Given:
The length of DE is 8 cm and the measure of ∠ADE is 60°.
We need to determine the surface area of the pyramid.
Length of AD:
The length of AD is given by
[tex]cos 60^{\circ}=\frac{FD}{8}[/tex]
[tex]4=FD[/tex]
Length of AD = 8 cm
Slant height:
The slant height EF can be determined using the trigonometric ratio.
Thus, we have;
[tex]sin \ 60^{\circ}=\frac{EF}{8}[/tex]
[tex]sin \ 60^{\circ} \times 8=EF[/tex]
[tex]\frac{\sqrt{3}}{2} \times 8=EF[/tex]
[tex]4\sqrt{3}=EF[/tex]
Thus, the slant height EF is 4√3
Surface area of the square pyramid:
The surface area of the square pyramid can be determined using the formula,
[tex]SA=Area \ of \ square + \frac{1}{2} (Perimeter \ of \ base ) (slant \ height)[/tex]
Substituting the values, we have;
[tex]SA=8^2+\frac{1}{2}(8+8+8+8)(4 \sqrt{3})[/tex]
[tex]SA=64+\frac{1}{2}(32)(4 \sqrt{3})[/tex]
[tex]SA=64+(16)(4 \sqrt{3})[/tex]
[tex]SA=64+64 \sqrt{3}[/tex]
The exact form of the area of the square pyramid is [tex]64+64 \sqrt{3}[/tex]
Substituting √3 = 1.732 in the above expression, we have;
[tex]SA = 64 + 110.848[/tex]
[tex]SA = 174.848[/tex]
Rounding off to one decimal place, we get;
[tex]SA = 174.8[/tex]
Thus, the area of the square pyramid is 174.8 cm²
