Question

what is the area of a regular octagon with a side of 8 in. Round to the nearest tenth, show work please

Answer 1

Length of each side of a regular octagon = 8 inches
We already know the formula for finding the area of a regular octagon as
Area = 2 (1 + square root 2) * (Side)^2
= 2 (1 + 1.414) * (8)^2
= 2 (2.414) * 64
= 308.99 inches ^2
= 309 inches ^2
So the area of the regular octagon having sides of 8 inches is 309 square inches. This is the easiest method for solving these kind of problems.

Answer 2

The area of the regular octagon having sides of 8 inches is 309 square inches.

Explanation:

It is given that Length of each side of a regular octagon is 8 inches.

Now, the formula for finding the area of a regular octagon is:

`{\text}{Area}=2(1+√(2)){*}(Side)^2`

Substituting the given values, we get

`A=2(1+1.414){*}(8)^2`

`A=2(2.414){*}64`

`A=308.99{in^2`

`A`≈
`309 in^2`

Thus, the area of the regular octagon having sides of 8 inches is 309 square inches.