IB Mathematics SL 3.4 length of an arc area of a sector AI HL Paper 1- Exam Style Questions- New Syllabus

(a)
The area of the shaded sector is \( \frac{1}{5} \) of the area of the annulus (ring) between the two circles.
Area of annulus = \( \pi (4)^2 – \pi (2.8)^2 = \pi(16 – 7.84) = 8.16\pi \, \text{m}^2 \).
Area of one sector = \( \frac{1}{5} \times 8.16\pi = 1.632\pi \, \text{m}^2 \approx 5.127 \, \text{m}^2 \).
Since \( 5.127 < 6 \), the available paint is sufficient.
Shown by calculation: \( \boxed{A = \frac{1}{5} (\pi (4)^2 – \pi (2.8)^2) \approx 5.13 \, \text{m}^2 < 6 \, \text{m}^2} \)

(b)
The path consists of:

1. The outer arc subtending \( \frac{3}{5} \) of the full circumference.
2. Two radii of length \( 4 \, \text{m} \) (to and from the centre).

Outer circumference = \( 2\pi(4) = 8\pi \, \text{m} \).
Length of outer arc = \( \frac{3}{5} \times 8\pi = \frac{24\pi}{5} \, \text{m} \).
Total distance = Arc length + \( 2 \times 4 = \frac{24\pi}{5} + 8 \approx 23.1 \, \text{m} \).
\( \boxed{\frac{24\pi}{5} + 8 \approx 23.1 \, \text{m}} \)