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Geometry and Trigonometry - Circle parts: radius, diameter, arc, sector, and segment-extended

Grade 10IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The fundamental parts of a circle include the radius (rr), which is the distance from the center to the edge, and the diameter (dd), which is twice the radius and passes through the center.

A circle showing the radius from the center O to the edge.
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An arc is a portion of the circumference, while a sector is a 'pie-slice' region bounded by two radii and an arc. The central angle θ\theta determines their size relative to the whole circle.

A circle highlighting a sector with central angle theta.
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A segment is the region between a chord and the corresponding arc. Its area is calculated by subtracting the area of the triangle formed by the radii and the chord from the area of the sector.

Diagram showing a segment as the area between a chord and an arc.
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For extended problems, remember that the perimeter of a sector includes the arc length PLUS two radii: P=l+2rP = l + 2r.

📐Formulae

Circumference(C)=2πr=πdCircumference (C) = 2\pi r = \pi d

Area of Circle(A)=πr2Area\ of\ Circle (A) = \pi r^2

Arc Length(l)=θ360×2πrArc\ Length (l) = \frac{\theta}{360} \times 2\pi r

Area of Sector=θ360×πr2Area\ of\ Sector = \frac{\theta}{360} \times \pi r^2

Area of Triangle in Segment=12r2sin⁡(θ)Area\ of\ Triangle\ in\ Segment = \frac{1}{2}r^2 \sin(\theta)

Area of Segment=θ360πr2−12r2sin⁡(θ)Area\ of\ Segment = \frac{\theta}{360} \pi r^2 - \frac{1}{2}r^2 \sin(\theta)

💡Examples

Problem 1:

A sector of a circle has a radius of 1212 cm and a central angle of 60∘60^{\circ}. Calculate the length of the arc.

Solution:

l=60360×2×π×12l = \frac{60}{360} \times 2 \times \pi \times 12 l=16×24πl = \frac{1}{6} \times 24\pi l=4π≈12.57 cml = 4\pi \approx 12.57\text{ cm}

Explanation:

Substitute the values r=12r = 12 and θ=60∘\theta = 60^{\circ} into the arc length formula. Simplify the fraction and solve.

Problem 2:

Find the area of a sector with a radius of 1010 cm and a central angle of 90∘90^{\circ}.

Solution:

A=90360×π×102A = \frac{90}{360} \times \pi \times 10^2 A=14×100πA = \frac{1}{4} \times 100\pi A=25π≈78.54 cm2A = 25\pi \approx 78.54\text{ cm}^2

Explanation:

The sector represents a quarter of the circle since 90/360=1/490/360 = 1/4. Multiply this fraction by the total area of the circle.

Problem 3:

Calculate the area of the segment formed by a chord in a circle of radius 88 cm where the central angle is 120∘120^{\circ}.

Solution:

Areasector=120360×π×82=64π3≈67.02 cm2Area_{sector} = \frac{120}{360} \times \pi \times 8^2 = \frac{64\pi}{3} \approx 67.02\text{ cm}^2 Areatriangle=12×82×sin⁡(120∘)=32×32≈27.71 cm2Area_{triangle} = \frac{1}{2} \times 8^2 \times \sin(120^{\circ}) = 32 \times \frac{\sqrt{3}}{2} \approx 27.71\text{ cm}^2 Areasegment=67.02−27.71=39.31 cm2Area_{segment} = 67.02 - 27.71 = 39.31\text{ cm}^2

Explanation:

First, find the area of the sector. Then, calculate the area of the triangle formed by the two radii and the chord using the formula 12absin⁡(C)\frac{1}{2}ab \sin(C). Subtract the triangle area from the sector area.

Problem 4:

A silver pendant is shaped like a sector of a circle with a radius of 55 cm and a central angle of 135∘135^{\circ}. Find the total perimeter of the pendant. (Take π≈3.142\pi \approx 3.142)

Sector with 135 degree angle and 5 cm radius.

Solution:

  1. Calculate Arc Length (ll): l=135360×2×π×5l = \frac{135}{360} \times 2 \times \pi \times 5 l=0.375×10×3.142≈11.78 cml = 0.375 \times 10 \times 3.142 \approx 11.78 \text{ cm}
  2. Calculate Total Perimeter (PP): P=l+2rP = l + 2r P=11.78+2(5)P = 11.78 + 2(5) P=11.78+10=21.78 cmP = 11.78 + 10 = 21.78 \text{ cm}

Explanation:

To find the total perimeter of a sector, you must add the curved arc length to the two straight radii that bound the shape.

Problem 5:

A circular garden has a radius of 66 m. A straight path is built as a chord that subtends an angle of 90∘90^{\circ} at the center. Find the area of the smaller segment created by this path.

Circle with a 90 degree sector and a chord forming a segment.

Solution:

  1. Area of Sector: Asector=90360×π×62=14×36π=9π≈28.27 m2A_{sector} = \frac{90}{360} \times \pi \times 6^2 = \frac{1}{4} \times 36 \pi = 9\pi \approx 28.27 \text{ m}^2
  2. Area of Triangle: Atriangle=12×r2×sin⁡(90∘)A_{triangle} = \frac{1}{2} \times r^2 \times \sin(90^{\circ}) Atriangle=12×36×1=18 m2A_{triangle} = \frac{1}{2} \times 36 \times 1 = 18 \text{ m}^2
  3. Area of Segment: Asegment=28.27−18=10.27 m2A_{segment} = 28.27 - 18 = 10.27 \text{ m}^2

Explanation:

The segment area is the difference between the sector area and the area of the triangle formed by the center and the chord endpoints.

Circle parts: radius, diameter, arc, sector, and segment-extended Grade 10 Notes & Examples