Find the area of a regular octagon with a side length of 10 cm. Round to the nearest tenth.

The question keeps telling me to STOP!  Never mind.

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

482.8 sq cm

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step-by-step explanation:

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