SOLUTION: How many square inches are there in the lateral surface of a regular octagonal pyramid with 12 and 18 inches basal and lateral edges. Round off the answer to the nearest integer.

Question 1198259: How many square inches are there in the lateral surface of a regular octagonal pyramid with 12 and 18 inches basal and lateral edges. Round off the answer to the nearest integer.
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
**1. Find the Slant Height**
* We can use the Pythagorean theorem to find the slant height (the height of each triangular face of the pyramid).
* Imagine a right triangle formed by:
* The base: Half the base edge of the octagon (12 inches / 2 = 6 inches)
* The height: The slant height (which we'll call 's')
* The hypotenuse: The lateral edge (18 inches)
* Using Pythagorean Theorem:
* s² = 18² - 6²
* s² = 324 - 36
* s² = 288
* s = √288
* s ≈ 16.97 inches
**2. Calculate the Area of One Triangular Face**
* Area of a triangle = (1/2) * base * height
* Area of one triangular face = (1/2) * base edge * slant height
* Area of one triangular face = (1/2) * 12 inches * 16.97 inches ≈ 101.82 square inches
**3. Calculate the Lateral Surface Area**
* The octagonal pyramid has 8 triangular faces.
* Lateral Surface Area = 8 * Area of one triangular face
* Lateral Surface Area = 8 * 101.82 square inches ≈ 814.58 square inches
**4. Round to the Nearest Integer**
* Lateral Surface Area ≈ 815 square inches
**Therefore, the lateral surface area of the regular octagonal pyramid is approximately 815 square inches.**

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