CBSE Important Questions Class 10 Maths Chapter 11 – Area Related to Circles

Class 10 Maths Chapter 11 Area Related to Circles is part of the CBSE Class 10 Maths syllabus. It explains how to find the perimeter and area of a circle, and then goes on to the area of a sector and a segment of a circle. Students also learn to find the area of combinations of plane figures, such as a circle inside a square or shapes made from semicircular arcs.

These CBSE important questions for Class 10 Maths Chapter 11 cover the area of a minor segment, the area of a quadrant, a circular park with a path around it, the shaded region between a circle and a square, a case study on a circular panel with a chord, and shapes made of semicircular arcs. Every question has a clear step-by-step solution in simple English with figures, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. In the given figure, a circle with centre O has a radius of 4 cm. The angle subtended by the minor arc APB at the centre is 60°. Calculate the area of the minor segment APB.

A circle with centre O and a minor segment APB subtending 60 degrees at the centre

Answer: Area of segment = Area of sector − Area of triangle OAB.
Area of sector = (θ/360) × πr²
= (60/360) × π × 4²
= (1/6) × 16π = 8π/3 cm²
Area of △OAB: Since OA = OB (radii) and the angle between them is 60°, the triangle is equilateral with side 4 cm.
= (√3/4) × 4² = 4√3 cm²
Area of minor segment = (8π/3 − 4√3) cm²
≈ 8.38 − 6.93 = 1.45 cm² (approximately)

Q2. What is the area of the minor segment if ∠AOC is 60° and the circumference of the circle is 44 cm?

Answer: First find the radius from the circumference.
2πr = 44
2 × (22/7) × r = 44, so r = 7 cm
Area of sector = (60/360) × (22/7) × 7²
= (1/6) × 154 = 25.67 cm²
Area of the triangle (equilateral with side 7 cm, since the angle is 60°):
= (√3/4) × 49 = 21.22 cm²
Area of minor segment = 25.67 − 21.22 = 4.45 cm² (approximately)

Q3. Find the area of a quadrant of a circle whose circumference is 22 cm.

Answer: First find the radius.
2πr = 22
2 × (22/7) × r = 22, so r = 7/2 = 3.5 cm
A quadrant is one-fourth of a circle, so its angle is 90°.
Area of quadrant = (1/4) × πr²
= (1/4) × (22/7) × (3.5)²
= (1/4) × (22/7) × 12.25
= 9.625 cm²

Q4. In the given figure, the inner circle is inscribed in a square, and the square is circumscribed by the outer circle. If the radius of the smaller circle is r, find the area of the shaded region.

A circle inscribed in a square which is itself inscribed in a bigger circle

Answer: The inner circle has radius r, so the side of the square = 2r (the diameter of the inner circle).
The diagonal of the square is the diameter of the outer circle:
Diagonal = √2 × side = 2√2 r
So the radius of the outer circle R = √2 r.
Area of outer circle = πR² = π(√2 r)² = 2πr²
Area of the square = (2r)² = 4r²
Shaded region (between the outer circle and the square) = 2πr² − 4r²
= 2r²(π − 2) square units

Q5. A circular park of radius 21 m has a path of uniform width 3.5 m running around it on the outside. Find the cost of paving the path at ₹25 per m². Also find the cost of fencing the outer boundary at ₹15 per metre. [Use π = 22/7]

Answer: Inner radius r = 21 m, outer radius R = 21 + 3.5 = 24.5 m.
Area of the path = π(R² − r²)
= (22/7)(24.5² − 21²)
= (22/7)(600.25 − 441)
= (22/7) × 159.25
= 500.5 m²
Cost of paving = 500.5 × 25 = ₹12,512.50
Outer boundary (circumference) = 2πR = 2 × (22/7) × 24.5 = 154 m
Cost of fencing = 154 × 15 = ₹2,310

Q6. A carpenter is working on a circular decorative panel. A straight strip AB is fixed as a chord of the circle. The chord AB is 12 cm long and subtends an angle of 120° at the centre O.
(i) Find the radius of the circular panel.
(ii) Find the area of the minor segment cut off by the chord AB.

A circular panel with a chord AB subtending 120 degrees at the centre

Answer: (i) Draw OM perpendicular to AB. This bisects both the chord and the angle.
So AM = 6 cm and ∠AOM = 60°.
In right △OAM: sin 60° = AM/OA
√3/2 = 6/r
r = 12/√3 = 4√3 cm ≈ 6.93 cm
(ii) Area of sector = (120/360) × πr² = (1/3) × π × 48 = 16π cm²
Area of △OAB = ½ × r² × sin 120° = ½ × 48 × (√3/2) = 12√3 cm²
Area of minor segment = 16π − 12√3
≈ 50.27 − 20.78 = 29.49 cm² (approximately)

Q7. In the figure, the boundary of the shaded region consists of four semicircular arcs, the two smallest being equal. If the diameter of the largest is 14 cm and that of the smallest is 3.5 cm, calculate the area of the shaded region. [Use π = 22/7]

A shaded region bounded by four semicircular arcs of different sizes

Answer: Radii: largest = 7 cm, middle = 3.5 cm, two smallest = 1.75 cm each.
Area of shaded region = [area of largest semicircle − 2 × area of smallest semicircle] + area of middle semicircle
= ½π(7)² − 2 × ½π(1.75)² + ½π(3.5)²
= ½π[49 − 2(3.0625) + 12.25]
= ½π[49 − 6.125 + 12.25]
= ½ × (22/7) × 55.125
= 86.625 cm²

Q8. Sia purchased a circular dining table with a radius of 50 cm. She covered it with a rectangular tablecloth with one side measuring 60 cm, and the corners of the cloth just touch the edge of the table. How much area of the table is not covered? [Take π = 3.14]

A circular dining table partly covered by a rectangular tablecloth

Answer: Since the corners of the rectangle touch the circle, the diagonal of the rectangle is the diameter of the circle = 100 cm.
Let the other side be b. By the Pythagoras theorem:
60² + b² = 100²
3600 + b² = 10000
b² = 6400, so b = 80 cm
Area of the table = πr² = 3.14 × 50² = 7850 cm²
Area of the tablecloth = 60 × 80 = 4800 cm²
Area not covered = 7850 − 4800 = 3050 cm²

Q9. Write the formulae for the area and perimeter of a sector of a circle of radius r with angle θ.

Answer: For a sector of angle θ (in degrees) in a circle of radius r:
• Length of arc = (θ/360) × 2πr
• Area of sector = (θ/360) × πr²
• Perimeter of sector = length of arc + 2r (the arc plus the two radii)
• Area of segment = area of sector − area of the triangle formed by the two radii and the chord

Q10. The minute hand of a clock is 14 cm long. Find the area swept by it in 5 minutes. [Use π = 22/7]

Answer: The minute hand covers 360° in 60 minutes.
So in 1 minute it covers 6°, and in 5 minutes it covers 5 × 6° = 30°.
The area swept is the area of a sector with θ = 30° and r = 14 cm.
Area = (30/360) × (22/7) × 14²
= (1/12) × (22/7) × 196
= (1/12) × 616
= 51.33 cm² (approximately)

Q11. The wheel of a car makes 1000 revolutions to cover a distance of 1.76 km. Find the diameter of the wheel. [Use π = 22/7]

Answer: In one revolution the wheel covers a distance equal to its circumference.
Total distance = 1.76 km = 176000 cm
Distance in one revolution = 176000 / 1000 = 176 cm
So the circumference = 176 cm.
2πr = 176
2 × (22/7) × r = 176
r = 176 × 7 / 44 = 28 cm
Diameter = 2 × 28 = 56 cm

Q12. Find the area of a sector of a circle of radius 21 cm whose central angle is 120°. Also find the length of the corresponding arc. [Use π = 22/7]

Answer: Area of sector = (θ/360) × πr²
= (120/360) × (22/7) × 21²
= (1/3) × (22/7) × 441
= (1/3) × 1386
= 462 cm²
Length of arc = (θ/360) × 2πr
= (1/3) × 2 × (22/7) × 21
= (1/3) × 132
= 44 cm

Related Study Material on Extramarks

Q.1 Draw a circle of radius 4 cm. From a point 12 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.

Marks:3
Ans

Steps of construction:

Step 1 : Draw a circle with Centre O and radius 4 cm.

Step 2 : Take a point P outside the circle such thatOP = 12 cm. join OP.

Step 3: Draw a perpendicular bisector of line segment OP. it intersect OP at S.

Step 4 : Taking S as a centre and OS or SP as a radius, draw a circle it intersect the previous circle at R and T.

Step 5 : Join P to R and P to T.

PR and PT are required tangents from point P to circle with centre O.

On measuring we get PT = PR = 11.3 cm

Q.2 Draw a right triangle in which the sides (other than hypotenuse) are of length 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.

Marks:3
Ans

Steps of Construction
Step 1: Draw a right angle triangle ABC right angled at B, AB = 3 cm and BC = 4 cm.
Step 2: Draw an acute angle BX and make five equal arcs on it A1, A2, A3, A4, A5 such that BA1 = A1A2 = A2A3 = A3A4 = A4A5.
Step 3: Join A3 to C.
Step 4: Through A5 draw a line parallel to A3C. This line intersects BC produced at point B.
Step 5: Through B draw a line parallel to AC. This line intersects the line AB produced at point A.
Thus, ABB is the required triangle which is 5/3 times of ABC.

Q.3 Draw a line segment of length 8.1 cm and divide it in the ratio 4:5. Measure the two parts.

Marks:5
Ans

Steps of Construction:
Step 1: Draw a line segment AB of 8.1 cm.
Step 2: Draw an acute angle BAX and draw BY AX.
Step 3: Mark A1, A2, A3, A4 on AX and B1, B2, B3, B4, B5 on BX such that A1A2 = A2A3 = A3A4 = B1B2 = B2B3 = B3B4 = B4B5.
Step 4: Join A4 to B5 to find out the point 4:5. It intersects AB at O.
Length of AO = 3.6 cm and length of OB = 4.5 cm
O is the point which divides AO in 4:5

Q.4 Draw a triangle ABC with sides BC = 7 cm, ?B = 45°, ?A = 105°. Then construct a triangle whose sides are 4/3 times the corresponding sides of ?ABC.

Marks:3
Ans

Steps of construction
Step 1: Calculate C = 30° (by angle sum property).
Step 2: Draw a line segment BC of 7 cm.
Step 3: Draw a ABC according to the given angles.
Step 4: Through B draw an acute angle AX and make arc A1, A2, A3, A4 on it such that AA1 = A1A2 = A2A3 = A3A4.
Step 5: Join A3 to C.
Step 6: Through A4 draw A4C parallel to A3C which intersects BC produced at C.
Step 7: Through C draw CB parallel to AC which intersects AB produced at B.
BBC is the required triangle which is 4/3 of ABC.

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