Model Test – Solved ഉത്തരസൂചിക
The same 19 questions as the model paper, in the same order, each with a full solution, a marking hint and the page of this guide to revisit. Questions 9–19 have step-by-step animations: press Next ▶.
c = 2πr = 2 × π × 7 = 14π cm (≈ 43.98 cm).
Marks: 1 mark for 14π (with unit).
Review: Page 7 · π, a new number
a = πr² = π × 3² = 9π sq cm (≈ 28.27 sq cm).
Marks: 1 mark for 9π.
Review: Page 9 · Area of a circle
Circumferences change in the same proportion as diameters: c = πd, so 2d gives 2c.
Marks: 1 mark for (b).
πd = 10π ⇒ d = 10 cm.
Marks: 1 mark.
Review: Page 7 · π, a new number
πr² = 49π ⇒ r² = 49 ⇒ r = 7 cm.
Marks: 1 mark.
Review: Page 9 · Area of a circle
π = 3.14159… ; the fifth decimal (9) rounds the fourth up: 3.1416.
Marks: 1 mark.
Review: Page 7 · π, a new number
Area = ½ × circumference × radius (a = ½cr), from the polygon argument.
Marks: 1 mark.
Review: Page 9 · Area of a circle
Madhavan of Sangamagrama (14th century) found that 1 − 1/3 + 1/5 − 1/7 + … approaches π/4.
Marks: 1 mark for (b).
Review: Page 7 · π, a new number
The wire becomes the circumference: πd = 31.4, so d = 31.4 ÷ 3.14 = 10 cm.
Marks: 1 mark for πd = 31.4; 1 mark for d = 10 cm.
Review: Page 7 · π, a new number
The perpendicular from the centre halves the chord: half-chord 4 cm. Radius² = 3² + 4² = 25, so r = 5 cm. Circumference 2π × 5 = 10π ≈ 31.42 cm; area π × 25 = 25π ≈ 78.54 sq cm.
Marks: 1 mark for r = 5; ½ mark each for 10π and 25π.
Review: Page 8 · Practice set 2, Page 3 · Chords and Pythagoras
Ring = big circle − small circle = π × 5² − π × 3² = 25π − 9π = 16π ≈ 50.27 sq cm.
Marks: 1 mark for π(5² − 3²); 1 mark for 16π.
Review: Page 9 · Area of a circle
Inner circle: diameter = side = 4, radius 2, area 4π. Outer circle: diameter = diagonal = 4√2, radius 2√2, area π × 8 = 8π.
Marks: 1 mark each.
Review: Page 10 · Practice set 3
Circumferences 6π and 12π: ratio 1 : 2 (same as the radii). Areas 9π and 36π: ratio 1 : 4 (the square of the radius ratio).
Marks: 1 mark each.
Review: Page 5 · Diameter and circumference, Page 9 · Area of a circle
The hexagon’s side equals the radius, so r = 3. Circle: 9π ≈ 28.27. Hexagon: 6 equilateral triangles of side 3, each (√3/4) × 9, total 13.5√3 ≈ 23.38. Difference 9π − 13.5√3 ≈ 4.89 sq cm.
Marks: 1 mark circle area; 1 mark hexagon area; 1 mark difference.
The centre lies on the height. Let it be k cm above the base; then R = 9 − k and, from the right triangle with half-base 3, R² = 3² + k². So (9 − k)² = 9 + k² ⇒ 81 − 18k = 9 ⇒ k = 4, R = 5. Circumference 2π × 5 = 10π ≈ 31.42 cm.
Marks: 1 mark for the equation; 1 mark for R = 5; 1 mark for 10π.
Review: Page 8 · Practice set 2, Page 3 · Chords and Pythagoras
The diagonal is a diameter: √(6² + 8²) = 10, so r = 5 and the circle’s area is 25π ≈ 78.54 sq cm. Rectangle 48 sq cm. Outside part 25π − 48 ≈ 30.54 sq cm.
Marks: 1 mark diameter 10; 1 mark 25π; 1 mark 30.54.
Review: Page 10 · Practice set 3
Circumferences of the small circles: 2π, 4π, 6π; sum 12π. Big circle: π × 12 = 12π. In general π(d₁ + d₂ + d₃) = π × (sum of diameters) = π × D.
Marks: 1 mark each small circumference; 1 mark the sum and conclusion.
Review: Page 8 · Practice set 2
- Using regular polygons drawn inside a circle, explain why the area of a circle is ½ × circumference × radius.
- Hence show that the area is πr².
- Show that the ring between circles of radii 5 cm and 4 cm (same centre) has the same area as a circle of radius 3 cm.
- Explain why this happens.
(i) Join the centre to the corners of a regular polygon: n triangles of area ½sh, total ½ph. With more sides p → c and h → r, while the polygon’s area → the circle’s area, so a = ½cr. (ii) c = 2πr ⇒ a = ½ × 2πr × r = πr². (iii) Ring = π(5² − 4²) = 9π = π × 3². (iv) 5² − 4² = 3²: a right triangle with hypotenuse 5 and side 4 has third side 3, so the ring equals the circle on that side.
Marks: 2 marks (i), 1 mark (ii), 1 mark (iii), 1 mark (iv).
Review: Page 9 · Area of a circle
- Find the area of the hexagon.
- Find the total area of the six parts cut off.
- Find the area of the star-shaped part left.
- Find the total length of the six arcs.
(i) Six equilateral triangles of side 4: 6 × (√3/4) × 16 = 24√3 ≈ 41.57 sq cm. (ii) Each interior angle is 120° = ⅓ of a full turn, so each cut-off part is ⅓ of a circle of radius 2: (4π)/3; six give 8π ≈ 25.13 sq cm. (iii) Star = 24√3 − 8π ≈ 16.44 sq cm. (iv) Each arc is ⅓ × 2π × 2 = 4π/3; six give 8π ≈ 25.13 cm.
Marks: 1 + 1 + 1 marks for (i)–(iii); 2 marks for (iv) with the 120° reasoning.
Review: Page 10 · Practice set 3
Marks summary
| Section | Questions | Marks each | Total |
|---|---|---|---|
| A | 1–8 | 1 | 8 |
| B | 9–13 | 2 | 10 |
| C | 14–17 | 3 | 12 |
| D | 18–19 | 5 | 10 |
| Total | 40 | ||
- Section A (10 min): one line each; write the formula first (2πr or πr²).
- Section B (20 min): draw a quick figure; for chords, always drop the perpendicular from the centre.
- Section C (25 min): keep π in the working and round only at the end.
- Section D (25 min): for 18 write the polygon argument in clear steps; for 19 note the 120° angle before computing.
- Last 10 min: check units – cm for lengths, sq cm for areas.
If you scored 32 or more, you are ready. Otherwise revisit the pages under each question you missed, then try the scored quiz again. Back to the home page.