Answer:
The lateral area of pyramid ABCDE is [tex]256\sqrt{3}[/tex] in².
Step-by-step explanation:
According to the properties of trigonometry,
[tex]\tan \theta=\frac{perpendicular}{base}[/tex]
[tex]\tan (60^{\circ})=\frac{height}{8}[/tex]
[tex]8\tan (60^{\circ})=height[/tex]
The area of a triangle is
[tex]A=\frac{1}{2}\times base \times height[/tex]
[tex]A=\frac{1}{2}\times 16 \times 8\tan (60^{\circ})[/tex]
[tex]A=8\times 8\sqrt{3}[/tex]
[tex]A_1=64\sqrt{3}[/tex]
Lateral surface area of a pyramid is the sum of area of all 4 triangles. So, the lateral area of pyramid ABCDE is
[tex]A=4\times A_1[/tex]
[tex]A=4\times 64\sqrt{3}[/tex]
[tex]A=256\sqrt{3}[/tex]
Therefore the lateral area of pyramid ABCDE is [tex]256\sqrt{3}[/tex] in².
