Question

find the surface area of a right octagonal pyramid with height 2.5 yards, and its base has apothem length 1.5 yards

Answer 1

Find the surface area of a right octagonal pyramid with height 2.5 yards, and its base has a apothem length 1.5 yards

Solution:

We have

$$A O = 2. 5 (y a r d s), O D = 1. 5 (y a r d s), O D \perp B C.$$

Denote $S$ as the surface area of the right octagonal pyramid. Then

$$S = 8 S _ {A B C}$$

where $S_{ABC}$ is the area of $\Delta ABC$ and

$$S _ {A B C} = \frac {1}{2} A D \cdot B C.$$

We have

$$A D ^ {2} = A O ^ {2} + O D ^ {2},$$ $$A D = \sqrt {A O ^ {2} + O D ^ {2}} = \sqrt {(2 . 5) ^ {2} + (1 . 5) ^ {2}} = \sqrt {6 . 2 5 + 2 . 2 5} = \sqrt {8 . 5} (y a r d s).$$

Because the pyramid is the right octagonal one then

$$\angle B O D = \frac {1}{2} \cdot \frac {2 \pi}{8} = \frac {\pi}{8} (r a d).$$

Then

$$B C = 2 \cdot B D = 2 \cdot O D \cdot \tan \angle B O D = 2 \cdot 1. 5 \cdot \tan \frac {\pi}{8} = 3 \tan \frac {\pi}{8} (y a r d s).$$

Thus we have

$$S = 8 S _ {A B C} = 8 \cdot \frac {1}{2} A D \cdot B C = 4 \cdot \sqrt {8 . 5} \cdot 3 \tan \frac {\pi}{8} = 1 2 \sqrt {8 . 5} \tan \frac {\pi}{8} \approx 1 4. 4 9 1 6 (y a r d s ^ {2}).$$

Answer:

$$S = 1 2 \sqrt {8 . 5} \tan \frac {\pi}{8} \approx 1 4. 4 9 1 6 (y a r d s ^ {2})$$