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

Question

A decorative circular pattern is created in a garden. Two concentric circles are designed around a central point, \( O \). The radius of the inner circle is \( 2.8 \text{ m} \), and the radius of the outer circle is \( 4 \text{ m} \). Radial paths are drawn from \( O \) to the circumference of the larger circle, partitioning each circle into \( 5 \) congruent sectors, as illustrated in the diagram below.
Assume the widths of the paths are negligible for all calculations.
 
Specific regions between the paths are to be filled with colored gravel. The shaded region shown below is designated to be filled with orange gravel.
 
Matt has a supply of orange gravel sufficient to cover an area of \( 6 \text{ m}^2 \).
(a) Show that Matt has enough gravel to cover the entire shaded region.
A visitor enters the garden at the “entrance” and follows the path indicated by the arrows in the following diagram.
 
(b) Calculate the total distance traveled along this specific path, starting from the entrance and returning to the entrance.

Most appropriate topic codes (IB Mathematics: applications and interpretation):

SL 3.4: Length of an arc; area of a sector — parts (a) and (b)
▶️ Answer/Explanation

(a)
The shaded region is a sector of an annulus. The area of the full annulus is:
\( \pi R^2 – \pi r^2 = \pi(4^2 – 2.8^2) = \pi(16 – 7.84) = 8.16\pi \ \text{m}^2 \).
Since the circle is divided into 5 equal sectors:
Area of one sector \( = \frac{1}{5} \times 8.16\pi \approx 5.127 \ \text{m}^2 \).
Since \( 5.127 < 6 \), Matt has enough dye.
Area \( \approx 5.13 \ \text{m}^2 \); yes.

(b)
The path consists of:
1. Outer arc through three sectors: angle \( = \frac{3}{5} \times 360^\circ = 216^\circ \)
Arc length \( = \frac{216}{360} \times 2\pi \times 4 = \frac{3}{5} \times 8\pi \approx 15.08 \ \text{m} \).
2. Return along a radial path of length \(4 \ \text{m} \).
Total distance \( \approx 15.08 + 4 = 19.08 \ \text{m} \).
\( \approx 19.1 \ \text{m} \).

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