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Mensuration - Arc Length and Sector Area

Grade 12IGCSE

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

🔑Concepts

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Understanding the definition of a radian: the angle subtended at the center of a circle by an arc equal in length to the radius.

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Relationship between degrees and radians: π\pi radians = 180∘180^\circ.

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Distinguishing between a Minor Sector (angle < π\pi) and a Major Sector (angle > π\pi).

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Calculating the perimeter of a sector, which includes the arc length plus two radii (s+2rs + 2r).

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Calculating the area of a segment by subtracting the area of the triangle from the area of the sector.

📐Formulae

s=rθs = r\theta (Arc length where θ\theta is in radians)

A=12r2θA = \frac{1}{2}r^2\theta (Sector area where θ\theta is in radians)

s=θ360×2πrs = \frac{\theta}{360} \times 2\pi r (Arc length where θ\theta is in degrees)

A=θ360×πr2A = \frac{\theta}{360} \times \pi r^2 (Sector area where θ\theta is in degrees)

Areasegment=12r2(θ−sin⁡θ)Area_{segment} = \frac{1}{2}r^2(\theta - \sin\theta) (Segment area where θ\theta is in radians)

💡Examples

Problem 1:

A sector of a circle has a radius of 66 cm and an angle of 1.51.5 radians. Calculate the arc length and the area of the sector.

Solution:

s=6×1.5=9s = 6 \times 1.5 = 9 cm; A=12×62×1.5=27A = \frac{1}{2} \times 6^2 \times 1.5 = 27 cm²

Explanation:

To find the arc length, use s=rθs = r\theta. To find the area, use A=12r2θA = \frac{1}{2}r^2\theta. Substitute r=6r=6 and θ=1.5\theta=1.5 directly into the radian-based formulas.

Problem 2:

A sector has a perimeter of 2424 cm and a radius of 55 cm. Find the area of the sector.

Solution:

s=24−(2×5)=14s = 24 - (2 \times 5) = 14 cm; θ=145=2.8\theta = \frac{14}{5} = 2.8 rad; A=12×52×2.8=35A = \frac{1}{2} \times 5^2 \times 2.8 = 35 cm²

Explanation:

First, find the arc length ss by subtracting the two radii from the total perimeter (P=s+2rP = s + 2r). Then, find the angle θ\theta using s=rθs = r\theta. Finally, use the area formula A=12r2θA = \frac{1}{2}r^2\theta.

Problem 3:

Find the area of the segment cut off by a chord in a circle of radius 1010 cm, where the chord subtends an angle of 22 radians at the center.

Solution:

Areasector=12(102)(2)=100Area_{sector} = \frac{1}{2}(10^2)(2) = 100 cm²; Areatriangle=12(102)sin⁡(2)≈45.47Area_{triangle} = \frac{1}{2}(10^2)\sin(2) \approx 45.47 cm²; Areasegment=100−45.47=54.53Area_{segment} = 100 - 45.47 = 54.53 cm²

Explanation:

The segment area is the difference between the sector area (0.5r2θ0.5r^2\theta) and the area of the triangle formed by the two radii and the chord (0.5r2sin⁡θ0.5r^2\sin\theta). Ensure the calculator is in Radian mode when calculating sin⁡(2)\sin(2).