Grade 10 · Core Mathematics · Lesson 9 of 12

Applications of Area of Parts of a Circle

A Grade 10 Core Mathematics CBE lesson from the sub-strand Area of a Part of a Circle. In it, learners calculate the area of an annulus and an annular sector in practical contextual problems.

Strand Measurements and GeometrySub-strand Area of a Part of a CircleLesson code 10CM-26-L940 minutes · 2 slides

KICD learning outcome (f)

Apply the area of a part of a circle in real-life situations.

By the end of this lesson, learners should be able to

Calculate the area of an annulus and an annular sector in practical contextual problems.
Determine the area of circular sectors in everyday engineering and design applications.

Key vocabulary

Annulus: The ring-shaped region bounded by two concentric circles of different radii.; Calculating the paved area around a circular pool requires finding the annulus area.
Annular Sector: A portion of an annulus bounded by two radii and two concentric arcs.; The curved flower garden forms an annular sector subtending seventy-two degrees at the centre.
Sector: A region of a circle bounded by two radii and the intercepted arc.; Each fabric panel of the umbrella forms a sector subtending an acute central angle.

Lesson preview

Slide 1 · Real-Life Applications of Annulus and Annular Sector AreasModel · 18 min

Figure 1: A diagram illustrating an annular path around a circular fountain with an inner radius of six metres and an outer radius of nine metres.

Real-life circular objects often involve concentric boundaries. A paved path surrounding a lawn, a steel nut specified by inner and outer diameters, or a baked doughnut forms an annulus between two concentric circles. When only a portion of such a ring is bounded by two radii, such as a curved flower garden, it forms an annular sector. To calculate these practical areas, subtract the inner circular or sector area from the outer one.

An annulus is the region enclosed between two concentric circles of different radii.
Subtracting the inner area from the outer area gives the net area of an annulus.
Diameters given in practical contexts like steel nuts must first be converted to radii.
An annular sector represents a proportional fraction of a full annulus subtending a central angle.

Worked example. Find the inner radius:

r = \frac{12}{2} = 6 m

. Determine the outer radius:

R = 6 + 3 = 9 m

. Apply the annulus area formula:

A = π R^{2} - π r^{2} = π (R^{2} - r^{2})

. Substitute the radii:

A = 3.142 \times (9^{2} - 6^{2}) = 3.142 \times (81 - 36)

. Multiply to find the final area:

A = 3.142 \times 45 = 141.39 m^{2}

. Answer: $141.39 m^{2}$ .

Check question: A flower garden is designed as an annular sector with inner radius $10 m$ , outer radius $15 m$ , and a central angle of $72^{\circ}$ . Calculate the area of this flower garden using $π = 3.142$ .

$15.71 m^{2}$
$62.84 m^{2}$
$78.55 m^{2}$
$141.39 m^{2}$

Show answer

$78.55 m^{2}$ . First, find the fraction of the circle: $\frac{72^{\circ}}{360^{\circ}} = 0.2$ . Then, find the difference between the squares of the outer and inner radii: $15^{2} - 10^{2} = 225 - 100 = 125$ . Finally, multiply to find the area of the annular sector: $Area = 0.2 \times 3.142 \times 125 = 78.55 m^{2}$ .

Slide 2 · Sector Areas in Design and Engineering ContextsPractice · 22 min

Figure 2: A circle partitioned into eight equal parts, featuring a shaded sector with a central angle of 45 degrees and an 84-centimetre radius.

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Teacher notes, SNE adaptations and assessment checks
Homework and the other 11 lessons in this sub-strand

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Applications of Area of Parts of a Circle

By the end of this lesson, learners should be able to

Key vocabulary

Lesson preview

Teach the full lesson on the projector

All 12 lessons in Area of a Part of a Circle