CBSE Class 10 · Boards 2027 · NCERT 2026-27 syllabus
NCERT Exemplar Solutions Class 10 Maths Chapter 11 Areas Related to Circles
Try each question first: pick an option or type your final answer, then see the full step-by-step solution. Stuck? Open the first step as a hint. Free, no account needed.
Boards begin February 2027
Pre-boards are in December
Areas Related to Circles is in Mensuration, which carries 10 of the 80 marks in the 2026-27 CBSE Maths paper
35 NCERT Exemplar questions from Chapter 11
Exercise 11.1
Exemplar Ex 11.1 Q11 mark
If the sum of the areas of two circles with radii R₁ and R₂ equals the area of a circle of radius R, then
If the circumference of a circle and the perimeter of a square are equal, then
Step 1[1 mark] For a fixed perimeter the circle encloses the maximum area; with 2πr = 4a one gets area of circle πr² and area of square π²r²/4, and since π < 4 the circle's area is larger — option (B).
Step-by-step solution
[1 mark] For a fixed perimeter the circle encloses the maximum area; with 2πr = 4a one gets area of circle πr² and area of square π²r²/4, and since π < 4 the circle's area is larger — option (B).
Answer:Area of the circle > Area of the square — option (B).
It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be
Step 1[1 mark] Radii are 8 m and 6 m; equating areas, πR² = π(8²) + π(6²) ⇒ R² = 64 + 36 = 100 ⇒ R = 10 m — option (A).
Step-by-step solution
[1 mark] Radii are 8 m and 6 m; equating areas, πR² = π(8²) + π(6²) ⇒ R² = 64 + 36 = 100 ⇒ R = 10 m — option (A).
Is it true to say that area of a segment of a circle is less than the area of its corresponding sector? Why?
Step 1[1 mark] For a minor segment, area = area of sector − area of the triangle, which is indeed less than the sector.
Step-by-step solution
[1 mark] For a minor segment, area = area of sector − area of the triangle, which is indeed less than the sector.
[1 mark] But for a major segment the triangle's area is added to the major sector, so the major segment exceeds its sector; hence the statement is not true in general (only for a minor segment).
Answer:Not always — it is true only for a minor segment, not a major one.
The numerical value of the area of a circle is greater than the numerical value of its circumference. Is this statement true? Why?
Step 1[1 mark] Area = πr² and circumference = 2πr; πr² > 2πr ⇔ r > 2.
Step-by-step solution
[1 mark] Area = πr² and circumference = 2πr; πr² > 2πr ⇔ r > 2.
[1 mark] So it holds only when r > 2 (for r < 2 the circumference is numerically greater, for r = 2 they are equal) — the statement is not always true; it depends on the radius.
Answer:Not always — it is true only when the radius r > 2; it depends on r.
With vertices A, B and C of a right triangle ABC (right-angled at B) as centres, arcs of radius 5 cm each are drawn. If AB = 14 cm, BC = 48 cm and CA = 50 cm, find the area of the shaded region inside the triangle but outside the three sectors.
Step 1[1 mark] Verification: 14² + 48² = 196 + 2304 = 2500 = 50², so ΔABC is right-angled at B.
Step-by-step solution
[1 mark] Verification: 14² + 48² = 196 + 2304 = 2500 = 50², so ΔABC is right-angled at B.
[1 mark] Area of ΔABC = (1/2) × AB × BC = (1/2) × 14 × 48 = 336 cm².
[1 mark] Each sector contributes (∠/360) × π × 5², and the total angle = ∠A + ∠B + ∠C = 180° (angle sum of a triangle).
The diameters of front and rear wheels of a tractor are 80 cm and 2 m respectively. Find the number of revolutions that the rear wheel will make in covering a distance in which the front wheel makes 1400 revolutions.
Step 1[1 mark] Distance covered by the front wheel = 1400 × π × 80 cm (circumference = π × diameter).
Step-by-step solution
[1 mark] Distance covered by the front wheel = 1400 × π × 80 cm (circumference = π × diameter).
[1 mark] Rear wheel diameter = 200 cm, so its revolutions = (1400 × π × 80) ÷ (π × 200).
A circular pond is 17.5 m in diameter. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate of Rs 25 per m². (Use π = 3.14)
Step 1[1 mark] Pond radius = 8.75 m; outer radius (pond + path) = 8.75 + 2 = 10.75 m.
Step-by-step solution
[1 mark] Pond radius = 8.75 m; outer radius (pond + path) = 8.75 + 2 = 10.75 m.
[1 mark] Area of path = π(10.75² − 8.75²) = 3.14 × (115.5625 − 76.5625) = 3.14 × 39 = 122.46 m².
The central angles of two sectors of circles of radii 7 cm and 21 cm are respectively 120° and 40°. Find the areas of the two sectors as well as the lengths of the corresponding arcs. What do you observe?
[1 mark] Observation: the two arc lengths are equal (44/3 cm each) although the sector areas (154/3 cm² and 154 cm²) are different — equal arcs need not give equal areas.
Answer:Areas 154/3 cm² and 154 cm²; both arcs 44/3 cm — equal arc lengths but unequal areas.
CBSE board pattern: Assertion-Reason and case-based questions
Written in the CBSE board format. These are not numbered NCERT/Exemplar questions; some case studies are built on textbook problems. Neither NCERT book has Assertion-Reason questions: they, and case-based questions, are a CBSE board-paper format, here on this chapter's 2026-27 syllabus, with solutions by LazyTopper.
CBSE board pattern · Assertion-Reason1 mark
Assertion (A): If two circles have equal circumferences, then their areas are equal.
Reason (R): Both circumference and area are determined by the radius of a circle.
Board examiners mark every step. Photograph your written answer and see the marks for each step, and exactly where they were lost. Your first answer check is free.
Questions from the NCERT Class 10 Mathematics Exemplar Problems (the CBSE board-pattern section is in the board format, not numbered book questions); the solutions are written by LazyTopper. LazyTopper is an independent platform and is not affiliated with CBSE or NCERT.