Answer:
Part 1) [tex]LA=108\sqrt{2}\ in^2[/tex]
Part 2) [tex]B=54\sqrt{3}\ in^2[/tex]
Step-by-step explanation:
Part 1) Find the lateral area of the pyramid
we know that
The lateral area of the pyramid, is equal to the area of six congruent isosceles triangles
so
[tex]LA=6[\frac{1}{2}(b)(h)][/tex]
we have
[tex]b=6\ in[/tex]
Applying Pythagorean Theorem calculate the height of triangle
[tex]9^2=(6/2)^2+h^2[/tex]
[tex]h^2=81-9\\h^2=72\\h=\sqrt{72}\ in[/tex]
simplify
[tex]h=6\sqrt{2}\ in[/tex]
Find the lateral area
[tex]LA=6[\frac{1}{2}(6)(6\sqrt{2})]\\\\LA=108\sqrt{2}\ in^2[/tex]
Part 2) Find the base area
we know that
The area of a regular hexagon is equal to the area of six equilateral triangles
so
Applying the law of sines to calculate the area of triangle
[tex]B=6[\frac{1}{2}b^2sin(60^o)][/tex]
we have
[tex]b=6\ in[/tex]
[tex]sin(60^o)=\frac{\sqrt{3}}{2}[/tex]
substitute
[tex]B=6[\frac{1}{2}6^2(\frac{\sqrt{3}}{2})][/tex]
[tex]B=54\sqrt{3}\ in^2[/tex]
