Radius incorrect.

you should draw a perpendicular line to AE from C first.

Name the foot of the perpendicular F.

Then realise that FCDE is a rectangle, so FC = ED = 8cm

(Note that ∠AED = ∠CDE = 90° since tangent is perpendicular to radius. And ACDE is a trapezium so angles betwen parallel lines add up to 180°)

△FAC is a right angled triangle so we can used Pythagoras' Theorem


Letting the radius of the small circle be r,

CD = radius of small circle = r

FE = CD = r (opposite sides of the same rectangle)

Since radius of big circle is twice of small circle,

AE = radius of big circle = 2r

AF = AE - FE = 2r - r = r

Then, AC = radius of small circle + radius of big circle

= r + 2r = 3r


Now use Pythagoras' Theorem.

AC² = AF² + FC²

(3r)² = r² + 8²

9r² = r² + 64

8r² = 64

r² = 8

r = √8 = √(4×2) = √4 × √2 = 2√2

Alternatively you can draw a perpendicular line from A to the tangent point where the two small circles meet. Then draw another line from that point to C.

You'll also get a right angled triangle (congruent to the one used in the previous working). Apply the same idea.

Next, apply trigo to find ∠FAC.

tan ∠FAC = opposite side / adjacent side

= FC/AF
= 8/r
= 8/√8
= √8 (or 2√2. Also realise that 8 = 4 × 2 = (2√2)²)

∠FAC = tan-¹(2√2)
(No need to find the approximated angle yet)


Since ACDE is a trapezium , and AE // CD,
then ∠EAC + ∠ACD = 180°

(angles between parallel lines add up to 180° OR interior angles are supplementary)

But ∠EAC = ∠FAC (common angle) so
∠FAC + ∠ACD = 180°


So ∠ ACD = 180° - tan-¹(2√2)

Lastly find arc length.


Arc length of bigger circle (bordering the shaded region)

= (angle subtended by the arc) / 360° × circumference (which is 2π × radius)

= ∠FAC / 360° × 2π(2r)

= tan-¹(2√2) / 360° × 2π(2 × √2)

= tan-¹(2√2) / 360° × 8π√2


Arc length of smaller circle (bordering the shaded region)

= (angle subtended by the arc) / 360° × circumference (which is 2π × radius)

= ∠ACD/ 360° × 2πr

= (180° - tan-¹(2√2) ) / 360° × 2π(2√2)

= (180° - tan-¹(2√2) ) / 360° × 4π√2


Perimeter


= The two arc lengths + ED

= (tan-¹(2√2) / 360° × 8π√2) cm
+ ((180° - tan-¹(2√2) ) / 360° × 4π√2) cm
+ 8cm

≈ 20.367 cm

= 20.4cm (3s.f)