Question

6. Find the surface area of the regular pyramid shown to the nearest whole number. The figure is (1 point)
not drawn to scale.
11 m
12 m
6√3
01,540 m²
0770 m²
0396 m²
0749 m²

Answer 1

Area of hexagon

• 6(1/2BH)
• 3BH
• 3(6√3)(12)
• 216√3m²

Now

area of upper 6 triangles

• 3BH
• 3(11)(12)
• 396m²

Total area

• 396+216√3
• 770.112m²
• 770m²

Answer 2

Answer:

770 m²

Step-by-step explanation:

The surface area of a regular pyramid comprises the area of the base (regular polygon) and the area of each of the slanted sides (triangles).

Apothem:  The line segment from the center of the regular polygon to the midpoint of one of its sides.

Area of the base

The base of the prism is a regular polygon with 6 sides (hexagon).

$\textsf{Area of a regular polygon}=\sf \dfrac{1}{2}nsa$

where:

• n = number of sides
• s = side length
• a = apothem

Given:

• n = 6
• s = 12 m
• a = 6√3

Substitute the given values into the formula:

$\begin{aligned}\implies \textsf{Base area}& =\sf \dfrac{1}{2}\cdot 6 \cdot 12 \cdot 6\sqrt{3}\\ & = \sf 216\sqrt{3}\:m^2\end{aligned}$

Area of one side

The sides of the regular pyramid are congruent triangles.

$\textsf{Area of a triangle} = \sf \dfrac{1}{2} \times base \times height$

Given:

• base = 12 m
• height = 11 m

Substitute the given values into the formula:

$\implies \textsf{Area of a triangle} = \sf \dfrac{1}{2} \times 12 \times 11=66\:m^2$

Total Surface Area

$\begin{aligned}\implies \textsf{Total Surface Area} & = \sf base \: area + 6 \times side \: area\\& = \sf 216\sqrt{3}+6 \cdot 66\\& = \sf 216\sqrt{3}+396\\& = \sf 770\:m^2\:(nearest\:whole\:number)\end{aligned}$