IB Mathematics AI SL length of an arc area of a sector MAI Study Notes - New Syllabus

IB Mathematics AI SL length of an arc area of a sector MAI Study Notes

LEARNING OBJECTIVE

The circle: length of an arc; area of a sector.

Key Concepts:

Arcs & Sectors

MAI HL and SL Notes – All topics

Consider a circle of radius $ r $. Let $ \theta $ be the angle shown below:

The length of the arc $ AB $ is given by:

$ L = \frac{\theta}{360} \times 2\pi r $

The area of the sector $ OAB $ is given by:

$ A = \frac{\theta}{360} \times \pi r^2 $
(where $ \theta $ is measured in degrees).

Example

Consider the following sector of a circle with $ r = 5 \, \text{m} $ and $ \theta = 36^\circ $:

Length of arc:
Area of sector:
Perimeter of sector:

▶️Answer/Explanation

Solution:

Length of arc:
$ L = \frac{36}{360} \times 2\pi \cdot 5 = \pi \approx 3.14 \, \text{m} $

Area of sector:
$ A = \frac{36}{360} \times \pi \cdot 5^2 = \frac{5\pi}{2} \approx 7.85 \, \text{m}^2 $

Moreover:
Perimeter of sector: $ L + r + r = 3.14 + 5 + 5 = 13.14 \, \text{m} $.

Example

Let $ r = 3 \, \text{cm} $ and $ \theta = 30^\circ $. Compare:

a) the lengths of the arc $ AB $ and the chord $ AB $ (distance $ AB $),
b) the areas of the sector $ OAB $ and the triangle $ OAB $.

▶️Answer/Explanation

Solution:

a)
$ L = \frac{30}{360} \times 2\pi \cdot 3 = \frac{\pi}{2} \approx 1.57 \, \text{cm} $

For $ AB $, we use COSINE RULE:
$ AB^2 = 3^2 + 3^2 – 2 \cdot 3 \cdot 3 \cos 30^\circ = 18 – 18 \cdot \frac{\sqrt{3}}{2} \approx 2.41 \Rightarrow AB \approx 1.55 \, \text{cm} $

The length of the arc $ AB $ is slightly greater than the chord $ AB $ (as expected).

b)
$ A_{\text{sector}} = \frac{30}{360} \times \pi \cdot 3^2 = \frac{3\pi}{4} \approx 2.36 \, \text{cm}^2 $
$ A_{\text{triangle}} = \frac{1}{2} \cdot 3 \cdot 3 \cdot \sin 30^\circ = 2.25 \, \text{cm}^2 $

An Informal Idea about Major/Minor Segment

The region between the chord AB and the arc AB is known as segment.

From Upper Example:

$ A_{\text{sector}} = \frac{30}{360} \times \pi \cdot 3^2 = \frac{3\pi}{4} \approx 2.36 \, \text{cm}^2 $
$ A_{\text{triangle}} = \frac{1}{2} \cdot 3 \cdot 3 \cdot \sin 30^\circ = 2.25 \, \text{cm}^2 $

The area of this segment is

$ A_{\text{sector}} – A_{\text{triangle}} = 2.36 – 2.25 = 0.11 $

NOTICE. The formulas hold for the major sector OACB and the major arc ACB.

We simply set $\theta = 360^\circ – 30^\circ = 330^\circ$.

NOTICE. The formulas hold for the major sector OACB and the major arc ACB.

We simply set $ \theta = 360^\circ – 30^\circ = 330^\circ $.