Question

A design for party favors are in the shape of a regular hexagonal pyramid. The hexagon has a side length of 6 inches. The lateral faces are isosceles triangles with equal sides of 9 inches.
The lateral area of the pyramid is:
A) 1082
B) 903
C) 965

And the base area is:
A) 362
B) 543
C) 245

Answer 1

Answer:

Part 1) $LA=108\sqrt{2}\ in^2$

Part 2) $B=54\sqrt{3}\ in^2$

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

$LA=6[\frac{1}{2}(b)(h)]$

we have

$b=6\ in$

Applying Pythagorean Theorem calculate the height of triangle

$9^2=(6/2)^2+h^2$

$h^2=81-9\\h^2=72\\h=\sqrt{72}\ in$

simplify

$h=6\sqrt{2}\ in$

Find the lateral area

$LA=6[\frac{1}{2}(6)(6\sqrt{2})]\\\\LA=108\sqrt{2}\ in^2$

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

$B=6[\frac{1}{2}b^2sin(60^o)]$

we have

$b=6\ in$

$sin(60^o)=\frac{\sqrt{3}}{2}$

substitute

$B=6[\frac{1}{2}6^2(\frac{\sqrt{3}}{2})]$

$B=54\sqrt{3}\ in^2$