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Mensuration: Areas Related to Circles - Calculate area of sectors and segments of circles for standard central angles

Grade 10CBSE

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

🔑Concepts

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A sector is the region bounded by two radii and an arc of a circle. The area is proportional to the central angle θ\theta. If θ=360∘\theta = 360^\circ, the area is πr2\pi r^2. For any other angle θ\theta, the area is θ360∘×πr2\frac{\theta}{360^\circ} \times \pi r^2.

A circle showing a sector with a 60 degree central angle and radius r.
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A segment is the region bounded by a chord and its corresponding arc. The area of a minor segment is found by subtracting the area of the triangle formed by the radii and the chord from the area of the corresponding sector.

A circle illustrating a segment formed by a chord and an arc.
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Standard central angles frequently used in Grade 10 include 30∘,45∘,60∘,90∘,120∘30^\circ, 45^\circ, 60^\circ, 90^\circ, 120^\circ, and 180∘180^\circ. For these angles, the sector area is a specific fraction of the circle (e.g., 90∘90^\circ is 14\frac{1}{4} of the circle).

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The area of the major sector or major segment is always calculated by subtracting the area of the minor part from the total area of the circle (πr2\pi r^2).

📐Formulae

Area of a circle = πr2\pi r^2

Circumference of a circle = 2πr2\pi r

Length of an arc of a sector with angle θ\theta = θ360∘×2πr\frac{\theta}{360^\circ} \times 2\pi r

Area of a sector of a circle with radius rr and angle θ\theta = θ360∘×πr2\frac{\theta}{360^\circ} \times \pi r^2

Area of a major sector = πr2−Area of minor sector\pi r^2 - \text{Area of minor sector}

Area of a triangle with two sides as radii rr and included angle θ\theta = 12r2sin⁡θ\frac{1}{2} r^2 \sin \theta

Area of a minor segment = Area of sector - Area of triangle = (θ360∘×πr2)−12r2sin⁡θ\left( \frac{\theta}{360^\circ} \times \pi r^2 \right) - \frac{1}{2} r^2 \sin \theta

Area of a major segment = πr2−Area of minor segment\pi r^2 - \text{Area of minor segment}

💡Examples

Problem 1:

Find the area of a sector of a circle with radius 66 cm if the angle of the sector is 60∘60^\circ. (Use π=227\pi = \frac{22}{7})

Solution:

  1. Given: Radius r=6r = 6 cm, Angle θ=60∘\theta = 60^\circ.
  2. Use the formula: Area of sector = θ360∘×πr2\frac{\theta}{360^\circ} \times \pi r^2.
  3. Substitute the values: Area = 60360×227×6×6\frac{60}{360} \times \frac{22}{7} \times 6 \times 6.
  4. Simplify: Area = 16×227×36\frac{1}{6} \times \frac{22}{7} \times 36.
  5. Area = 22×67=1327\frac{22 \times 6}{7} = \frac{132}{7} cm2^2.
  6. In decimal form: Area ≈18.86\approx 18.86 cm2^2.

Explanation:

To find the area of the sector, we determine the fraction of the total circle area using the ratio of the central angle to 360∘360^\circ and multiply it by the full area πr2\pi r^2.

Problem 2:

A chord of a circle of radius 1010 cm subtends a right angle at the center. Find the area of the corresponding minor segment. (Use π=3.14\pi = 3.14)

Solution:

  1. Given: Radius r=10r = 10 cm, Central Angle θ=90∘\theta = 90^\circ.
  2. Step 1: Calculate Area of Sector = 90360×3.14×10×10=14×314=78.5\frac{90}{360} \times 3.14 \times 10 \times 10 = \frac{1}{4} \times 314 = 78.5 cm2^2.
  3. Step 2: Calculate Area of Triangle OAB=12×base×height=12×r×r=12×10×10=50OAB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times r \times r = \frac{1}{2} \times 10 \times 10 = 50 cm2^2 (since the angle is 90∘90^\circ).
  4. Step 3: Area of Minor Segment = Area of Sector - Area of Triangle = 78.5−50=28.578.5 - 50 = 28.5 cm2^2.

Explanation:

The minor segment is the region between the chord and the arc. We find the area of the entire 'slice' (sector) and subtract the triangular part formed by the radii and the chord to leave only the segment area.

Problem 3:

In a circle of radius 2121 cm, an arc subtends an angle of 60∘60^\circ at the centre. Find the area of the sector formed by the arc. (Use π=227\pi = \frac{22}{7})

A sector with radius 21 cm and central angle 60 degrees.

Solution:

Given: Radius (rr) = 2121 cm Angle (θ\theta) = 60∘60^\circ

Area of sector = θ360∘×πr2\frac{\theta}{360^\circ} \times \pi r^2 Area = 60∘360∘×227×21×21\frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 21 \times 21 Area = 16×22×3×21\frac{1}{6} \times 22 \times 3 \times 21 Area = 12×22×21\frac{1}{2} \times 22 \times 21 Area = 11×2111 \times 21 Area = 231231 cm2^2

Explanation:

To find the area of the sector, we identify the radius and the central angle. We substitute these into the formula for sector area and simplify the fraction 60360\frac{60}{360} to 16\frac{1}{6} to make the calculation easier.

Problem 4:

A chord of a circle of radius 1414 cm subtends an angle of 60∘60^\circ at the centre. Find the area of the minor segment. (Use π=227\pi = \frac{22}{7} and 3=1.73\sqrt{3} = 1.73)

A circle with a 60 degree sector and a chord forming a minor segment.

Solution:

Given: r=14r = 14 cm, θ=60∘\theta = 60^\circ

  1. Area of Sector = 60360×227×14×14\frac{60}{360} \times \frac{22}{7} \times 14 \times 14 Area of Sector = 16×22×2×14=3083≈102.67\frac{1}{6} \times 22 \times 2 \times 14 = \frac{308}{3} \approx 102.67 cm2^2

  2. Area of Triangle = 12r2sin⁡θ\frac{1}{2} r^2 \sin \theta Since θ=60∘\theta = 60^\circ and OA=OBOA=OB, △OAB\triangle OAB is equilateral. Area of Triangle = 34×142=1.734×196=1.73×49=84.77\frac{\sqrt{3}}{4} \times 14^2 = \frac{1.73}{4} \times 196 = 1.73 \times 49 = 84.77 cm2^2

  3. Area of Minor Segment = Area of Sector - Area of Triangle Area = 102.67−84.77=17.9102.67 - 84.77 = 17.9 cm2^2

Explanation:

Since the central angle is 60∘60^\circ and the two bounding sides are radii, the triangle formed is equilateral. We find the area of the 60∘60^\circ sector and subtract the area of this equilateral triangle to find the area of the segment.