NCERT Solutions for Class 9th Maths Chapter 6 Exercise Set 6.1 — Problems, Puzzles, and Paradoxes on Perimeter

Book page 129–130 Updated on2026-09-08

Q1.

Unless stated otherwise, use the approximation 22/7 for π. The perimeter of a circle is 44 cm. What is its radius?

Answer

r = 7 cm.

C = 2πr
44 = 2 × ²²⁄₇ × r
44 = ⁴⁴⁄₇ r
r = 44 × ⁷⁄₄₄ = 7 cm

Tip: with π ≈ 22/7 the circumference of a circle of radius 7 is exactly 44. Radii of 7, 14, 21, 3.5 and 10.5 cm keep the arithmetic clean, which is why they appear so often in this exercise.

Q2.

Calculate, correct to 3 significant figures, the circumference of a circle with: (i) radius 7 cm (ii) radius 10 cm (iii) radius 12 cm.

Answer

Using C = 2πr with π ≈ ²²⁄₇:

Radius	2 × 22/7 × r	Exact value	To 3 s.f.
(i) 7 cm	2 × 22/7 × 7	44 cm	44.0 cm
(ii) 10 cm	2 × 22/7 × 10 = 440/7	62.857… cm	62.9 cm
(iii) 12 cm	2 × 22/7 × 12 = 528/7	75.428… cm	75.4 cm

Why it happens: "3 significant figures" counts from the first non-zero digit, not from the decimal point. So 62.857… keeps 6, 2 and 8 and rounds to 62.9; and 44 must be written 44.0 to show that three figures are being claimed.

Tip: the Hindi edition of this question asks for the answer correct to 3 decimal places instead, giving 44.000 cm, 62.857 cm and 75.429 cm.

Q3.

Calculate the length of the arc of a circle if: (i) the radius is 3.5 cm and the angle at the centre is 60°, and (ii) the radius is 6.3 m and the angle at the centre is 120°.

Answer

An arc that turns through θ° at the centre is the fraction θ/360 of the whole circle.

l = 2πr × ^θ°⁄_360°

(i) r = 3.5 cm, θ = 60°

whole circumference = 2 × ²²⁄₇ × 3.5 = 22 cm
l = 22 × ⁶⁰⁄₃₆₀ = ²²⁄₆ = ¹¹⁄₃ = 3.67 cm (2 d.p.)

(ii) r = 6.3 m, θ = 120°

whole circumference = 2 × ²²⁄₇ × 6.3 = ⁴⁴⁄₇ × 6.3 = 44 × 0.9 = 39.6 m
l = 39.6 × ¹²⁰⁄₃₆₀ = 39.6 × ¹⁄₃ = 13.2 m

Why it happens: the circle looks exactly the same after a rotation about its centre through any angle. So if you cut the circle into 360 equal one-degree arcs, they are all congruent. An arc of θ° is made of θ of them, hence θ/360 of the whole. No new idea is needed — the formula is pure proportion.

Q4.

Find the perimeter of a sector (i.e., the curved portion as well as the two straight portions) of a circle of radius 14 cm and sector angle 75°.

Answer

Perimeter = 46.33 cm (exactly ¹³⁹⁄₃ cm).

whole circumference = 2 × ²²⁄₇ × 14 = 88 cm
arc = 88 × ⁷⁵⁄₃₆₀ = 88 × ⁵⁄₂₄ = ⁵⁵⁄₃ = 18.33 cm

perimeter = arc + 2 radii
= ⁵⁵⁄₃ + 2 × 14
= 18.33 + 28
= 46.33 cm

Tip: the commonest slip in this question is to give only the arc. A sector is bounded by three lines — one curved and two straight — so both radii must be added.

Q5.

Find the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate) (Fig. 6.14i to 6.14ix): (i) a rectangle 80 m by 60 m with a semicircle on each of the two shorter sides; (ii) a half-ring whose outer semicircle has diameter 12 cm and inner semicircle diameter 8 cm; (iii) a square of side 10 cm with a semicircle drawn outwards on each side; (iv) an equilateral triangle of side 12 cm with a semicircle drawn outwards on each side; (v) a square divided into a 3 × 3 grid of cells of side 14 cm, with a semicircle drawn outwards on the middle third of each side and a quarter circle of radius 14 cm at each corner; (vi) a semicircle of diameter 28 cm, with the diameter divided into four equal parts and a semicircle of diameter 7 cm drawn on each part, alternately below and above the diameter; (vii) semicircles drawn outwards on the three sides of a right-angled triangle whose legs are 8 cm and 6 cm; (viii) a semicircle of diameter 12 cm with three semicircles of diameter 4 cm each drawn on its diameter; (ix) a semicircle of diameter 20 cm, together with a semicircle of diameter 10 cm above the left half of the diameter and a semicircle of diameter 10 cm below the right half.

Answer

Every one of these is built from arcs whose lengths you already know. Recall that a semicircle of radius r has arc length πr, and a quarter circle has arc length πr/2. Use π ≈ ²²⁄₇ throughout.

Shape	Made of	In terms of π	Perimeter
(i)	2 straight sides of 80 m + 2 semicircles of diameter 60 m	160 + 60π	348.57 m
(ii)	semicircle r = 6 + semicircle r = 4 + 2 straight bits of 2 cm	10π + 4	35.43 cm
(iii)	4 semicircles of radius 5 cm	20π	62.86 cm
(iv)	3 semicircles of radius 6 cm	18π	56.57 cm
(v)	4 semicircles of radius 7 cm + 4 quarter circles of radius 14 cm	28π + 28π = 56π	176 cm
(vi)	1 semicircle of radius 14 cm + 4 semicircles of radius 3.5 cm	14π + 14π = 28π	88 cm
(vii)	semicircles on 6 cm, 8 cm and 10 cm	3π + 4π + 5π = 12π	37.71 cm
(viii)	1 semicircle of radius 6 cm + 3 semicircles of radius 2 cm	6π + 6π = 12π	37.71 cm
(ix)	1 semicircle of radius 10 cm + 2 semicircles of radius 5 cm	10π + 10π = 20π	62.86 cm

The working, shape by shape.

(i) 2(80) + 2 × ½ × π × 60 = 160 + 60 × ²²⁄₇ = 160 + ¹³²⁰⁄₇ = ²⁴⁴⁰⁄₇ = 348.57 m

(ii) The flat edge is not one straight piece: the inner semicircle leaves two gaps of (12 − 8) ÷ 2 = 2 cm each.
π(6) + π(4) + 2 + 2 = 10π + 4 = ²²⁰⁄₇ + 4 = 35.43 cm

(iii) Each side of the square is the diameter of one semicircle, so r = 5.
4 × π(5) = 20π = ⁴⁴⁰⁄₇ = 62.86 cm

(iv) Each side of the triangle is a diameter, so r = 6.
3 × π(6) = 18π = ³⁹⁶⁄₇ = 56.57 cm

(v) Semicircle on a 14 cm segment: π(7) = 22 cm each, four of them = 88 cm.
Quarter circle of radius 14: ¼ × 2π(14) = 7π = 22 cm each, four of them = 88 cm.
Total = 176 cm

(vi) Big semicircle: π(14) = 44 cm. Each small semicircle: π(3.5) = 11 cm; four of them = 44 cm.
Total = 88 cm

(vii) The hypotenuse is √(8² + 6²) = √100 = 10 cm.
½π(10 + 8 + 6) = 12π = ²⁶⁴⁄₇ = 37.71 cm

(viii) π(6) + 3 × π(2) = 6π + 6π = 12π = 37.71 cm

(ix) π(10) + π(5) + π(5) = 20π = 62.86 cm

Shape (vi): the four small semicircles alternate below and above the diameter, but each still contributes π(3.5) = 11 cm — which side they bulge to makes no difference to length.

Why it happens: in every one of these figures the straight edges are either absent or easy, and the curved edges are always whole semicircles or quarter circles. So the total is a sum of terms of the form πr or πr/2. The neat surprises — (vi) coming out as exactly 28π, (viii) as 12π — happen because the small semicircles sit on pieces of a diameter, and half of (sum of the pieces) equals half of the whole diameter. Length is additive along the diameter, so cutting a semicircle into several smaller semicircles on the parts does not change the total arc length at all.

Check it yourself: compare (viii) with (vi). Both have small semicircles sitting on a divided diameter, and in both the small arcs add up to exactly the same length as one big semicircle on the whole diameter. Try it with 5 or 10 pieces — you always get the same total.

Q6.

If the diameter of a car tyre is 56 cm, then: (i) How far does the car need to travel for the tyre to complete one revolution? (ii) How many revolutions does the tyre make if the car travels 10 km?

Answer

(i) 176 cm = 1.76 m. (ii) About 5682 revolutions.

(i) distance in one revolution = circumference = πd
= ²²⁄₇ × 56
= 22 × 8
= 176 cm = 1.76 m

(ii) 10 km = 10 × 1000 × 100 = 1 000 000 cm
number of revolutions = ^{1 000 000}⁄₁₇₆ = 5681.81…
≈ 5682 revolutions

Why it happens: a wheel that rolls without slipping lays its rim down on the road exactly once per turn, like unrolling a measuring tape. So one turn advances the car by one circumference. This is also why a bigger wheel needs fewer turns for the same journey — and why cycle computers must be told the wheel size before they can measure distance.

Tip: the tyre completes 5681 full revolutions and is part way through the 5682nd. If a question asks "how many complete revolutions", answer 5681.

Q7.

Find the total perimeter of all the petals in each of the given flowers. (i) Fig. 6.15A: a square of side 14 cm, the centres of the arcs being the midpoints of the sides of the square. (ii) Fig. 6.15B: a regular hexagon of side 42 cm, the centres of the arcs being the vertices of the hexagon.

Answer

(i) 88 cm. (ii) 528 cm.

(i) The four-petalled flower in a square of side 14 cm.

Each arc has its centre at the midpoint of a side and passes through two corners of the square, so its radius is half the side: r = 7 cm. Take the midpoint of the bottom side as centre. The arc runs from the bottom-left corner to the centre of the square — from a point along the side to a point perpendicular to it — so it turns through 90°.

each arc = ¼ × 2π(7) = ^7π⁄₂
each petal is bounded by 2 such arcs
4 petals ⟹ 8 arcs

total = 8 × ^7π⁄₂ = 28π = 28 × ²²⁄₇ = 88 cm

Four petals, eight quarter-circle arcs of radius 7 cm, centred at the midpoints of the sides.

(ii) The six-petalled flower in a regular hexagon of side 42 cm.

In a regular hexagon the distance from the centre to a vertex equals the side, 42 cm. An arc centred at a vertex, of radius 42, therefore passes through the centre of the hexagon and through the two neighbouring vertices. Each petal runs from the centre O out to a vertex V, and is bounded by the arc centred at the vertex before V and the arc centred at the vertex after V.

Take the arc from O to V centred at the neighbouring vertex W.
WO = 42 (centre to vertex), WV = 42 (a side), OV = 42 (centre to vertex)
∴ triangle OWV is equilateral ⟹ ∠OWV = 60°

each arc = ⁶⁰⁄₃₆₀ × 2π(42) = 14π = 44 cm
each petal = 2 arcs = 88 cm
6 petals ⟹ 12 arcs

total = 12 × 44 = 528 cm

Why it happens: both flowers are drawn by putting the compass point at a special point of the polygon and swinging an arc of a length that is already present in the figure. That forces a nice angle — 90° from a midpoint of a square's side, 60° from a vertex of a hexagon — and a nice angle is what makes the arc length a simple fraction of a circumference. Rangoli and jāli patterns are built this way for exactly the same reason: a single compass setting produces the whole design.

Q8.

The ratio of the perimeters of two circles is 5:4. What is the ratio of their radii?

Answer

5 : 4 — the same ratio.

P₁ : P₂ = 5 : 4
2πr₁ : 2πr₂ = 5 : 4
^2πr₁⁄_2πr₂ = ⁵⁄₄
the factor 2π cancels
^r₁⁄_r₂ = ⁵⁄₄
∴ r₁ : r₂ = 5 : 4

Why it happens: perimeter is directly proportional to radius, with the fixed constant of proportionality 2π. Any two quantities in direct proportion keep the same ratio. Contrast this with area: since A = πr², areas would be in the ratio 5² : 4² = 25 : 16. Lengths scale like k, areas like k².

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