PLEASE HELP - Find the area of a regular octagon with a side length of 5 cm. Round to the nearest tenth. (Please explain in full detail the process, im trying to learn how to do these) thank you!!

Answer: 482.8 sq cm
Step-by-step explanation:
The octagon is 8 isosceles triangles. Each has common side r and base 10. The central angle is 360/8 = 45°. By the Law of cosines,
10² = r² + r² - 2 r² cos 45° = 2r²(1 - √2/2) = r²(2-√2)
r² = 100/(2-√2)
The area of a triangle with sides a,b and included angle C is (1/2)ab sin C. Our 8 octagon isosceles triangles each have sides r and r and included angle 45° so our total area, 8 triangles, is
A = 8(1/2) r² sin 45° = (400/(2-√2)) (√2/2) × (2+√2)/(2+√2) = 200(1+√2)
A ≈482.84271
Answer: 482.8 sq cm
Answer:
120.7
Step-by-step explanation:
The formula for finding the area of a regular octagon is: A = 2(1+[tex]\sqrt{2}[/tex])[tex]a^{2}[/tex].
Using the formula I got 120.7(rounded).