Three guided lessons with explanations, visual steps, practice and a final check.
Original learning companion, not full textbook exercise solutions. Covers sectors, right-angle minor segments and selected composite areas. Other central-angle triangle calculations and all textbook exercise figures are not claimed complete.
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LESSON 1
Find a sector area as a fraction
Use the central angle fraction for area and arc length.
A full turn is 360°. A full circle has area πr² and circumference 2πr.
Identify the fraction of a turn
A 90° sector is 90/360 = 1/4 of a circle.
Use the circle area
For radius 14 cm, full area is 196π cm². One quarter is 49π cm².
Keep arc and area separate
Its arc length is one quarter of 2π × 14 = 28π, hence 7π cm. Area uses square units; arc length uses length units.
Evaluate only when requested
If π = 22/7 is specified, area is 154 cm² and arc length is 22 cm. Otherwise keep the exact π forms.
Worked example: A 60° sector of radius 12 cm has area (1/6) × 144π = 24π cm² and arc length (1/6) × 24π = 4π cm. Its perimeter is 24 + 4π cm.
Watch for this: The sector perimeter includes two radii as well as the arc. Arc length alone is not the whole perimeter.
Discuss: Choose a radius and compare 60°, 120° and 180° sectors. Predict the area ratio before calculating.
LESSON 2
Subtract a triangle to get a segment
Distinguish a circular segment from a sector.
A chord joins two points on a circle. A segment lies between a chord and its arc.
Identify the boundaries
A 90° sector of radius 14 contains a right triangle formed by the two radii and the chord.
Find the sector area
The sector is one quarter of the circle: area = 49π cm².
Find the triangle area
The two perpendicular radii are legs of length 14, so triangle area = ½ × 14 × 14 = 98 cm².
Subtract the triangle
Minor segment area = 49π − 98 cm². With π = 22/7, it is 154 − 98 = 56 cm².
Worked example: For radius 6 and a 90° central angle, sector area = 9π and triangle area = 18. The segment area is 9π − 18 square units, which is positive because π > 2.
Watch for this: A segment is not the same region as a sector. Draw the chord before deciding what to subtract.
Discuss: On the same circle draw the sector, its triangle and its segment using different shading. Explain how their areas fit together.
LESSON 3
Combine circular regions without double counting
Choose addition or subtraction from a clearly partitioned diagram.
The area of disjoint parts adds; overlapping area must not be counted twice.
Identify the outer boundary
A circular path lies between radii 7 m and 5 m with the same centre. The outer disk has area 49π.
Identify what is removed
The inner disk, area 25π, is not part of the path.
Subtract the areas
Path area = 49π − 25π = 24π m². Squaring the width 2 m would not give the ring area.
Check a different partition
For a rectangle with a semicircle attached along one side, add the rectangle area and half a disk only when their interiors do not overlap.
Worked example: A 10 m by 6 m rectangle has a semicircle attached externally along its 6 m side. Radius = 3 m. Total area = 60 + ½π×9 = 60 + 4.5π m².
Watch for this: Area of a ring is π(R² − r²), not π(R − r)². A radius is half a diameter.
Discuss: Draw a playground consisting of a rectangle and two external semicircles on opposite ends. Express its area using the rectangle dimensions.
Original learning companion, not full textbook exercise solutions. Covers sectors, right-angle minor segments and selected composite areas. Other central-angle triangle calculations and all textbook exercise figures are not claimed complete.
Original HelioLearn teaching and questions. Source-aligned companion; not an official NCERT/CBSE publication or full textbook solution bank. Independent teacher review and narrated video are pending.
