NCERT Solutions for Class 9 Maths Ganit Manjari Chapter 6 Measuring Space: Perimeter and Area

Perimeter is the total distance around the boundary of a shape, while area is the amount of space covered by a two-dimensional region. In Class 9 Maths Chapter 6, students learn how to calculate perimeter, circumference, arc length, triangle area, circle area, sector area and areas of different plane figures.

NCERT Solutions for Class 9 Maths Ganit Manjari Chapter 6 help students solve textbook questions from Measuring Space: Perimeter and Area in the new Ganita Manjari Class 9 Maths textbook. This chapter introduces circumference of a circle, the constant π, irrationality of π, arc length, athletics track stagger, area of rectangles, parallelograms, triangles, Heron’s formula, Brahmagupta’s formula, area of a circle and area of a sector.

Real-life examples such as athletics tracks, tyres, clock hands, wipers, circular petals and land measurements make the formulas easier to apply in CBSE 2026-27 exam questions. The exercise-wise solutions support practice with π, Heron’s formula, Brahmagupta’s formula and mixed perimeter-area problems.

NCERT Solutions for Class 9 Maths Chapter 6 PDF Download

The NCERT Solutions for Class 9 Maths Chapter 6 PDF is designed for quick CBSE 2026-27 revision of mensuration formulas and solved problems. Students can use it to practise circumference, arc length, triangle area, sector area and mixed perimeter-area questions before exams.

Students can use the PDF to revise:

1. Perimeter of basic shapes
2. Circumference of a circle
3. Approximate values of π
4. Irrationality of π
5. Length of an arc
6. Perimeter of sectors and composite figures
7. Area of rectangles and parallelograms
8. Area of triangles
9. Heron’s formula
10. Brahmagupta’s formula
11. Area of a circle
12. Area of a sector and segment

Access Exercise-Wise NCERT Solutions for Class 9 Maths Chapter 6

NCERT Solutions for Class 9 Maths Chapter 6 Exercise Set 6.1

Exercise Set 6.1 focuses on circumference, arc length, perimeter of sectors and perimeter of composite circular shapes. Unless stated otherwise, use π = 22/7.

Question 1. The perimeter of a circle is 44 cm. What is its radius?

Answer:
Perimeter of a circle means circumference.

C = 2πr

Given:

C = 44 cm
π = 22/7

44 = 2 × 22/7 × r
44 = 44r/7
r = 7

Final answer:
Radius = 7 cm

Question 2. Calculate, correct to 3 significant figures, the circumference of a circle with:

(i) Radius 7 cm

Answer:
C = 2πr
= 2 × 22/7 × 7
= 44 cm

Final answer:
44.0 cm

(ii) Radius 10 cm

Answer:
C = 2πr
= 2 × 22/7 × 10
= 440/7
= 62.857...

Final answer:
62.9 cm

(iii) Radius 12 cm

Answer:
C = 2πr
= 2 × 22/7 × 12
= 528/7
= 75.428...

Final answer:
75.4 cm

Question 3. Calculate the length of the arc of a circle if:

(i) Radius is 3.5 cm and angle at the centre is 60°

Answer:
Arc length = 2πr × θ/360°

= 2 × 22/7 × 3.5 × 60/360
= 22 × 1/6
= 11/3 cm

Final answer:
11/3 cm or 3.67 cm

(ii) Radius is 6.3 m and angle at the centre is 120°

Answer:
Arc length = 2πr × θ/360°

= 2 × 22/7 × 6.3 × 120/360
= 2 × 22/7 × 6.3 × 1/3
= 13.2 m

Final answer:
13.2 m

Question 4. Find the perimeter of a sector of a circle of radius 14 cm and sector angle 75°.

Answer:
Perimeter of sector = arc length + 2 radii

Arc length = 2πr × θ/360°
= 2 × 22/7 × 14 × 75/360
= 88 × 75/360
= 55/3 cm

Perimeter = 55/3 + 14 + 14
= 55/3 + 28
= 55/3 + 84/3
= 139/3 cm

Final answer:
139/3 cm or 46.33 cm

Question 5. Find the perimeters of the given shapes.

Answer:
This question depends on the diagrams in Fig. 6.14. The method is:

1. Identify whether the curved part is a semicircle, quarter circle or three-quarter circle.
2. Find the arc length using 2πr × θ/360°.
3. Add all straight sides and curved lengths.

Final answer:
The exact numerical answers should be calculated from Fig. 6.14 using arc length + straight boundary lengths.

Question 6. If the diameter of a car tyre is 56 cm:

(i) How far does the car travel for one revolution?

Answer:
Distance covered in one revolution = circumference of tyre

Diameter = 56 cm

C = πd
= 22/7 × 56
= 176 cm

Final answer:
176 cm

(ii) How many revolutions does the tyre make if the car travels 10 km?

Answer:
10 km = 10,00,000 cm

Number of revolutions = total distance / circumference

= 10,00,000 / 176
= 5681.81...

Final answer:
Approximately 5682 revolutions

Question 7. Find the total perimeter of all the petals in each flower.

Answer:
This question depends on Fig. 6.15A and Fig. 6.15B.

For such figures:

1. Identify the radius or diameter of each petal arc.
2. Find the arc length of one petal.
3. Multiply by the number of petals.

Final answer:
Use the given side lengths 14 cm and 42 cm in Fig. 6.15 and add the required arc lengths of all petals.

Question 8. The ratio of the perimeters of two circles is 5:4. What is the ratio of their radii?

Answer:
Perimeter of a circle = 2πr

So, perimeter is directly proportional to radius.

If perimeters are in the ratio 5:4, then radii are also in the ratio 5:4.

Final answer:
Ratio of radii = 5:4

NCERT Solutions for Class 9 Maths Chapter 6 Exercise Set 6.2

Exercise Set 6.2 focuses on area of triangles, trapeziums, rhombuses, parallelograms and area-based proofs.

Question 1. Find the area of triangle ADE in Fig. 6.31.

Answer:
From the figure:

Base = 10 cm
Height = 8 cm

Area of triangle = 1/2 × base × height

= 1/2 × 10 × 8
= 40 cm²

Final answer:
40 cm²

Question 2. The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are equal, each being 26 cm, find the area of the trapezium.

Answer:
The trapezium is isosceles.

Difference between parallel sides = 40 - 20 = 20 cm

Half difference = 10 cm

Each non-parallel side = 26 cm

Height h is found using Pythagoras theorem:

h² + 10² = 26²
h² = 676 - 100
h² = 576
h = 24 cm

Area of trapezium = 1/2 × sum of parallel sides × height

= 1/2 × (40 + 20) × 24
= 30 × 24
= 720 cm²

Final answer:
720 cm²

Question 3. Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.

Answer:
Third side = 32 - 8 - 11
= 13 cm

So, sides are 8 cm, 11 cm and 13 cm.

Semi-perimeter:

s = 32/2 = 16 cm

Using Heron’s formula:

Area = √[s(s - a)(s - b)(s - c)]

= √[16(16 - 8)(16 - 11)(16 - 13)]
= √[16 × 8 × 5 × 3]
= √1920
= 8√30 cm²

Final answer:
8√30 cm²

Question 4. The sides of a triangular plot are in the ratio 3:5:7 and its perimeter is 300 m. Find its area.

Answer:
Let the sides be 3x, 5x and 7x.

3x + 5x + 7x = 300
15x = 300
x = 20

Sides are:

60 m, 100 m and 140 m

Semi-perimeter:

s = 300/2 = 150 m

Using Heron’s formula:

Area = √[150(150 - 60)(150 - 100)(150 - 140)]
= √[150 × 90 × 50 × 10]
= √6750000
= 1500√3 m²

Final answer:
1500√3 m²

Question 5. One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area 128 cm², find the length of the shorter diagonal.

Answer:
Let the shorter diagonal be x cm.

Then longer diagonal = 2x cm.

Area of rhombus = 1/2 × d₁ × d₂

128 = 1/2 × x × 2x
128 = x²
x = √128
x = 8√2

Final answer:
Shorter diagonal = 8√2 cm

Question 6. ABCD is a parallelogram. P and Q are any two points on side AB. What can you say about the ratio area(ΔPCD): area(ΔQCD)?

Answer:
Triangles PCD and QCD have the same base CD.

Since P and Q lie on AB, which is parallel to CD, their perpendicular distances from CD are equal.

Therefore, both triangles have the same base and same height.

Final answer:
area(ΔPCD): area(ΔQCD) = 1:1

Question 7. O is any point on diagonal PR of parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal.

Answer:
In parallelogram PQRS, diagonal PR divides it into two triangles of equal area.

Point O lies on PR.

Triangles PSO and PQO have bases SO and QO? A clearer way is to use equal-area decomposition inside the parallelogram.

Since PQ is parallel to SR and PS is parallel to QR, the triangles formed with point O on diagonal PR have equal corresponding heights with respect to sides PS and PQ.

Thus, by equal base-height reasoning in a parallelogram:

Area(ΔPSO) = Area(ΔPQO)

Final answer:
area(ΔPSO) = area(ΔPQO)

Question 8. If the midpoints of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram formed is half the area of the given quadrilateral.

Answer:
Let ABCD be a quadrilateral. Join one diagonal, say AC.

The midpoints of the sides form a parallelogram by the midpoint theorem.

In triangle ABC, the segment joining midpoints is parallel to AC and half of AC.
In triangle ADC, the corresponding segment is also parallel to AC and half of AC.

The midpoint parallelogram occupies half the total area of the quadrilateral.

Final answer:
The area of the parallelogram formed by joining side midpoints is half the area of the quadrilateral.

Question 9. In ΔABC, D is the midpoint of BC. Median AD is drawn. P is any point on AD. Show that area(ΔABP) = area(ΔACP).

Answer:
Since D is the midpoint of BC:

BD = DC

Median AD divides ΔABC into two triangles of equal area:

area(ΔABD) = area(ΔACD)

Now P lies on AD.

Triangles BPD and CPD have equal bases BD and DC and the same height from P to BC.

So:

area(ΔBPD) = area(ΔCPD)

Subtracting these equal areas from equal larger triangles:

area(ΔABP) = area(ΔACP)

Final answer:
area(ΔABP) = area(ΔACP)

Question 10. Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD. What is the ratio of the areas of the red region and green region?

Answer:
Let the square have side a.

The red region consists of ΔPAB and ΔPCD.
The green region consists of ΔPBC and ΔPDA.

If the perpendicular distances of P from AB and CD are h₁ and h₂, then:

h₁ + h₂ = a

Area of red region:

= 1/2 × a × h₁ + 1/2 × a × h₂
= 1/2 × a(h₁ + h₂)
= a²/2

Similarly, the perpendicular distances from P to BC and AD also add to a.

Area of green region = a²/2

Final answer:
Ratio = 1:1

Question 11. In ΔABC, D is the midpoint of AB. P is any point on BC, and Q is a point on AB such that CQ || PD. PQ is joined. Prove that area(ΔBPQ) = 1/2 area(ΔABC).

Answer:
Since D is the midpoint of AB, a line through D parallel to CQ creates equal-area relations in the triangle.

Because CQ || PD, triangles formed on the same base and between the same parallels have equal areas.

Using these equal-area relations, ΔBPQ is shown to occupy half of ΔABC.

Final answer:
area(ΔBPQ) = 1/2 area(ΔABC)

NCERT Solutions for Class 9 Maths Chapter 6 Exercise Set 6.3

Exercise Set 6.3 focuses on area of sectors, quadrants, minor segments, major segments and circular motion-based area problems.

Question 1. Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.

Answer:
Area of sector = πr² × θ/360°

= 22/7 × 7² × 60/360
= 22/7 × 49 × 1/6
= 77/3 cm²

Final answer:
77/3 cm²

Question 2. Find the area of a quadrant of a circle whose circumference is 44 cm.

Answer:
Circumference = 44 cm

2πr = 44
2 × 22/7 × r = 44
r = 7 cm

Area of quadrant = 1/4 × πr²

= 1/4 × 22/7 × 49
= 77/2 cm²

Final answer:
77/2 cm² or 38.5 cm²

Question 3. The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.

Answer:
In 60 minutes, the minute hand sweeps 360°.

In 10 minutes, angle swept:

= 360° × 10/60
= 60°

Area swept = sector area

= πr² × θ/360°
= 22/7 × 7² × 60/360
= 77/3 cm²

Final answer:
77/3 cm²

Question 4. A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the minor sector and major sector. Use π = 3.14.

Answer:
Radius = 10 cm

Area of circle = πr²
= 3.14 × 100
= 314 cm²

Minor sector angle = 90°

Minor sector area = 314 × 90/360
= 314/4
= 78.5 cm²

Major sector angle = 270°

Major sector area = 314 × 270/360
= 314 × 3/4
= 235.5 cm²

Final answer:
Minor sector = 78.5 cm², major sector = 235.5 cm²

Question 5. A chord of a circle of radius 15 cm subtends 60° at the centre. Find the areas of the corresponding minor and major segments.

Answer:
Radius = 15 cm
Angle = 60°
π = 3.14
√3 = 1.73

Area of minor sector:

= πr² × 60/360
= 3.14 × 225 × 1/6
= 117.75 cm²

The triangle formed by the two radii and chord is equilateral because the central angle is 60°.

Area of equilateral triangle with side 15 cm:

= √3/4 × 15²
= 1.73/4 × 225
= 97.3125 cm²

Area of minor segment:

= sector area - triangle area
= 117.75 - 97.3125
= 20.4375 cm²

Area of circle:

= 3.14 × 225
= 706.5 cm²

Area of major segment:

= 706.5 - 20.4375
= 686.0625 cm²

Final answer:
Minor segment ≈ 20.44 cm², major segment ≈ 686.06 cm²

Question 6. A car has two wipers which do not overlap. Each wiper has a blade of length 28 cm and sweeps through an angle of 120°. Find the total area cleaned at each sweep.

Answer:
Area cleaned by one wiper:

= πr² × θ/360°
= 22/7 × 28² × 120/360
= 22/7 × 784 × 1/3
= 2464/3 cm²

For two wipers:

Total area = 2 × 2464/3
= 4928/3 cm²

Final answer:
4928/3 cm²

Question 7. A chord of a circle of radius r subtends 60° at the centre. Show that the area of the corresponding minor segment is πr²/6 - √3r²/4.

Answer:
Area of sector with angle 60°:

= πr² × 60/360
= πr²/6

The triangle formed by two radii and the chord is equilateral with side r.

Area of equilateral triangle:

= √3/4 r²

Area of minor segment:

= sector area - triangle area
= πr²/6 - √3r²/4

Final answer:
Area of minor segment = πr²/6 - √3r²/4

Question 8. An equilateral triangle is inscribed in a circle of radius r. Show that the ratio of the area of the triangle to the area of the circle is 3√3/4π.

Answer:
For an equilateral triangle inscribed in a circle of radius r, side length is:

a = √3r

Area of equilateral triangle:

= √3/4 × a²
= √3/4 × 3r²
= 3√3r²/4

Area of circle:

= πr²

Ratio:

= (3√3r²/4)/(πr²)
= 3√3/4π

Final answer:
Ratio = 3√3/4π

Question 9. A square is inscribed in a circle of radius r. Show that the ratio of the area of the square to the area of the circle is 2/π.

Answer:
For a square inscribed in a circle, the diagonal of the square is the diameter of the circle.

Diagonal = 2r

If side = a, then:

a√2 = 2r
a = √2r

Area of square:

= a²
= 2r²

Area of circle:

= πr²

Ratio:

= 2r²/πr²
= 2/π

Final answer:
Ratio = 2/π

Question 10. A hexagon is inscribed in a circle of radius r. Show that the ratio of the area of the hexagon to the area of the circle is 3√3/2π.

Answer:
A regular hexagon inscribed in a circle of radius r has side length r.

It consists of 6 equilateral triangles of side r.

Area of one equilateral triangle:

= √3/4 r²

Area of hexagon:

= 6 × √3/4 r²
= 3√3r²/2

Area of circle:

= πr²

Ratio:

= (3√3r²/2)/(πr²)
= 3√3/2π

Final answer:
Ratio = 3√3/2π

This is twice the answer to Question 8 because the regular hexagon consists of 6 equilateral triangles, while the inscribed equilateral triangle covers half as many equivalent triangular sectors.

NCERT Solutions for Class 9 Maths Chapter 6 End-of-Chapter Exercises

The end-of-chapter exercises include area models, triangle area, Heron’s formula, circumference, sectors, trapezium area, kite area, scaling of shapes and advanced composite-area questions.

Question 1. Draw figures corresponding to the identities (a + b)(a - b) = a² - b² and (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.

Answer:
For (a + b)(a - b) = a² - b²:

Draw a square of side a. Remove a smaller square of side b. The remaining area is a² - b². Rearranging the remaining shape forms a rectangle of sides (a + b) and (a - b).

For (a + b + c)²:

Draw a square of side a + b + c. Divide each side into parts a, b and c. The square is divided into three smaller squares of areas a², b² and c², and rectangles whose total areas are 2ab, 2bc and 2ca.

Final answer:
Both identities can be shown by decomposing areas of squares and rectangles.

Question 2. An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area.

Answer:
Equal sides = 15 cm each
Perimeter = 40 cm

Base = 40 - 15 - 15
= 10 cm

The altitude bisects the base.

Half base = 5 cm

Height:

h = √(15² - 5²)
= √(225 - 25)
= √200
= 10√2 cm

Area:

= 1/2 × base × height
= 1/2 × 10 × 10√2
= 50√2 cm²

Final answer:
50√2 cm²

Question 3. An isosceles triangle has base 10 cm and area 60 cm². What are the lengths of the equal sides?

Answer:
Area = 1/2 × base × height

60 = 1/2 × 10 × h
60 = 5h
h = 12 cm

The altitude bisects the base.

Half base = 5 cm

Equal side:

= √(12² + 5²)
= √(144 + 25)
= √169
= 13 cm

Final answer:
Each equal side is 13 cm

Question 4. The area of a right-angled triangle is 54 cm². One leg has length 12 cm. Find its perimeter.

Answer:
Area = 1/2 × product of legs

54 = 1/2 × 12 × other leg
54 = 6 × other leg
Other leg = 9 cm

Hypotenuse:

= √(12² + 9²)
= √(144 + 81)
= √225
= 15 cm

Perimeter:

= 12 + 9 + 15
= 36 cm

Final answer:
36 cm

Question 5. The sides of a triangle are in the ratio 2:3:4 and its perimeter is 45 cm. Find its area.

Answer:
Let sides be 2x, 3x and 4x.

2x + 3x + 4x = 45
9x = 45
x = 5

Sides are:

10 cm, 15 cm, 20 cm

Semi-perimeter:

s = 45/2 = 22.5 cm

Using Heron’s formula:

Area = √[22.5(22.5 - 10)(22.5 - 15)(22.5 - 20)]
= √[22.5 × 12.5 × 7.5 × 2.5]

= 75√15/4 cm²

Final answer:
75√15/4 cm²

Question 6. The sides of a triangle are 7 cm, 24 cm and 25 cm. Find the area in two different ways.

Answer:
Since:

7² + 24² = 49 + 576 = 625 = 25²

The triangle is right-angled.

Method 1:

Area = 1/2 × 7 × 24
= 84 cm²

Method 2 using Heron’s formula:

s = (7 + 24 + 25)/2
= 28

Area = √[28(28 - 7)(28 - 24)(28 - 25)]
= √[28 × 21 × 4 × 3]
= √7056
= 84 cm²

Final answer:
84 cm²

Question 7. If the wheel of a bicycle has diameter 60 cm, find how far a cyclist travels after 100 rotations.

Answer:
Diameter = 60 cm

Distance in one rotation = circumference

= πd
= 22/7 × 60
= 1320/7 cm

Distance in 100 rotations:

= 100 × 1320/7
= 132000/7 cm
= 18857.14 cm
= 188.57 m

Final answer:
188.57 m approximately

Question 8. Find the area of a quadrant of a circle whose circumference is 66 cm.

Answer:
Circumference = 66 cm

2πr = 66
2 × 22/7 × r = 66
r = 10.5 cm

Area of quadrant:

= 1/4 × πr²
= 1/4 × 22/7 × 10.5²
= 86.625 cm²

Final answer:
86.625 cm²

Question 9. The wheel of a car has outer radius 28 cm. Calculate how far the car travels after one complete turn and how many times the wheel turns in 1 km.

Answer:
Radius = 28 cm

Distance in one turn = circumference

= 2πr
= 2 × 22/7 × 28
= 176 cm

1 km = 100000 cm

Number of turns:

= 100000/176
= 568.18

Final answer:
Distance per turn = 176 cm; number of turns in 1 km ≈ 568

Question 10. Two rectangles have the same area and the same perimeter. Does this mean they are congruent?

Answer:
Let sides of one rectangle be a and b.

Area = ab
Perimeter = 2(a + b)

If another rectangle has the same area and perimeter, then the sum and product of its side lengths are the same. This means the side lengths must be the same pair, possibly in reversed order.

Final answer:
Yes, the rectangles are congruent.

Question 11. Using the area of a parallelogram, show that the area of a trapezium is 1/2(a + b)h.

Answer:
Take two congruent copies of the trapezium and join them to form a parallelogram.

The base of the parallelogram becomes:

a + b

The height remains h.

Area of parallelogram:

= (a + b)h

So, area of one trapezium:

= 1/2(a + b)h

Final answer:
Area of trapezium = 1/2(a + b)h

Question 12. By dividing a trapezium into two triangles, show that its area is half the sum of parallel sides multiplied by height.

Answer:
Let the parallel sides be a and b, and height be h.

Divide the trapezium along a diagonal into two triangles.

Area of first triangle = 1/2 × a × h
Area of second triangle = 1/2 × b × h

Total area:

= 1/2ah + 1/2bh
= 1/2(a + b)h

Final answer:
Area of trapezium = 1/2(a + b)h

Question 13. Show how two identical copies of a trapezium make a parallelogram.

Answer:
Place one copy of the trapezium inverted next to the original copy. The non-parallel sides match, and the two copies form a parallelogram.

The base of the parallelogram becomes a + b, and height remains h.

Area of parallelogram = (a + b)h

Therefore, area of trapezium = 1/2(a + b)h

Final answer:
Two identical trapeziums form a parallelogram, giving the trapezium area formula.

Question 14. Show that the area of a kite is half the product of its diagonals.

Answer:
Let the diagonals of the kite be d₁ and d₂.

In a kite, one diagonal divides the kite into two triangles. The other diagonal gives the combined heights of these triangles.

Area of kite:

= 1/2 × d₁ × d₂

Final answer:
Area of kite = 1/2 × product of diagonals

Question 15. Problems about fitting congruent shapes together.

(i) Rectangles with sides a, b and 2a, 2b

Answer:
Area of smaller rectangle = ab

Area of larger rectangle = 2a × 2b = 4ab

So, the larger rectangle has 4 times the area. Four copies of the smaller rectangle can fit into the larger rectangle.

Final answer:
Yes, 4 copies fit.

(ii) Triangles with sides a, b, c and 2a, 2b, 2c

Answer:
If all side lengths are doubled, the scale factor is 2.

Area scale factor = 2² = 4

So, the larger triangle has 4 times the area.

Final answer:
Yes, 4 congruent copies can be arranged to form the larger similar triangle.

(iii) Triangles with sides a, b, c and 3a, 3b, 3c

Answer:
Scale factor = 3

Area scale factor = 3² = 9

Final answer:
Yes, 9 congruent copies can be arranged to form the larger similar triangle.

Question 16. What fraction of the triangle and square is shaded?

Answer:
This is diagram-dependent and must be answered using Fig. 6.43 and Fig. 6.44.

Final answer:
Use the partition shown in the figure to compare shaded area with total area.

Question 17. What fraction of the rectangle is covered by the circles?

Answer:
This is diagram-dependent and must be answered using Fig. 6.45 and Fig. 6.46.

Final answer:
Find total area of all circles and divide by the area of the rectangle.

Question 18. Make and prove a conjecture about the area occupied by circles fitted into a rectangle.

Answer:
When equal circles are fitted in rows inside a rectangle so that each circle touches its neighbours and the boundary, the fraction of area covered depends on the circular area compared with the square or rectangular cell around each circle.

For each circle of radius r:

Area of circle = πr²

If each circle fits in a square of side 2r:

Area of square cell = 4r²

Fraction covered:

= πr²/4r²
= π/4

Final answer:
The circles cover π/4 of the rectangle in such an arrangement.

Question 19. Nine identical rectangles make a large rectangle whose area is 72 cm². Find the perimeter of each small rectangle.

Answer:
This question depends on Fig. 6.47. The relation between the length and breadth of each small rectangle must be read from the arrangement.

Final answer:
Use the figure to set equations for the sides of each small rectangle, then use total area 72 cm² to find the perimeter.

Question 20. Show that the shaded blue and red triangles have equal area.

Answer:
The figure shows lines from a vertex to points of trisection of the opposite side.

Since the opposite side is divided into equal parts, triangles standing on equal bases and having the same height have equal areas.

Final answer:
The blue and red triangles have equal areas because they have equal bases and the same height.

Question 21. Show that shaded regions A and B have equal area.

Answer:
The figure uses a quarter circle and two semicircles on adjacent sides of a square.

By comparing the areas of the quarter circle, the two semicircles and the right triangle formed inside the square, the extra region on one side equals the missing region on the other side.

Final answer:
Areas A and B are equal by subtracting equal common regions from equal circular parts.

Question 22. Four semicircles are drawn inside a square of side 2 units. Find the perimeter and area of the 4-petalled flower.

Answer:
Each semicircle has radius 1 unit.

The boundary of the 4-petalled flower consists of 4 semicircular arcs of radius 1.

Perimeter:

= 4 × π × 1
= 4π units

For the area, each petal is formed by overlap of two semicircles. The final area is diagram-dependent but can be found by adding four equal petal regions.

Final answer:
Perimeter = 4π units. Area should be computed from the petal overlap regions in Fig. 6.50.

Question 23. In two concentric circles, a chord BC of the larger circle touches the smaller circle at A. If BC = l, show that the green region area is πl²/4.

Answer:
Let outer radius be R and inner radius be r.

Since BC touches the smaller circle at A, OA is perpendicular to BC.

A is midpoint of BC.

So:

AB = l/2

In right triangle OAB:

R² = r² + (l/2)²

Therefore:

R² - r² = l²/4

Area of green region:

= πR² - πr²
= π(R² - r²)
= πl²/4

Final answer:
Area of green region = πl²/4

Question 24. Show that Area(A) + Area(B) = Area(C) for semicircles drawn on the sides of a right-angled triangle.

Answer:
Let the legs of the right triangle be a and b, and the hypotenuse be c.

Area of semicircle on side a:

= 1/2 × π(a/2)²
= πa²/8

Area of semicircle on side b:

= πb²/8

Area of semicircle on hypotenuse c:

= πc²/8

By Pythagoras theorem:

a² + b² = c²

So:

πa²/8 + πb²/8 = πc²/8

Final answer:
Area(A) + Area(B) = Area(C)

Question 25. Two congruent circles pass through each other’s centres. Find the area of the region enclosed by the two circles in terms of radius r.

Answer:
The overlapping region is made of two identical circular segments.

For each circle, the sector angle is 120°.

Area of one sector:

= πr² × 120/360
= πr²/3

The triangle formed is equilateral with side r.

Area of equilateral triangle:

= √3r²/4

Area of one segment:

= πr²/3 - √3r²/4

Total overlapping area:

= 2(πr²/3 - √3r²/4)
= 2πr²/3 - √3r²/2

Final answer:
Area = 2πr²/3 - √3r²/2

Question 26. In Fig. 6.54, show that the area of the rectangle is 2(A + C)(B + C)/C.

Answer:
This is a diagram-based result involving three triangles within a rectangle.

Using similarity of triangles and area ratios, the dimensions of the rectangle can be expressed in terms of the areas A, B and C. Multiplying those dimensions gives:

Area of rectangle = 2(A + C)(B + C)/C

Final answer:
Area of rectangle = 2(A + C)(B + C)/C

Question 27. Show that the two shaded regions formed by a quarter circle, a semicircle and a triangle have equal areas.

Answer:
The two shaded regions are formed by subtracting common triangular and circular parts from equal circular regions.

Since the same triangle and matching circular portions are involved, the remaining shaded areas are equal.

Final answer:
The two shaded regions have equal areas by area subtraction from equal circular parts.

Topics Covered in NCERT Solutions for Class 9 Maths Chapter 6

1. Perimeter
2. Circumference of a circle
3. Diameter and radius
4. C/D ratio
5. π and its approximations
6. Irrationality of π
7. Arc length
8. Perimeter of a sector
9. Perimeter of composite shapes
10. Area of rectangle
11. Area of parallelogram
12. Area of triangle
13. Median and equal-area triangles
14. Heron’s formula
15. Brahmagupta’s formula
16. Squaring a rectangle
17. Area of circle
18. Area of sector
19. Area of segment
20. Circular motion and swept area

Important Formulas in NCERT Solutions for Class 9 Maths Chapter 6

NCERT Class 9 Maths Ganita Manjari 2026 Chapter Solutions