex]. Therefore, [tex] PC=4 [/tex]
As we can see [tex] \Delta APC [/tex] is a right triangle where, AC which is the hypotenuse of the triangle is the slant height of the pyramid.
Thus, applying the Pythagoras' Theorem in [tex] \Delta APC [/tex] we get:
[tex] AC=\sqrt{(AP)^2+(PC)^2}=\sqrt{10^2+4^2}=\sqrt{116}\approx10.77 m [/tex]
Therefore the slant height of the pyramid is represented by 10.77 m. Thus, Option A is the correct option.
