Areas Related to Circles Class 10 Maths Notes

This chapter lives in Unit VI Mensuration, which carries 10 of the 80 theory marks. It is a small chapter with a very short formula list, and almost every question is a substitution followed by one subtraction or one addition.The 2024-25 retained content is: the area of sectors and segments of a circle, and problems based on the areas and the perimeter or circumference of those plane figures. The plane figures involved are triangles, simple quadrilaterals and circles — nothing else is brought in.

• Area of a circle, and its circumference.
• Area of a sector of given central angle and radius, and the length of its arc.
• Area of a segment of given central angle and radius.
• Combined problems on the areas and the perimeter of triangles, simple quadrilaterals and circles.

Two measurements of a circle are used everywhere in this chapter. The circumference C is the distance once round the circle, and the area A is the region the circle encloses. Both come from the radius r. The diameter d is just 2r, which is why the circumference is sometimes written using d.

• r — the radius, the distance from the centre O to any point on the circle.
• d — the diameter, always 2r.
• C = 2πr, so C = πd. Use whichever of the two the question gives you.
• A = πr². Square the radius, never the diameter — but you may write A = π(d/2)² if only d is given.
• Throughout this chapter take π = 22/7 unless the question states otherwise, and choose a radius that is a multiple of 7 so the arithmetic stays exact.

A sector is the region bounded by two radii and the arc between them. Its share of the whole circle is fixed by its central angle θ. Since θ out of 360° gives that share of the circle, both the arc length and the sector area take the same fraction.

• θ is the central angle in degrees, measured at the centre O between the two radii.
• θ = 90° gives a quadrant, which is one quarter of the circle, and a quarter of the circumference.
• θ = 60° gives one sixth of the circle, and one sixth of the circumference.
• The perimeter of a sector is the two radii plus the arc, so perimeter = 2r + arc length.
• θ is not always given directly. If the question gives the area of the sector and the radius, find θ first by inverting the formula.

The chord of a sector is the straight line joining the two ends of the arc. A segment is the smaller region between that chord and the arc — the part of the sector that is not the triangle. So a segment is never found directly; it is always a subtraction.At 90° the two radii are perpendicular, so the triangle is right-angled with both legs equal to r. At 60° the chord equals the radius, because all three sides of the triangle are radii, so the triangle is equilateral.

• The triangle is formed by the two radii and the chord, and its base is the chord with the height from the centre.
• At 90°: area of triangle = ½ × r × r, so the segment is (πr²/4) − (r²/2).
• At 60°: the triangle is equilateral with side r, so its area is (√3/4)r² and the segment is (πr²/6) − (√3/4)r².
• If there are two segments asked for, the minor plus the major equals the area of the whole circle, which is a useful check.

A large share of the questions in this chapter combine two plane figures. The rule is always the same: get each figure separately, then add or subtract according to whether the second figure was added to, or cut out of, the first. Perimeter questions need the boundary, so an internal edge is not counted at all.

• Triangle — area = ½ × base × height.
• Square — area = a², perimeter = 4a.
• Rectangle — area = length × breadth, perimeter = 2 × (length + breadth).
• Rhombus — area = d₁ × d₂ ÷ 2, where d₁ and d₂ are the two diagonals.
• Parallelogram — area = base × height.
• Trapezium — area = ½ × (sum of the parallel sides) × distance between them.
• Circle — add its area and its arc where the circle actually appears in the figure, and never count a shared edge twice in the perimeter.

Find the length of the arc, the area of the sector and the area of the segment of a circle of radius 14 cm in which a sector has a central angle of 90°. Take π = 22/7.This is the cleanest segment question there is, because at 90° the triangle is right-angled with two sides equal to r, so its area is exactly ½ × 14 × 14 with no square root anywhere.

• Arc length = (90/360) × 2 × 22/7 × 14 = (1/4) × 88 = 22 cm.
• Area of sector = (90/360) × 22/7 × 14² = (1/4) × 22 × 28 = (1/4) × 616 = 154 cm².
• The triangle OAB has OA = OB = 14 cm and ∠AOB = 90°, so its area = ½ × 14 × 14 = 98 cm².
• Area of the segment = area of sector − area of triangle = 154 − 98 = 56 cm².
• Check with the formula: (πr²/4) − (r²/2) = 154 − 98 = 56 cm². The two routes agree.

A rectangular lawn is 70 m long and 50 m broad. A circular flower bed of radius 14 m is cut out from the middle of it. Find the area of the lawn that is left, and the cost of levelling it at ₹12 per m². Take π = 22/7.This is the shape of a real board question: two plane figures, one subtraction for the area, then a rate multiplied by the area.

• Area of the rectangular lawn = 70 × 50 = 3500 m².
• Area of the circular flower bed = (22/7) × 14² = (22/7) × 196 = 22 × 28 = 616 m².
• Area of lawn remaining = 3500 − 616 = 2884 m².
• Cost of levelling = 2884 × ₹12 = ₹34,608.
• Answer: 2884 m² of lawn is left, and the cost of levelling it is ₹34,608.

Find the length of the arc, the area of the sector, and the perimeter of the sector of a circle of radius 21 cm in which the central angle is 60°. Take π = 22/7.The perimeter of a sector is a length, not an area, so the ½ × πr² formula never appears in it. It is simply the two radii plus the arc.

• Arc length = (60/360) × 2 × 22/7 × 21 = (1/6) × 132 = 22 cm.
• Area of sector = (60/360) × 22/7 × 21² = (1/6) × 22 × 63 = (1/6) × 1386 = 231 cm².
• Perimeter of the sector = radius + radius + arc = 21 + 21 + 22 = 64 cm.
• If a segment were asked for, the triangle here is equilateral with side 21 cm, so its area is (√3/4) × 21² = 110.25√3 cm² and the segment is 231 − 110.25√3 cm².

The 2024-25 paper is 80 marks in 3 hours with 38 questions. Section A holds 18 MCQs of 1 mark each and 2 assertion-reason questions of 1 mark each, Section B holds 5 very short answer questions of 2 marks, Section C holds 6 short answer questions of 3 marks, Section D holds 4 long answer questions of 5 marks, and Section E holds 3 case-study questions of 4 marks. There is internal choice in two questions each of Sections B, C and D. Calculators are not allowed, so take π = 22/7 and do the arithmetic exactly.Note that Unit VI Mensuration is only 10 marks, so a three-mark question here usually asks for one or two quantities, not a chain of four.

• Find the area of a sector of a circle of radius 14 cm and central angle 90°.
• Find the area of the segment cut off by a chord that subtends a right angle at the centre, radius 21 cm.
• A 90° sector of a circle of radius 7 cm and a square of side 7 cm are joined along one side. Find the combined perimeter.
• A sector of area 231 cm² has radius 21 cm. Find the angle at the centre.
• Find the area of the sector of a circle of radius 21 cm whose arc length is 22 cm.
• MCQ trap: the area of a sector is always less than the area of the circle, and the area of a segment is always less than the area of the sector.
• MCQ trap: doubling the radius multiplies the area by four, not by two.