Let's tackle each problem step-by-step.

Problem 1: Lateral Surface Area of a Regular Hexagonal Pyramid

Given:

• Edge of the pyramid (slant height): $s = 20$ cm
• Radius of the inscribed circle in the base: $r = 9\sqrt{3}$ cm

1. Calculate the side length of the hexagon base.
   For a regular hexagon, the radius of the inscribed circle is equal to the apothem ($a$) of the hexagon. So, we have: $a = 9\sqrt{3} \text{ cm}$ For a regular hexagon, the relationship between the apothem $a$ and the side length $b$ is: $a = \frac{\sqrt{3}}{2} b$ Solving for $b$: $b = \frac{2a}{\sqrt{3}} = \frac{2 \cdot 9\sqrt{3}}{\sqrt{3}} = 18 \text{ cm}$

2. Calculate the lateral surface area.
   The lateral surface area of a pyramid with a regular polygonal base is given by: $\text{Lateral Surface Area} = \frac{1}{2} \times \text{Perimeter of Base} \times \text{Slant Height}$

   • The perimeter of a hexagon with side length $b = 18$ cm is: $\text{Perimeter} = 6 \times b = 6 \times 18 = 108 \text{ cm}$
   • The slant height $s$ is given as 20 cm.

   Substituting these values: $\text{Lateral Surface Area} = \frac{1}{2} \times 108 \times 20 = 1080 \text{ cm}^2$

Solution for Problem 1:

The lateral surface area of the hexagonal pyramid is 1080 cm².

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Problem 2: Volume of a Right Pyramid with a Square Base

Given:

• Lateral area: $144 \, \text{in}^2$
• Slant height: twice the base edge length ($s = 2b$)

1. Find the base edge length.
   The lateral area of a pyramid with a square base is given by: $\text{Lateral Area} = 2 \times \text{Base Edge} \times \text{Slant Height}$ Substitute the values: $144 = 2 \times b \times (2b)$ Simplify: $144 = 4b^2$ Solving for $b$: $b^2 = 36 \Rightarrow b = 6 \, \text{in}$

2. Calculate the volume of the pyramid.
   The volume $V$ of a pyramid is given by: $V = \frac{1}{3} \times \text{Base Area} \times \text{Height}$

   • The base area of a square with edge $b = 6$ in is: $\text{Base Area} = b^2 = 6^2 = 36 \, \text{in}^2$
   • To find the height $h$, use the Pythagorean theorem in the right triangle formed by the height, half of the base edge, and the slant height $s = 2b = 12$ in: $h = \sqrt{s^2 - \left(\frac{b}{2}\right)^2} = \sqrt{12^2 - 3^2} = \sqrt{144 - 9} = \sqrt{135} = 3\sqrt{15} \, \text{in}$

   Substitute into the volume formula: $V = \frac{1}{3} \times 36 \times 3\sqrt{15} = 36\sqrt{15} \, \text{in}^3$

Solution for Problem 2:

The volume of the pyramid is $36\sqrt{15} \, \text{in}^3$ (approximately 139.1 in³).

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Would you like further explanations, or have any questions?

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Related Questions:

1. How would the lateral surface area change if the slant height of the hexagonal pyramid was increased?
2. What would be the effect on the volume of the square pyramid if the base edge was doubled?
3. How does the radius of the inscribed circle impact the calculations in polygonal bases?
4. What is the lateral area formula for other polygonal-based pyramids?
5. How would you determine the total surface area of the hexagonal pyramid?

Tip: For pyramids with polygonal bases, knowing the apothem simplifies many calculations involving the base.