(a) 2a2 (1 + √2)
Let ABCDEFGH be the regular octagon of side a cm.
Now if we produce the sides of the octagon on both the sides, we get a square PQRS.
Given, BC = DE = FG = HA = a cm.
Also,
BQ = QC = DR = RE = FS = SG = HP = PA = \(\frac{a}{\sqrt2}\)
(∵ BQC, DRE, FSG, HPA are rt. ∠d Δs)
∴ Side of square = \(\frac{a}{\sqrt2}\) + a + \(\frac{a}{\sqrt2}\) = a (1 + √2) cm.
∴ Area of square = a2(1 + √2)2 cm2 = a2(3 + 2√2) cm2
Each of the shaded Δs, APH, BCQ, DER, FGS is an isosceles right angled Δ, whose area
= \(\frac{1}{2}\times\frac{a}{\sqrt2}\) x \(\frac{a}{\sqrt2}\) cm2 = \(\frac{a^2}{4}\) cm2
∴ Area of octagon = Area of square PQRS – Total area of shaded isosceles Δs = a2 (3 + 2√2) - a2
= 2a2 (1 + √2) cm2.