Question 1
↑ Jump to NavigatorThe perimeter of a circle is 44 cm. What is its radius?
Solution
Given:
We know,
Therefore,
Using:
Hence,
Answer:r = 7 cm
Chapter 6: Measuring Space: Perimeter and AreaPage 129–130All 8 Questions on This Page
Class 9 Maths Chapter 6 Exercise Set 6.1 Solutions
Step-by-step solutions for Class 9 Mathematics Ganita Manjari Chapter 6 Exercise Set 6.1 (Questions 1–8, Page 129–130). Covers circumference of circles, arc lengths, sector perimeter, composite shapes with semicircles and quarter-circles, tyre revolutions, floral petal perimeters, using π = 22/7, and perimeter calculations.
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Key Concepts for Exercise Set 6.1
Jump to Question 11
Perimeter of a Composite Shape
The perimeter of any composite geometric figure is the total length of its continuous outer boundary. Internal construction lines, diameters, and grid lines lie inside the shape and must never be added to the perimeter.
Perimeter = Sum of all exterior boundary segments
2
Circumference of a Circle
The total distance around a complete circle of radius r is given by 2πr. For all circular calculations in this exercise, start from this fundamental formula using the textbook default approximation π ≈ 22/7.
Circumference = 2πr
3
Radius and Diameter
When a question or diagram gives the diameter (d) of a circle or semicircle, always find the radius first using r = d/2 before substituting into the circumference formula.
r = d / 2 or d = 2r
4
Fractional Circular Arcs
The curved boundary of symmetrical shapes often consists of fractions of a circle: • Semicircle: half of a circle (2πr × 1/2 = πr) • Quarter-circle: one-fourth of a circle (2πr × 1/4 = πr/2) • Three-quarter circle: three-fourths of a circle (2πr × 3/4)
Arc length = 2πr × fraction
5
Length of an Arc Subtending Angle θ
A circular arc subtending an angle θ at the centre represents the fraction θ/360° of the full circle's circumference 2πr.
Length of arc = 2πr × (θ / 360°)
6
Perimeter of a Sector
A sector is bounded by a curved arc and two straight radii. Its perimeter is the sum of the curved arc length and the two straight radii.
Perimeter of sector = Arc length + 2r
7
Identifying the Outer Boundary
Only the exposed exterior curved arcs and straight edges form the boundary. Dotted construction lines (such as square or hexagon sides, triangle hypotenuses, and internal baselines) are internal aids and must never be included in the perimeter.
Actual Boundary = Outer curved arcs (+ exposed straight segments)
8
Distance and Revolutions of a Rolling Wheel
In one complete revolution, a circular wheel or tyre travels a distance equal to its circumference (2πr). To find the total revolutions over a given distance, divide the total distance by the distance in one revolution.
Distance in 1 revolution = 2πr, Revolutions = Total distance / 2πr
9
Perimeter of Symmetrical Flower Petals
In floral patterns framed by polygons, each petal is enclosed by 2 congruent circular arcs meeting at their ends. Find the perimeter of 1 petal first (2 × arc length), then multiply by the total number of petals.
Perimeter of all petals = Number of petals × (2 × Arc length)
10
Ratio of Circumferences and Radii
Because the circumference of a circle is directly proportional to its radius (P = 2πr), the constant factor 2π cancels in ratios. The ratio of the perimeters of two circles is therefore strictly equal to the ratio of their radii.
P₁ : P₂ = 2πr₁ : 2πr₂ = r₁ : r₂
Question 2
↑ Jump to NavigatorCalculate, correct to 3 significant figures, the circumference of a circle with: (i) radius 7 cm (ii) radius 10 cm (iii) radius 12 cm
Solution
We know,
Using:
(i) Radius 7 cm:
Given:
Therefore,
Correct to 3 significant figures:
(ii) Radius 10 cm:
Given:
Therefore,
Correct to 3 significant figures:
(iii) Radius 12 cm:
Given:
Therefore,
Correct to 3 significant figures:
Answer:(i) 44.0 cm | (ii) 62.9 cm | (iii) 75.4 cm
Question 3
↑ Jump to NavigatorCalculate 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 cm and the angle at the centre is 120°.
Solution
We know,
Using:
(i) Radius 3.5 cm and central angle 60°:
Given:
Therefore,
Since:
we get,
Hence,
(ii) Radius 6.3 cm and central angle 120°:
Given:
Therefore,
Since:
we get,
= 2 × 22/7 × 6.3 × 1/3
Hence,
Answer:(i) 113 cm ≈ 3.67 cm | (ii) 13.2 cm
Question 4
↑ Jump to NavigatorFind 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°.
Sector of a Circle (Radius 14 cm, Angle 75°)Fig. 6.1
Fig. 6.1: A circle sector with radius r = 14 cm and central angle θ = 75°, showing that the perimeter consists of the curved arc AB and the two straight radii OA and OB.
Solution
Given:
Using:
The perimeter of a sector consists of the curved arc and the two radii.
Length of the curved arc
We know,
Substituting the given values,
= 2 × 22/7 × 14 × 75/360
Simplifying,
= 2 × 22 × 2 × 5/24
So, the length of the curved portion is:
Perimeter of the sector
The two straight portions are the two radii:
Therefore,
Hence,
Perimeter of the sector = 139/3 cm
or
Answer:Perimeter of the sector = 1393 cm (or 46 13 cm ≈ 46.33 cm)
Question 5
↑ Jump to NavigatorFind the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate). The figures are: (i)–(ix), Fig. 6.14.
- (i)Two straight portions of 80 m each and two semicircular ends with diameter 60 m
- (ii)Outer semicircle diameter 12 cm, inner semicircle diameter 8 cm, and two short straight connecting portions
- (iii)Four semicircular arcs of diameter 10 cm each on a central square
- (iv)Three semicircles constructed outward on equilateral triangle sides (d = 12 cm each)
- (v)Four quarter-circles (radius 14 cm) and four semicircles (diameter 14 cm)
- (vi)Large semicircle (diameter 28 cm) and four small semicircles (diameter 7 cm each)
- (vii)Three semicircles constructed on the sides of a right-angled triangle (sides 8 cm and 6 cm)
- (viii)Large semicircle (diameter 12 cm) with three small semicircles (diameter 4 cm each) along the base
- (ix)Large semicircle (diameter 20 cm) and two small semicircles (diameter 10 cm each) forming a monad contour
Solution
(i) Two straight portions of 80 m each and two semicircular ends (d = 60 m):
Given:
Diameter of each semicircle:
Therefore,
The two semicircular arcs together make one complete circle.
Therefore,
The two straight portions together are:
Therefore,
Perimeter of shape
Therefore,
(ii) Outer semicircle (d = 12 cm) and inner semicircle (d = 8 cm):
Given:
Outer diameter:
Therefore,
Inner diameter:
Therefore,
Outer semicircular arc:
2πr₁ × 1/2
= 2 × 22/7 × 6 × 1/2
Inner semicircular arc:
2πr₂ × 1/2
= 2 × 22/7 × 4 × 1/2
The two short straight portions together are:
Therefore,
Perimeter of shape
Therefore,
(iii) Four semicircular arcs (d = 10 cm each):
Given:
Therefore,
Length of one semicircular arc:
2πr × 1/2
= 2 × 22/7 × 5 × 1/2
There are four such semicircular arcs.
Therefore,
Perimeter of shape
Therefore,
The dotted square/construction lines are NOT part of the perimeter.
(iv) Three semicircles on equilateral triangle sides (d = 12 cm each):
Given:
Therefore,
Length of one semicircular arc:
2πr × 1/2
= 2 × 22/7 × 6 × 1/2
There are three such semicircular arcs.
Therefore,
Perimeter of shape
Therefore,
The dotted sides of the equilateral triangle are internal construction lines and are NOT part of the perimeter.
(v) Four quarter-circles (r = 14 cm) and four semicircles (d = 14 cm):
From the figure:
4 quarter-circles of radius 14 cm
4 semicircles of diameter 14 cm
Four quarter-circles
For one quarter-circle:
2πr × 1/4
= 2 × 22/7 × 14 × 1/4
Therefore, for four quarter-circles:
Four semicircles
Each semicircle has:
Therefore,
Length of one semicircle:
2πr × 1/2
= 2 × 22/7 × 7 × 1/2
Therefore, for four semicircles:
Total perimeter
Perimeter of shape
Therefore,
The dotted grid and construction lines are NOT part of the perimeter.
(vi) Large semicircle (d = 28 cm) and four small semicircles (d = 7 cm each):
For the large upper semicircle:
Therefore,
Length of large semicircular arc:
2πr₁ × 1/2
= 2 × 22/7 × 14 × 1/2
Along the dotted baseline, the 28 cm diameter is divided into four equal sections:
For each of the four small semicircles:
Therefore,
Length of one small semicircular arc:
2πr₂ × 1/2
= 2 × 22/7 × 7/2 × 1/2
There are four such small semicircular arcs along the baseline (two curving downwards and two curving upwards):
Total length of four small arcs:
Total perimeter of the shape:
Therefore,
The dotted horizontal line is an internal construction line and is NOT part of the perimeter.
Note
Alternative View:
• The shape is bounded by 1 large semicircle of radius 14 cm and 4 small semicircles of radius 7/2 cm.
• 4 small semicircles equal 2 complete circles of radius 7/2 cm (or 1 circle of radius 7 cm).
• Total perimeter = (1/2 × 2π × 14) + (2 × 2π × 7/2) = 44 + 44 = 88 cm.
(vii) Three semicircles on sides of right-angled triangle (8 cm and 6 cm):
Given:
Two perpendicular sides of the right-angled triangle are:
By Pythagoras theorem, the hypotenuse is:
The three semicircles are constructed on the three sides (diameters 10 cm, 8 cm, and 6 cm).
For semicircle on hypotenuse (d₁ = 10 cm):
Length of semicircular arc:
2πr₁ × 1/2
= 2 × 22/7 × 5 × 1/2
For semicircle on 8 cm side (d₂ = 8 cm):
Length of semicircular arc:
2πr₂ × 1/2
= 2 × 22/7 × 4 × 1/2
For semicircle on 6 cm side (d₃ = 6 cm):
Length of semicircular arc:
2πr₃ × 1/2
= 2 × 22/7 × 3 × 1/2
Total perimeter of the shape:
Therefore,
The dotted sides of the right-angled triangle are internal construction lines and are NOT part of the perimeter.
(viii) Large semicircle (d = 12 cm) with three small semicircles (d = 4 cm each):
Given:
The base has three equal segments of 4 cm each.
Diameter of large semicircle:
d₁ = 4 + 4 + 4 = 12 cm
Therefore,
Length of large semicircular arc:
2πr₁ × 1/2
= 2 × 22/7 × 6 × 1/2
For each of the three small semicircles:
Therefore,
Length of one small semicircular arc:
2πr₂ × 1/2
= 2 × 22/7 × 2 × 1/2
There are three such small semicircular arcs.
Total length of three small arcs:
3 × 44/7
Total perimeter of the shape:
Therefore,
The dotted baseline is a construction line and is NOT part of the perimeter.
(ix) Large semicircle (d = 20 cm) and two small semicircles (d = 10 cm each):
Given:
The base has two equal segments of 10 cm each.
Diameter of large upper semicircle:
d₁ = 10 + 10 = 20 cm
Therefore,
Length of large semicircular arc:
2πr₁ × 1/2
= 2 × 22/7 × 10 × 1/2
For the two small semicircles (one curving upward, one curving downward):
Diameter of each:
Therefore,
Length of one small semicircular arc:
2πr₂ × 1/2
= 2 × 22/7 × 5 × 1/2
There are two such small semicircular arcs.
Total length of both small arcs:
2 × 110/7
Total perimeter of the shape:
Therefore,
The dotted horizontal line is a construction line and is NOT part of the perimeter.
Answer:(i) Perimeter = 24407 m ≈ 348.57 m | (ii) Perimeter = 2487 cm ≈ 35.43 cm | (iii) Perimeter = 4407 cm ≈ 62.86 cm | (iv) Perimeter = 3967 cm ≈ 56.57 cm | (v) Perimeter = 176 cm | (vi) Perimeter = 88 cm | (vii) Perimeter = 2647 cm ≈ 37.71 cm | (viii) Perimeter = 2647 cm ≈ 37.71 cm | (ix) Perimeter = 4407 cm ≈ 62.86 cm
Question 6
↑ Jump to NavigatorIf 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?
Solution
(i) Distance covered in one complete revolution:
Given:
Diameter of tyre:
Therefore,
We know that the distance covered by a tyre in one complete revolution is equal to its circumference.
Using:
Distance in 1 revolution:
Converting into metres:
176 cm = 176/100 m = 1.76 m
Therefore,
Distance in 1 revolution = 176 cm = 1.76 m
(ii) Number of revolutions if the car travels 10 km:
Given:
From part (i):
Distance covered in 1 revolution = 176 cm
Convert 10 km into centimetres:
10 km = 10 × 1000 m = 10,000 m
10,000 m = 10,000 × 100 cm = 1,000,000 cm
We know,
Number of revolutions = Total distance / Distance in 1 revolution
Therefore,
Dividing numerator and denominator by 16:
Rounding off to the nearest whole revolution:
Therefore,
Number of revolutions = 62,500/11 ≈ 5681.82 revolutions ≈ 5682 revolutions (rounded off)
Answer:(i) Distance = 176 cm = 1.76 m | (ii) Number of revolutions = 62,50011 ≈ 5681.82 revolutions ≈ 5682 revolutions (rounded off)
Question 7
↑ Jump to NavigatorFind the total perimeter of all the petals in each of the given flowers.
- (i)Fig. 6.15A: The centres of the arcs are the midpoints of the sides of the square (side = 14 cm)
- (ii)Fig. 6.15B: The centres of the arcs are the vertices of the hexagon (side = 42 cm)
💡Hint from textbook:
Each petal is enclosed by 2 identical curved arcs meeting at their endpoints. • Find the perimeter of 1 petal by finding the length of one arc and multiplying by 2. • Then multiply the perimeter of 1 petal by the total number of petals. • The dotted lines form the construction frame (square or hexagon) and are NOT part of the petals.
Solution
(i) Four petals in a square of side 14 cm (Fig. 6.15A):
Understanding the geometry:
•The square has side length = 14 cm.
•The centres of the circular arcs are the midpoints of the sides of the square.
•The distance from each midpoint to an adjacent corner gives the radius:
•Each arc sweeps an angle of 90° from a corner to the centre of the square (a quarter-circle).
•Each petal is enclosed by two identical quarter-circle arcs meeting at their endpoints.
Method 1: Step-by-Step Method (Standard for School Exams):
Length of one quarter-circle arc:
= 2 × 22/7 × 7 × 1/4
Perimeter of one petal:
Each petal consists of 2 identical quarter-circle arcs:
Perimeter of 1 petal = 2 × 11 = 22 cm
Total perimeter of all four petals:
Since the flower has 4 identical petals:
Total perimeter = 4 × 22 = 88 cm
Therefore,
The square's dotted sides are internal construction lines and are NOT part of the perimeter of the petals.
Note
Method 2 — Quick Symmetry Method (Circle-Combining Shortcut):
• Across all 4 petals, there are 4 × 2 = 8 quarter-circle arcs in total.
• Since 4 quarter-circles make 1 complete circle, 8 quarter-circles equal 2 complete circles of radius 7 cm (8/4 = 2).
• Total perimeter = 2 × (2πr) = 2 × (2 × 22/7 × 7) = 2 × 44 = 88 cm.
Therefore, Total perimeter = 88 cm.
(ii) Six petals in a regular hexagon of side 42 cm (Fig. 6.15B):
Understanding the geometry:
•The flower is framed inside a regular hexagon of side length = 42 cm.
•The centres of the circular arcs are the 6 vertices of the hexagon.
•In a regular hexagon, segments connecting each vertex to the centre form 6 equilateral triangles of side 42 cm. Thus, the radius is:
•Each equilateral triangle has an angle of 60° at each vertex. Therefore, each arc subtends a central angle of θ = 60°.
•Each petal is enclosed by two identical 60° circular arcs meeting at a vertex and at the centre.
Method 1: Step-by-Step Method (Standard for School Exams):
Length of one 60° circular arc:
= 2 × 22/7 × 42 × 1/6
Perimeter of one petal:
Each petal consists of 2 identical 60° circular arcs:
Perimeter of 1 petal = 2 × 44 = 88 cm
Total perimeter of all six petals:
Since the flower has 6 identical petals:
Total perimeter = 6 × 88 = 528 cm
Therefore,
The hexagon's dotted sides are internal construction lines and are NOT part of the perimeter of the petals.
Note
Method 2 — Quick Symmetry Method (Circle-Combining Shortcut):
• Across all 6 petals, there are 6 × 2 = 12 circular arcs of 60° each.
• Since 12 × 60° = 720° = 2 × 360°, all 12 arcs combine to equal the circumference of 2 complete circles of radius 42 cm.
• Total perimeter = 2 × (2πr) = 2 × (2 × 22/7 × 42) = 2 × 264 = 528 cm.
Therefore, Total perimeter = 528 cm.
Answer:(i) Perimeter = 88 cm | (ii) Perimeter = 528 cm
Question 8
↑ Jump to NavigatorThe ratio of the perimeters of two circles is 5:4. What is the ratio of their radii?
Solution
Let the radii of the two circles be:
r₁ and r₂
We know,
Perimeter of a circle = 2πr
Therefore, the perimeters of the two circles are:
and
Given:
Therefore,
Cancelling the common factor 2π:
Hence,
Ratio of their radii = 5 : 4
Final answer:
Answer:Ratio of radii = 5 : 4
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