Question

⦁ Find the area of a regular Octagon with side lengths 5in. Round your answer to the nearest tenth.

apparently, this is incomplete so there's more to it...someone plz help me
A=2(1+√2)a^2=2*(1+√2)*5^2=120.71068

Answer 1

You forgot to round to the nearest tenth. 120.7

Answer 2

Answer:

Area of octagon = 120.7 $in^{2}$

Step-by-step explanation:

We know that,

Area of a regular octagon = $2 \times (1+\sqrt{2}) \times a^{2}$, where a is the length of the side.

Now, we have a = 5 inches.

Substituting the value of 'a' in the above formula, we have,

Area of a regular octagon = $2 \times (1+\sqrt{2}) \times 5^{2}$

i.e. Area of a regular octagon = $50 \times (1+\sqrt{2})$

i.e. Area of a regular octagon = $120.71068$

So, we get the area of the octagon is 120.71068 $in^{2}$.

But, it is required to round the answer to the nearest tenth.

Hence, area of the octagon = 120.7 $in^{2}$.