Question Bank — Measuring Space: Perimeter and Area

NCERT Class 9 Mathematics · Chapter 6 · CBSE Board Pattern

Class 9 Mathematics Chapter 6 Question Bank CBSE Board Pattern

QUESTION BANK

CBSE Class 09 Mathematics

Chapter: Measuring Space – Perimeter and Area
(Questions strictly based on NCERT chapter content and terminology)

Section A — MCQs (10 questions, 1 mark each)

  1. The perimeter of a square of side \(a\) units is
    (a) \(a\) (b) \(2a\) (c) \(3a\) (d) \(4a\)

  2. The circumference of a circle of radius \(r\) is
    (a) \(\pi r\) (b) \(2\pi r\) (c) \(\pi r^2\) (d) \(2r\)

  3. If the diameter of a circle is doubled, its circumference becomes
    (a) double (b) half (c) four times (d) remains same

  4. The length of an arc of a circle subtending an angle \(\theta^\circ\) at the centre is
    (a) \(2\pi r \times \frac{\theta}{360}\) (b) \(\pi r \times \frac{\theta}{180}\) (c) \(2\pi r \times \frac{\theta}{180}\) (d) \(\pi r \times \frac{\theta}{360}\)

  5. The area of a triangle with base \(b\) and height \(h\) is
    (a) \(bh\) (b) \(\frac{1}{2}bh\) (c) \(2bh\) (d) \(b+h\)

  6. A median of a triangle divides it into two triangles of
    (a) equal perimeter (b) equal area (c) equal sides (d) equal angles

  7. Which of the following is not an approximation for \(\pi\) mentioned in the chapter?
    (a) \(\frac{22}{7}\) (b) \(\frac{355}{113}\) (c) 3.1416 (d) 3.5

  8. The area of a sector of a circle with central angle \(\theta^\circ\) is
    (a) \(\pi r^2 \times \frac{\theta}{360}\) (b) \(2\pi r \times \frac{\theta}{360}\) (c) \(\pi r \times \frac{\theta}{180}\) (d) \(\pi r^2 \times \frac{\theta}{180}\)

Assertion-Reason Questions

  1. Assertion (A): The ratio of circumference to diameter (C/D) is the same for all circles.
    Reason (R): This constant ratio is denoted by \(\pi\) and is irrational.
    (a) Both A and R are true and R is the correct explanation of A.
    (b) Both A and R are true but R is not the correct explanation of A.
    (c) A is true but R is false.
    (d) A is false but R is true.

  2. Assertion (A): A median of a triangle divides it into two triangles of equal area.
    Reason (R): The two triangles have equal bases and the same height.
    (a) Both A and R are true and R is the correct explanation of A.
    (b) Both A and R are true but R is not the correct explanation of A.
    (c) A is true but R is false.
    (d) A is false but R is true.

Section B — Very Short Answer (6 questions, 2 marks each)

  1. The perimeter of a circle is 44 cm. Find its radius. (Use \(\pi = \frac{22}{7}\))

  2. Find the circumference of a circle of radius 7 cm. (Use \(\pi = \frac{22}{7}\))

  3. An arc of a circle of radius 3.5 cm subtends an angle of 60° at the centre. Find the length of the arc.

  4. State the formula for the length of an arc of a circle in terms of radius \(r\) and central angle \(\theta^\circ\).

  5. Write the formula for the area of a parallelogram. How is it related to the area of a rectangle?

  6. What is the value of \(\pi\) given by Āryabhaṭa? How did he describe it?

Section C — Short Answer (5 questions, 3 marks each)

  1. Find the length of an arc of a circle of radius 6.3 m that subtends an angle of 120° at the centre. (Use \(\pi = \frac{22}{7}\))

  2. A sector of a circle of radius 14 cm has a central angle of 75°. Find the perimeter of the sector (curved part + two radii).

  3. The parallel sides of a trapezium are 40 cm and 20 cm and the non-parallel sides are each 26 cm. Find its area.

  4. The sides of a triangle are 8 cm, 11 cm and the perimeter is 32 cm. Find its area using Heron’s formula.

  5. Find the area of a quadrant of a circle whose circumference is 44 cm. (Use \(\pi = \frac{22}{7}\))

Section D — Long Answer (3 questions, 5 marks each)

  1. A 400 m athletics track consists of two straight sections of 84.39 m each and two semicircular ends of innermost radius 36.5 m. An athlete runs 0.3 m from the inner border. Verify that the total distance covered in one complete circuit is 400 m. (Use \(\pi \approx 3.1416\))

  2. Prove that a median of a triangle divides it into two triangles of equal area. (Use the chapter’s theorem statement and reasoning with equal bases and common height.)

  3. Describe Baudhāyana’s construction to square a rectangle of sides \(a\) and \(b\) (\(a > b\)) and justify why the constructed square has the same area as the rectangle. (Multi-step geometric justification required.)

Section E — Case/Source-Based (2 questions, 4 marks each)

Case 1: Athletics Track Stagger

In a standard 400 m track, the innermost lane has semicircular ends of radius 36.5 m and straight sections of 84.39 m. The width of each lane is 1.22 m.

(i) Calculate the radius for an athlete running in the second lane (0.3 m from inner border of that lane).
(ii) Find the extra distance run by the second-lane athlete on the curved portions compared to the first lane.
(iii) Why is a stagger given to the outer-lane athletes?
(iv) If the stagger between first and second lane is calculated, will an equal stagger be required between second and third lane? Give reason.

Case 2: Flower Petal Design

A square garden of side 14 cm has semicircular arcs drawn with midpoints of sides as centres, forming petals.

(i) Find the radius of each semicircular arc.
(ii) Find the perimeter contribution of one petal (curved part only).
(iii) Calculate the total perimeter of all petals if there are four such petals.
(iv) How does this design relate to the perimeter of combined semicircles?

▶ Answer Key Attempt all questions first,
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Section A

  1. (d)
  2. (b)
  3. (a) — circumference scales linearly with diameter/radius.
  4. (a)
  5. (b)
  6. (b)
  7. (d)
  8. (a)
  9. (a) — R correctly explains the constant nature of the C/D ratio.
  10. (a) — Equal bases (halves of the same side) and common height give equal area (1 mark each for base equality and height).

Section B

  1. \(C = 2\pi r = 44\) → \(r = 7\) cm (correct formula 1 mark, substitution & answer 1 mark).
  2. \(C = 2 \times \frac{22}{7} \times 7 = 44\) cm.
  3. Arc length = \(2\pi r \times \frac{60}{360} = 3.67\) cm (approx.).
  4. \(l = 2\pi r \times \frac{\theta^\circ}{360^\circ}\).
  5. Area = base × height; obtained by transforming parallelogram into rectangle of same base and height.
  6. \(\frac{62832}{20000} = 3.1416\); described as asanna (approaching/approximate).

Section C

  1. Arc = \(2 \times \frac{22}{7} \times 6.3 \times \frac{120}{360} = 13.2\) m.
  2. Perimeter of sector = \(2r + \frac{75}{360} \times 2\pi r = 46.4\) cm (approx.).
  3. Height via Pythagoras on non-parallel sides; area = \(\frac{1}{2}(40+20) \times h = 480\) cm².
  4. \(s = 16\), area = \(\sqrt{16(8)(5)(5)} = 6\sqrt{55}\) cm².
  5. Radius = 7 cm; quadrant area = \(\frac{1}{4}\pi r^2 = 38.5\) cm².

Section D

  1. Straight parts = 168.78 m; curved parts form full circle of radius 36.8 m → circumference ≈ 231.22 m; total = 400 m (step-wise addition with marking for each component).
  2. Median creates two triangles with equal bases and common height → equal area (full proof steps).
  3. Full construction steps + algebraic verification using \(ab = \left(\frac{a+b}{2}\right)^2 - \left(\frac{a-b}{2}\right)^2\) (Baudhāyana identity).

Section E

Case 1

(i) Radius = 36.5 + 1.22 + 0.3 = 38.02 m.
(ii) Extra curved distance = \(2\pi \times 1.22\) m.
(iii) Outer lane has larger radius on curves.
(iv) Yes, each successive lane requires its own stagger (different radius increment).

Case 2

(i) Radius = 7 cm.
(ii) One petal curved length = \(\pi \times 7\) cm.
(iii) Total curved perimeter of four petals = \(4\pi \times 7 = 88\) cm.
(iv) Design uses four semicircles whose perimeters combine as shown in chapter examples.

All answers use only chapter-derived formulas, values of \(\pi\), and terminology.

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All questions are answerable from the NCERT chapter text. Reviewed by GFIS faculty.