📖 About Class 9 Maths Chapter 6 End of Chapter Exercises
Class 9 Maths Chapter 6 End of Chapter Exercises Solutions cover the complete End of Chapter Exercises from Chapter 6 – Measuring Space: Perimeter and Area in the latest NCERT Ganita Manjari (2026–27) textbook. This page provides complete, reliable and easy-to-follow NCERT Class 9 Maths solutions for students searching for Class 9 Maths Chapter 6 Solutions, Class 9 Maths End of Chapter Exercises Solutions and Class 9 Maths Chapter 6 answers in one place.
The Class 9 Maths Chapter 6 End of Chapter Exercises Solutions are arranged question-wise in a clear, student-friendly CBSE answer-writing format. The complete set contains 27 questions covering important concepts from the chapter, including algebraic identities through area models, areas and perimeters of triangles, Heron’s formula, circumference and area of circles, quadrants and semicircles, trapeziums, kites, rectangles, congruent figures, area ratios, shaded regions and challenging circle and geometry area problems. These solutions are useful for homework, classroom practice, self-study, revision and CBSE examination preparation.
📚 Learn Before You Solve
✨ Class 9 Maths Chapter 6 End of Chapter Exercises Solutions — पहले concept समझें, फिर solve करें
Before solving
Class 9 Maths Chapter 6 End of Chapter Exercises Solutions,
first revise the key ideas needed across
perimeter, area, triangles, quadrilaterals, circles,
sectors, segments and composite figures.
This
Learn Before You Solve section connects the
End of Chapter Exercises with the concepts already developed in
Exercise Sets 6.1, 6.2 and 6.3.
You do not need to relearn every formula here —
simply open the relevant concept when you need a deeper explanation.
🔷 Block 1 — Area Models & Algebraic Identities
Use areas of simple figures to understand and apply algebraic identities
In some End-of-Chapter questions, an algebraic expression can be
understood more easily through the
area of a square or rectangle.
Instead of treating an identity as a rule to memorise,
think of it as a relationship between different parts of an area.
📐 The Area Connection
If a square has side a + b, its total area is
represented by the product of its side lengths.
Dividing the square into smaller rectangles and squares
gives the same area in another form.
This geometric idea leads naturally to the identity:
Important Identity
(a + b)² = a² + 2ab + b²
Another Useful Area Identity
(a + b)(a − b) = a² − b²
🧠 When Will This Help?
Use the area interpretation when a question connects lengths, areas, squares, rectangles or algebraic expressions. It can help you recognise the correct identity before doing the calculation.
📚 Need the Full Concept Explanation?
See the detailed area-model explanation, diagrams and step-by-step identity development before returning to the End-of-Chapter question.
Open Detailed Concept →
🎯 Quick Exam Tip
Do not expand blindly. First identify the pattern of the expression and then choose the appropriate identity.
🔺 Block 2 — Triangle Area & Heron’s Formula
Choose the correct triangle-area method from the information given
Triangle-area questions in the
End-of-Chapter Exercises may give a
base and perpendicular height, or they may give
all three sides.
The important skill is to recognise which formula matches the
information provided.
📐 When Base and Height Are Known
If the base and its corresponding perpendicular height are known, use half of the product of base and height.
Area of a Triangle
Area = 1/2 × base × perpendicular height
The height must be perpendicular to the chosen base
🧮 When All Three Sides Are Known — Heron’s Formula
When the three sides of a triangle are known but the perpendicular height is not given, Heron’s Formula gives the area directly. First find the semi-perimeter.
Semi-perimeter = (a + b + c) / 2
Area = √[s(s − a)(s − b)(s − c)]
Area = √[s(s − a)(s − b)(s − c)]
🧠 Which Formula Should You Use?
Base + perpendicular height given:
use Area = 1/2 × base × height.
Three sides given:
find the semi-perimeter first, then use Heron’s Formula.
Side ratio + perimeter given:
first convert the ratio into the actual side lengths,
then choose the appropriate area method.
🎯 Quick Exam Tip
Before substituting values, identify exactly what information the question gives. Do not use Heron’s Formula when the perpendicular height is already available.
🔵 Block 3 — Perimeter, Circumference & Circular Motion
Connect circular boundary, circumference and wheel revolutions
For a circle, the boundary is its circumference.
The same idea connects directly to wheels:
one complete revolution covers exactly one circumference.
This concept is used in Q7 and Q9, and also supports
the circle-ratio reasoning in Q8.
⭕ Circumference — The Boundary of a Circle
If the radius is r and the diameter is d, the length of the complete circular boundary is called the circumference.
C = 2πr
C = πd
Radius and diameter determine the circumference
🛞 One Revolution = One Circumference
Imagine a wheel rolling without slipping. In one complete revolution, every point on the outer edge travels through exactly one circumference. Therefore, if a wheel makes n complete revolutions, the total distance travelled is n × circumference.
Distance = n × C
Number of revolutions = Distance ÷ Circumference
Number of revolutions = Distance ÷ Circumference
🧠 Connect the Three Ideas
Circle:
Its complete boundary is the circumference.
Wheel:
One complete revolution covers one circumference.
Repeated revolutions:
Multiply the circumference by the number of revolutions.
Known distance:
Divide the distance by the circumference to find the number
of revolutions.
🎯 Quick Exam Tip
Before solving a wheel question, first find the circumference. Then decide whether the question asks for distance or number of revolutions. Also check whether the given measurement is the radius or diameter.
🟠 Block 4 — Circle Area, Sectors, Segments & Circular Regions
Identify the circular region first, then choose the correct area formula
In Class 9 Maths Chapter 6, questions involving
area of a circle, sectors, segments and other circular regions
require you to identify the exact part of the circle shown in the figure.
Once the region is recognised, the appropriate formula can be applied
directly and the calculation becomes much easier.
📐 First Identify the Circular Region
Before calculating area, look carefully at the figure. A complete circle, quadrant, sector and segment are different regions, so the formula used depends on the region asked for.
Circle
Complete circular region
Quadrant
One-fourth of a circle
Sector
Region between two radii and an arc
Segment
Region between a chord and its arc
📌 Important Area Formulas
Area of a Circle
A = πr²
Area of a Sector
A =
θ
360°
× πr²
Area of a Quadrant
A =
1
4
× πr²
🟣 Sector and Segment — Know the Difference
A sector is the region enclosed by two radii and the corresponding arc. A segment is the region enclosed by a chord and its corresponding arc.
Area of Segment = Area of Sector − Area of Triangle
🧠 Which Formula Should You Use?
Complete circle:
use A = πr².
Quadrant:
use one-fourth of the area of the circle.
Sector:
use the angle at the centre to find the required fraction of the
circle.
Segment:
subtract the area of the triangle from the area of the sector.
📚 Explore the Detailed Concepts
Area of Circle →
A = πr² →
Area of Sector →
Chords, Sectors & Segments →
🎯 Quick Exam Tip
Do not start calculating immediately. First identify the region shown in the figure and write the corresponding formula. This prevents confusion between the area of a circle, sector, quadrant and segment.
🔶 Block 5 — Trapezium, Kite & Area Relationships
Understand how shapes can be divided, rearranged and compared to find area
In Class 9 Maths Chapter 6 End-of-Chapter Exercises,
several area questions can be solved by recognising the relationship
between familiar shapes. A trapezium can be understood through its
parallel sides and perpendicular height, while a kite can be divided
into triangles using its diagonals.
🔶 Area of a Trapezium
For a trapezium, identify the two parallel sides and the perpendicular distance between them. Let the parallel sides be a and b, and the perpendicular height be h.
Formula for Area of a Trapezium
Area =
1
2
× (a + b) × h
Two parallel sides and the perpendicular height
🪁 Area of a Kite
A kite can be divided into triangles using its diagonals. If the lengths of the two diagonals are d₁ and d₂, its area is half the product of the diagonals.
Formula for Area of a Kite
Area =
1
2
× d₁ × d₂
The diagonals divide the kite into triangular regions
🧠 Area Relationships Can Save Calculation
Same base + same perpendicular height:
the two triangles have equal areas.
Same base + different heights:
their areas are in the same ratio as their perpendicular heights.
Same perpendicular height + different bases:
their areas are in the same ratio as their bases.
Rearrangement:
dividing a figure into simpler parts can make its area easier
to determine.
🔎 How to Approach Area Questions
First identify the shape and the measurements given. If the figure is complicated, look for triangles, parallel sides, diagonals or smaller familiar shapes. Then choose the formula that matches the information available.
🎯 Quick Exam Tip
Do not rush to substitute numbers. First identify the shape, the required dimensions and any useful area relationship. A complicated figure can often be reduced to two or more familiar shapes.
🔷 Block 6 — Composite Shapes & Advanced Area Reasoning
Learn how to break complex figures into familiar shapes and find the required area or boundary
In Class 9 Maths Chapter 6 End-of-Chapter Exercises,
some questions combine two or more familiar shapes. To solve these
composite area problems, trace the required region
carefully, divide the figure into simpler parts, and use addition,
subtraction, symmetry or equal-area relationships wherever useful.
🧠 The Main Idea: Break the Figure into Familiar Parts
A complicated figure does not always require a new formula. Look for rectangles, triangles, circles, semicircles, quadrants and other familiar regions. Find the area of each useful part and combine the results according to the question.
Required Area
Required Area = Total Area − Unwanted Area
A complex region can be understood by separating its parts
📏 When the Question Asks for Perimeter
Trace the required boundary first.
Do not automatically add every side or curve visible inside
the figure.
For a composite boundary:
add only the lengths of the boundary parts that actually enclose
the required region.
For circular parts:
identify whether the boundary contains a full circle,
semicircle, quadrant or another arc before calculating its length.
Composite Perimeter
Composite Perimeter
= Sum of the required boundary parts
⚖️ Use Symmetry and Equal Areas
Some figures contain repeated or symmetrical regions. Instead of calculating every small region separately, identify equal parts and calculate one part first. Then multiply by the number of equal regions. This idea is especially useful in questions involving equal-area triangles, repeated circular regions and symmetrical arrangements.
🌸 Petals, Semicircles and Overlapping Circular Regions
In figures made from circles or parts of circles, first identify the radius or diameter and the type of circular region involved. A petal may be formed by overlapping arcs, while a shaded region may require the area of one circular part to be subtracted from another.
Think in Parts
Required Area
= Area Added − Area Removed
🔍 Where This Strategy Is Useful
This approach supports the more involved questions of the End-of-Chapter Exercises, including arrangements of rectangles, equal-area triangles, quadrants with semicircles, four-petalled figures, concentric circles, semicircles on triangle sides, overlapping circles and combinations of rectangles, triangles and circular regions.
📚 Go Deeper Into These Concepts
Composite Curved Shapes →
Chords, Sectors & Segments →
Inscribed Figures →
Equal-Area Reasoning →
Area-Ratio Reasoning →
🎯 Quick Exam Tip
Before calculating, mark the exact region asked for. Then decide whether you should add areas, subtract an unwanted area, use equal areas, or use a circle relationship. A clear breakdown of the figure usually makes the calculation much easier.
⚡ Quick Revision Dashboard
Class 9 Maths Chapter 6 — Important Concepts, Formulas & Problem-Solving Ideas
Before going to the Class 9 Maths Chapter 6 End of Chapter Exercises Solutions, quickly revise
the important formulas and concepts of
Class 9 Maths Chapter 6.
Identify the shape, choose the correct formula, and then
substitute the given values carefully.
🔵 Circle Basics & π
π — Circumference to Diameter Ratio
π is the constant ratio of the circumference of a
circle to its diameter.
π = Circumference ÷ Diameter
π — Common Approximate Values
In calculations, commonly used values are:
π ≈
22
7
or 3.14
π Is an Irrational Number
π cannot be expressed exactly as a terminating or
recurring decimal.
Archimedes’ Estimate of π
3
10
71
< π <
3
1
7
Chongzhi’s Estimate
π ≈
355
113
Aryabhata’s Estimate
π ≈ 3.1416
Madhava’s Exact Formula
π = 4
(
1 −
1
3
+
1
5
−
1
7
+ ⋯
)
📐 Circle, Circumference & Arc Length
Circumference of a Circle
C = 2πr = πd
r = radius | d = diameter
Arc Length
l = 2πr ×
θ°
360°
θ is the central angle.
Wheel & Repeated Distance
Distance = Number of revolutions × Circumference
Number of Revolutions
Number of revolutions =
Distance
Circumference
📊 Area Formulas
Area of a Triangle
Area =
1
2
× base × perpendicular height
Heron’s Formula
Area = √[s(s − a)(s − b)(s − c)]
where
s =
a + b + c
2
Area of a Circle
A = πr²
Area of a Sector
Area = πr² ×
θ°
360°
θ is the central angle.
Area of a Quadrant
Area =
1
4
πr²
Area of a Trapezium
Area =
1
2
× (a + b) × h
Area of a Kite
Area =
1
2
× d₁ × d₂
Brahmagupta’s Formula
Area = √[(s − a)(s − b)(s − c)(s − d)]
for a cyclic quadrilateral, where
s = a + b + c + d 2
s = a + b + c + d 2
🟣 Circular Regions & Segments
Area of a Segment
Segment Area =
Sector Area − Triangle Area
Region Between Two Concentric Circles
Required Area = π(R² − r²)
Quadrant
A quadrant is one-fourth of a circle.
Its central angle is 90°.
Sector
A sector is the region bounded by two radii
and the corresponding arc.
🧩 Composite Areas, Ratios & Scaling
Composite Area
Required Area =
Total Area − Unwanted Area
Composite Perimeter
Perimeter =
Sum of required boundary parts
Area Ratio
Compare corresponding areas using the given
lengths, sides, bases, heights or radii.
Scaling of Area
If every length is multiplied by k,
the area is multiplied by k².
🎯 Remember Before You Solve
Identify the shape → choose the formula →
substitute carefully → write the correct square units
🧠 Composite-Figure Strategy
For a complicated figure, trace the required region first. Divide it into familiar shapes such as triangles, rectangles, trapeziums, sectors and semicircles. Then add the required areas or subtract the unwanted regions. For circular figures, identify the radius, diameter, central angle and arc before applying a formula.
Chapter 6 • End-of-Chapter Exercises • Question 1
Class 9 Maths Chapter 6 End of Chapter Exercises Question 1 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
Identities in algebra can sometimes be shown as area
relationships. For example, the figure shown corresponds to
the identity
(a + b)² = a² + 2ab + b².
Draw figures corresponding to the identities (a + b)(a − b) = a² − b² and (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.
Draw figures corresponding to the identities (a + b)(a − b) = a² − b² and (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.
Solution
(i) Identity:
(a + b)(a − b) = a² − b²
Draw a square of side a. Therefore, its area is a².
From this square, remove a smaller square of side b.
Area model showing how a² − b² can be rearranged into a rectangle
Area of the original square = a².
Area of the removed square = b².
Therefore, area of the remaining region is
a² − b²
The two remaining parts can be rearranged to form a rectangle whose dimensions are (a + b) and (a − b).
Hence, the area of the rearranged rectangle is
(a + b)(a − b)
Since both figures represent the same area,
(a + b)(a − b) = a² − b²
(ii) Identity:
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Draw a large square whose side is (a + b + c).
Divide each side into three parts of lengths a, b and c.
Area model for (a + b + c)²
The large square is divided into nine smaller regions:
- one square of area a²
- one square of area b²
- one square of area c²
- two rectangles of area ab
- two rectangles of area bc
- two rectangles of area ca
Therefore, the total area of the large square is
(a + b + c)²
= a² + b² + c² + ab + ab + bc + bc + ca + ca
Combining the equal pairs,
(a + b + c)²
= a² + b² + c² + 2ab + 2bc + 2ca
∴ The required area figures represent the identities:
(a + b)(a − b) = a² − b²
and
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
(a + b)(a − b) = a² − b²
and
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
End-of-Chapter Exercises • Question 2
Class 9 Maths Chapter 6 End-of-Chapter Exercises Question 2 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
An isosceles triangle has perimeter 40 cm; the equal sides are
15 cm each. Find the area of the triangle.
Isosceles triangle ABC with height AD
Given
Perimeter of the isosceles triangle = 40 cm
Equal sides = 15 cm each
To Find: Area of the triangle
Solution
Let the base of the isosceles triangle be BC.
Since the perimeter is 40 cm,
BC + 15 + 15 = 40
Therefore,
BC = 40 − 30 = 10 cm
Hence, the three sides are
15 cm, 15 cm and 10 cm.
Since the perimeter is 40 cm,
BC + 15 + 15 = 40
Therefore,
BC = 40 − 30 = 10 cm
Hence, the three sides are
15 cm, 15 cm and 10 cm.
Method 1 — Using Pythagoras’ Theorem
Draw the perpendicular AD from A to BC.
Since ABC is an isosceles triangle,
AB = AC
Therefore, AD bisects the base BC.
Hence,
BD = DC = 10 ÷ 2 = 5 cm
In right-angled triangle ABD,
By Pythagoras’ theorem,
AB² = AD² + BD²
Therefore,
AD² = AB² − BD²
= 15² − 5²
= 225 − 25
= 200
Therefore,
AD = √200 = 10√2 cm
Now,
Area of triangle ABC
= 1 2 × base × height
= 1 2 × 10 × 10√2
= 50√2 cm²
Since ABC is an isosceles triangle,
AB = AC
Therefore, AD bisects the base BC.
Hence,
BD = DC = 10 ÷ 2 = 5 cm
In right-angled triangle ABD,
By Pythagoras’ theorem,
AB² = AD² + BD²
Therefore,
AD² = AB² − BD²
= 15² − 5²
= 225 − 25
= 200
Therefore,
AD = √200 = 10√2 cm
Now,
Area of triangle ABC
= 1 2 × base × height
= 1 2 × 10 × 10√2
= 50√2 cm²
Method 2 — Using Heron’s Formula for an Isosceles Triangle
For an isosceles triangle with equal sides a and
base b,
Area = b 4 √(4a² − b²)
Here,
a = 15 cm, b = 10 cm
Therefore,
Area = 10 4 √(4 × 15² − 10²)
= 10 4 √(4 × 225 − 100)
= 10 4 √(900 − 100)
= 10 4 √800
= 10 4 × 20√2
= 50√2 cm²
Area = b 4 √(4a² − b²)
Here,
a = 15 cm, b = 10 cm
Therefore,
Area = 10 4 √(4 × 15² − 10²)
= 10 4 √(4 × 225 − 100)
= 10 4 √(900 − 100)
= 10 4 √800
= 10 4 × 20√2
= 50√2 cm²
Method 3 — Using Heron’s Formula
The three sides of the triangle are
a = 15 cm, b = 15 cm, c = 10 cm
Semi-perimeter,
s = 15 + 15 + 10 2
= 40 2
= 20 cm
By Heron’s formula,
Area = √[s(s − a)(s − b)(s − c)]
= √[20(20 − 15)(20 − 15)(20 − 10)]
= √(20 × 5 × 5 × 10)
= √5000
= √(2500 × 2)
= 50√2 cm²
a = 15 cm, b = 15 cm, c = 10 cm
Semi-perimeter,
s = 15 + 15 + 10 2
= 40 2
= 20 cm
By Heron’s formula,
Area = √[s(s − a)(s − b)(s − c)]
= √[20(20 − 15)(20 − 15)(20 − 10)]
= √(20 × 5 × 5 × 10)
= √5000
= √(2500 × 2)
= 50√2 cm²
∴ The area of the isosceles triangle is
50√2 cm²
.
End-of-Chapter Exercises • Question 3
Class 9 Maths Chapter 6 End-of-Chapter Exercises Question 3 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
An isosceles triangle has base 10 cm, and its area is 60 cm².
What are the lengths of the equal sides?
Isosceles triangle ABC with perpendicular height AD
Solution
Given
Base of the isosceles triangle = 10 cm
Area of the triangle = 60 cm²
To Find: Lengths of the equal sides
Step 1 — Find the height of the triangle
Let the perpendicular height be AD = h.
We know,
Area of triangle = 1 2 × base × height
Therefore,
60 = 1 2 × 10 × h
60 = 5h
Therefore,
h = 12 cm
Step 2 — Use the property of an isosceles triangle
Draw the perpendicular AD from A to BC.
Since ABC is an isosceles triangle,
AB = AC.
Therefore, the perpendicular from the vertex A to the base BC bisects the base.
Hence,
BD = DC = 10 2 = 5 cm
Step 3 — Find the equal side
In right-angled triangle ABD,
By Pythagoras’ theorem,
AB² = AD² + BD²
Substituting the values,
AB² = 12² + 5²
= 144 + 25
= 169
Therefore,
AB = √169 = 13 cm
Since AB = AC,
AC = 13 cm
∴ The lengths of the equal sides are
13 cm each.
End-of-Chapter Exercises • Question 3
Class 9 Maths Chapter 6 End-of-Chapter Exercises Question 3 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
An isosceles triangle has base 10 cm, and its area is 60 cm².
What are the lengths of the equal sides?
Isosceles triangle ABC with perpendicular height AD
Solution
Given
Base of the isosceles triangle = 10 cm
Area of the triangle = 60 cm²
To Find: Lengths of the equal sides
Step 1 — Find the height of the triangle
Let the perpendicular height be AD = h.
We know,
Area of triangle = 1 2 × base × height
Therefore,
60 = 1 2 × 10 × h
60 = 5h
Therefore,
h = 12 cm
Step 2 — Use the property of an isosceles triangle
Draw the perpendicular AD from A to BC.
Since ABC is an isosceles triangle,
AB = AC.
Therefore, the perpendicular from the vertex A to the base BC bisects the base.
Hence,
BD = DC = 10 2 = 5 cm
Step 3 — Find the equal side
In right-angled triangle ABD,
By Pythagoras’ theorem,
AB² = AD² + BD²
Substituting the values,
AB² = 12² + 5²
= 144 + 25
= 169
Therefore,
AB = √169 = 13 cm
Since AB = AC,
AC = 13 cm
∴ The lengths of the equal sides are
13 cm each.
End-of-Chapter Exercises • Question 6
Class 9 Maths Chapter 6 End-of-Chapter Exercises Question 6 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
The sides of a triangle have lengths 7 cm, 24 cm, 25 cm.
Find the area of the triangle in two different ways.
Triangle ABC with sides 7 cm, 24 cm and 25 cm
Solution
Given
The sides of the triangle are
7 cm, 24 cm and 25 cm
To Find: Area of the triangle by two different methods
Method 1 — Using base × height
First, check whether the triangle is right-angled.
Taking 25 cm as the largest side,
7² + 24² = 49 + 576 = 625
and
25² = 625
Therefore,
7² + 24² = 25²
Hence, by the converse of Pythagoras’ theorem, the triangle is right-angled.
So, the two perpendicular sides are 7 cm and 24 cm.
Therefore,
Area = 1 2 × base × height
= 1 2 × 24 × 7
= 12 × 7
= 84 cm²
Method 2 — Using Heron’s Formula
Let the three sides be
a = 7 cm, b = 24 cm, c = 25 cm
Semi-perimeter,
s = a + b + c 2
= 7 + 24 + 25 2 = 28 cm
By Heron’s formula,
Area = √[s(s − a)(s − b)(s − c)]
= √[28(28 − 7)(28 − 24)(28 − 25)]
= √[28 × 21 × 4 × 3]
= √7056
= 84 cm²
∴ The area of the triangle is
84 cm².
Both methods give the same area: 84 cm².
Both methods give the same area: 84 cm².
End-of-Chapter Exercises • Question 7
Class 9 Maths Chapter 6 End-of-Chapter Exercises Question 7 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
Question 7
If the wheel of a bicycle has a diameter of 60 cm, find how far a cyclist will have travelled after the wheel has rotated 100 times.
If the wheel of a bicycle has a diameter of 60 cm, find how far a cyclist will have travelled after the wheel has rotated 100 times.
Bicycle wheel of diameter 60 cm
Given
Diameter of the bicycle wheel = 60 cm
Number of rotations = 100
To Find: Distance travelled by the cyclist
Solution
One complete rotation of the wheel covers a distance
equal to the circumference of the wheel.
We know,
Circumference of a circle = π × diameter
Here,
Diameter = 60 cm
Therefore,
Circumference = π × 60
Using π = 22 7 ,
Circumference = 22 7 × 60
= 1320 7 cm
= 1320 7 cm
= Distance covered in one rotation × Number of rotations
= 1320 7 × 100 cm
= 132000 7 cm
Converting centimetres into metres,
Distance travelled
= 132000 7 ÷ 100 m
= 1320 7 m
= 188 4 7 m
We know,
Circumference of a circle = π × diameter
Here,
Diameter = 60 cm
Therefore,
Circumference = π × 60
Using π = 22 7 ,
Circumference = 22 7 × 60
= 1320 7 cm
Step 1 — Distance covered in one rotation
Distance covered in one rotation
= 1320 7 cm
Step 2 — Distance covered in 100 rotations
Distance travelled
= Distance covered in one rotation × Number of rotations
= 1320 7 × 100 cm
= 132000 7 cm
Converting centimetres into metres,
Distance travelled
= 132000 7 ÷ 100 m
= 1320 7 m
= 188 4 7 m
∴ The cyclist will have travelled
188
4
7
m
after 100 rotations.
End-of-Chapter Exercises • Question 8
Class 9 Maths Chapter 6 End-of-Chapter Exercises Question 8 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
Find the area of a quadrant of a circle whose circumference is
66 cm.
Quadrant of a circle with radius r
Given
Circumference of the circle = 66 cm
For π, use 22 7
To Find: Area of the quadrant
Solution
We know that the circumference of a circle is
Circumference = 2πr
Therefore,
2πr = 66
Substituting 22 7 for π,
2 × 22 7 × r = 66
Therefore,
r = 66 × 7 44
r = 10.5 cm
r = 10.5 cm
Therefore,
Area of quadrant = 1 4 πr²
= 1 4 × 22 7 × (10.5)²
= 1 4 × 22 7 × 110.25
= 346.5 4
= 86.625 cm²
Circumference = 2πr
Therefore,
2πr = 66
Substituting 22 7 for π,
2 × 22 7 × r = 66
Therefore,
r = 66 × 7 44
r = 10.5 cm
Step 1 — Find the radius
Thus, the radius of the circle is
r = 10.5 cm
Step 2 — Find the area of the quadrant
A quadrant is one-fourth of a circle.
Therefore,
Area of quadrant = 1 4 πr²
= 1 4 × 22 7 × (10.5)²
= 1 4 × 22 7 × 110.25
= 346.5 4
= 86.625 cm²
∴ The area of the quadrant is
86.625 cm²
.
End-of-Chapter Exercises • Question 9
Class 9 Maths Chapter 6 End-of-Chapter Exercises Question 9 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
The wheel of a car has an outer radius of 28 cm.
Calculate how far the car travels after one complete turn
of the wheel, and how many times the wheel turns during
a journey of 1 km.
Car wheel of radius 28 cm
Given
Outer radius of the wheel,
r = 28 cm
For π, use 22 7
Distance of journey = 1 km
To Find:
(i) Distance travelled after one complete turn
(ii) Number of turns in 1 km
Solution
Step 1 — Distance travelled in one complete turn
The distance travelled by the car in one complete turn
of the wheel is equal to the circumference of the wheel.
We know that,
Circumference = 2πr
Therefore,
Circumference = 2 × 22 7 × 28
= 2 × 22 × 4
= 176 cm
∴ Distance travelled in one complete turn
= 176 cm
Step 2 — Convert 1 km into centimetres
We have,
1 km = 1000 m
and
1 m = 100 cm
Therefore,
1 km = 1000 × 100 cm
1 km = 1,00,000 cm
Step 3 — Find the number of turns
Number of turns
= Total distance Distance travelled in one turn
= 1,00,000 176
= 6250 11
= 568 2 11 turns
∴ Distance travelled in one complete turn
= 176 cm
∴ Number of turns in 1 km
=
568
2
11
turns
End-of-Chapter Exercises • Question 10
Class 9 Maths Chapter 6 End-of-Chapter Exercises Question 10 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
Two rectangles have the same area and the same perimeter.
Does this mean that they are congruent to each other?
Given
Let the length and breadth of the first rectangle be
l₁ and b₁.
Let the length and breadth of the second rectangle be l₂ and b₂.
The two rectangles have the same area and the same perimeter.
To Find:
Whether the two rectangles are congruent.
Figure
Two rectangles having the same area and perimeter
Solution
Step 1 — Use the condition of equal perimeter
Perimeter of a rectangle
= 2 × (length + breadth)
Since the two rectangles have the same perimeter,
2(l₁ + b₁) = 2(l₂ + b₂)
Dividing both sides by 2,
l₁ + b₁ = l₂ + b₂
Let this common value be S.
Therefore,
l₁ + b₁ = l₂ + b₂ = S
Step 2 — Use the condition of equal area
Area of a rectangle
= length × breadth
Since the two rectangles have the same area,
l₁ × b₁ = l₂ × b₂
Let this common area be P.
Therefore,
l₁ × b₁ = l₂ × b₂ = P
Step 3 — Show that the difference between the sides is also fixed
We use the identity
(l − b)² = (l + b)² − 4lb
For either rectangle,
(l − b)² = S² − 4P
Here, S and P are already fixed because the perimeter and area are the same.
Therefore,
(l − b)² is also fixed.
Hence the difference between the length and breadth is fixed.
Thus, both
l + b and l − b have fixed values.
Step 4 — The length and breadth are uniquely determined
Since
l + b = S and
l − b also has a fixed value, the values of l and b are fixed.
In fact,
2l = (l + b) + (l − b)
Therefore,
l = [(l + b) + (l − b)] ÷ 2
Similarly,
2b = (l + b) − (l − b)
Therefore,
b = [(l + b) − (l − b)] ÷ 2
Hence, the same fixed perimeter and the same fixed area determine one fixed pair of side lengths.
The second rectangle can only have the same pair of side lengths, possibly with length and breadth interchanged.
Therefore,
l₁ = l₂, b₁ = b₂
or
l₁ = b₂, b₁ = l₂
Interchanging length and breadth only changes the orientation of the rectangle, not its shape or size.
∴ Yes. Two rectangles having the same area and
the same perimeter are congruent to each other.
End-of-Chapter Exercises • Question 11
Class 9 Maths Chapter 6 End-of-Chapter Exercises Question 11 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
You know that the area of a parallelogram is base × height.
Using this and the figure, show that the area of a trapezium is
half the sum of the parallel sides × height, i.e.
1
2
(a + b)h
Trapezium divided into a parallelogram and a triangle
Given
The trapezium has parallel sides
a and b.
Its height is h.
The trapezium is divided into a parallelogram and a triangle.
To Show:
Area of trapezium 1 2 (a + b)h
Figure
Trapezium ABCD divided into parallelogram ABED and triangle BCE
Solution
Step 1 — Name the two parts
Let the trapezium be ABCD, where
AB = a
and
DC = b.
Let E be the point on DC where the dividing line meets the base.
Thus, trapezium ABCD is divided into:
1. Parallelogram ABED
2. Triangle BCE
Step 2 — Find the area of parallelogram ABED
In parallelogram ABED,
Base = AB = a
Height = h
We know,
Area of a parallelogram = base × height
Therefore,
Area of parallelogram ABED = ah
Step 3 — Find the base of triangle BCE
Since ABED is a parallelogram,
DE = AB = a
But,
DC = b
Therefore,
EC = DC − DE
EC = b − a
Hence, the base of triangle BCE is (b − a) and its height is h.
Step 4 — Find the area of triangle BCE
We know,
Area of a triangle = 1 2 × base × height
Therefore,
Area of triangle BCE = 1 2 × (b − a) × h
Step 5 — Find the area of the trapezium
The trapezium ABCD is made up of
parallelogram ABED and triangle BCE.
Therefore,
Area of trapezium ABCD = Area of parallelogram ABED + Area of triangle BCE
= ah + 1 2 (b − a)h
Taking h common,
= h 2a + b − a 2
= h a + b 2
Hence,
Area of trapezium ABCD = 1 2 (a + b)h
∴ Area of the trapezium is
1
2
(a + b)h.
End-of-Chapter Exercises • Question 12
Class 9 Maths Chapter 6 End-of-Chapter Exercises Question 12 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
By dividing a trapezium into two triangles show that its area is,
half the sum of the parallel sides multiplied by the height
(the same formula as the one given above).
Given
Let ABCD be a trapezium such that
AB ∥ CD
Let the lengths of the parallel sides be
AB = a and CD = b.
Let the perpendicular height of the trapezium be h.
To Show:
Area of trapezium ABCD
= 1 2 (a + b)h
Figure
Trapezium ABCD divided into triangles ABC and ACD
Solution
Step 1 — Divide the trapezium into two triangles
Draw the diagonal AC of trapezium ABCD.
The diagonal AC divides the trapezium into two triangles:
△ABC and △ACD.
Therefore,
Area of trapezium ABCD
= Area of △ABC + Area of △ACD
Step 2 — Find the areas of the two triangles
Since AB ∥ CD, the perpendicular distance
between these parallel sides is the height h.
For △ABC,
Base = AB = a
Height = h
Therefore,
Area of △ABC
= 1 2 × a × h
Similarly, for △ACD,
Base = CD = b
Height = h
Therefore,
Area of △ACD
= 1 2 × b × h
Step 3 — Add the areas of the two triangles
We know that
Area of trapezium ABCD
= Area of △ABC + Area of △ACD
= 1 2 ah + 1 2 bh
Taking 1 2 h common,
= 1 2 h(a + b)
Step 4 — Write the required formula
Hence,
Area of trapezium ABCD
= 1 2 (a + b)h
Thus, the area of a trapezium is half the sum of its parallel sides multiplied by its height.
∴
Area of trapezium ABCD =
1
2
(a + b)h
End-of-Chapter Exercises • Question 13
Class 9 Maths Chapter 6 End-of-Chapter Exercises Question 13 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
Show how we can use two identical copies of a trapezium to make
a parallelogram. How will this give us the formula for the area
of a trapezium?
Given
Consider a trapezium whose parallel sides are
a and b,
and whose height is h.
Take two identical copies of this trapezium.
Turn one copy upside down and join the two trapeziums along their equal slanting sides.
To Find:
The formula for the area of the trapezium.
Figure
One trapezium is turned upside down and joined to the other.
Solution
Step 1 — Take two identical trapeziums
Let the parallel sides of each trapezium be
a and b,
and its height be h.
Take another copy of the same trapezium.
Turn the second trapezium upside down and join it to the first trapezium along their equal slanting sides.
The two trapeziums together form a parallelogram.
Step 2 — Find the base and height of the parallelogram
The two parallel sides of the original trapezium are
a and b.
After joining the two identical trapeziums, the base of the parallelogram is the sum of these two parallel sides.
Therefore,
Base of parallelogram = a + b
The height remains the same as the height of each trapezium.
Therefore,
Height of parallelogram = h
Step 3 — Find the area of the parallelogram
Area of a parallelogram
= base × height
Therefore,
Area of parallelogram = (a + b) × h
Step 4 — Relate the areas
The parallelogram is made from
two identical trapeziums.
Therefore,
Area of parallelogram = 2 × Area of one trapezium
Hence,
2 × Area of trapezium = (a + b) × h
Dividing both sides by 2,
Area of trapezium = 1 2 (a + b)h
Thus, the area of a trapezium is half the sum of its parallel sides multiplied by its height.
∴ The area of a trapezium is
1
2
(a + b)h
.
End-of-Chapter Exercises • Question 14
Class 9 Maths Chapter 6 End-of-Chapter Exercises Question 14 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
Show that the area of a kite is half the product of its
diagonals. Show this: (i) using algebra, and (ii) using geometry.
Given
Consider a kite ABCD.
Let its diagonals AC and BD intersect at O.
In a kite, the diagonals are perpendicular and one diagonal bisects the other.
Let AC = d₁ and BD = d₂.
To Prove:
Area of kite 1 2 × product of its diagonals.
Figure
Kite ABCD with perpendicular diagonals AC and BD
Solution
(i) Using Algebra
Let
AO = x
,
OC = y
,
BO = p
and
OD = q
.
Therefore, the two diagonals are
AC = x + y
and
BD = p + q .
The kite is divided by its diagonals into four right-angled triangles.
Therefore, its area is
Area of kite = Area of △AOB + Area of △BOC + Area of △COD + Area of △DOA
Using the area formula for a triangle,
Area of △AOB = 1 2 xp
Similarly,
Area of △BOC = 1 2 yp
Area of △COD = 1 2 yq
Area of △DOA = 1 2 xq
Hence,
Area of kite = 1 2 (xp + yp + yq + xq)
= 1 2 [p(x + y) + q(x + y)]
= 1 2 (x + y)(p + q)
Since
x + y = AC and p + q = BD, therefore,
Area of kite = 1 2 × AC × BD
(ii) Using Geometry
The diagonals AC and BD
of the kite intersect at O.
Since the diagonals of a kite are perpendicular,
AC ⟂ BD
The kite is divided into two triangles: △ABC and △ADC.
Both triangles have the same base AC. Their corresponding heights are BO and DO.
Therefore,
Area of kite = Area of △ABC + Area of △ADC
= 1 2 × AC × BO + 1 2 × AC × DO
Taking AC common,
Area of kite = 1 2 × AC × (BO + DO)
But
BO + DO = BD
Therefore,
Area of kite = 1 2 × AC × BD
Hence, the area of a kite is half the product of its diagonals.
∴
Area of kite
=
1
2
× product of its diagonals
or Area of kite = 1 2 × AC × BD .
or Area of kite = 1 2 × AC × BD .
End-of-Chapter Exercises • Question 15
Class 9 Maths Chapter 6 End-of-Chapter Exercises Question 15 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
Three problems about fitting congruent shapes together:
(i) Rectangle ABCD has sides a, b, and rectangle PQRS has sides 2a, 2b. Show that PQRS has 4 times the area of ABCD. Does this mean that 4 copies of rectangle ABCD will fit into rectangle PQRS? Check and see!
(ii) △ABC has sides a, b, c, and △PQR has sides 2a, 2b, 2c. Show that △PQR has 4 times the area of △ABC. Does this mean that 4 copies of △ABC will fit into △PQR? Check and see!
(iii) △ABC has sides a, b, c, and △PQR has sides 3a, 3b, 3c. Show that △PQR has 9 times the area of △ABC. Does this mean that 9 copies of △ABC will fit into △PQR? Check and see!
(i) Rectangle ABCD has sides a, b, and rectangle PQRS has sides 2a, 2b. Show that PQRS has 4 times the area of ABCD. Does this mean that 4 copies of rectangle ABCD will fit into rectangle PQRS? Check and see!
(ii) △ABC has sides a, b, c, and △PQR has sides 2a, 2b, 2c. Show that △PQR has 4 times the area of △ABC. Does this mean that 4 copies of △ABC will fit into △PQR? Check and see!
(iii) △ABC has sides a, b, c, and △PQR has sides 3a, 3b, 3c. Show that △PQR has 9 times the area of △ABC. Does this mean that 9 copies of △ABC will fit into △PQR? Check and see!
Given
(i)
Rectangle ABCD has sides a and b.
Rectangle PQRS has sides 2a and 2b.
(ii) △ABC has sides a, b, c. △PQR has sides 2a, 2b, 2c.
(iii) △ABC has sides a, b, c. △PQR has sides 3a, 3b, 3c.
To Find:
Compare the areas and check whether the required number of congruent copies can fit exactly into the larger shape.
(i) Four Congruent Rectangles Fit Exactly
Rectangle PQRS divided into four congruent copies of ABCD
(ii) Four Congruent Triangles Fit Exactly
Joining the midpoints divides △PQR into four congruent triangles
(iii) Nine Congruent Triangles Fit Exactly
Trisecting each side and drawing parallel lines gives nine congruent triangles
Solution
(i) Rectangles
Area of rectangle ABCD
= length × breadth
= a × b
Therefore,
Area of ABCD = ab
Area of rectangle PQRS
= 2a × 2b
= 4ab
Therefore,
Area of PQRS = 4ab
Since
Area of ABCD = ab,
we get
Area of PQRS = 4 × Area of ABCD.
Now check whether four copies of ABCD fit into PQRS.
Along the length of PQRS,
2a ÷ a = 2
copies fit.
Along the breadth of PQRS,
2b ÷ b = 2
copies fit.
Hence, the total number of copies is
2 × 2 = 4.
Thus, four congruent copies of ABCD fit exactly into PQRS.
(ii) Triangles with sides doubled
Let the semiperimeter of △ABC be
a + b + c
2
= s
By Heron’s formula,
Area of △ABC
=
√[s(s − a)(s − b)(s − c)]
For △PQR, the sides are
2a, 2b, 2c.
Therefore, its semiperimeter is
2a + 2b + 2c
2
= a + b + c = 2s
Applying Heron’s formula,
Area of △PQR
=
√[(2s)(2s − 2a)(2s − 2b)(2s − 2c)]
Taking 2 from each of the four factors,
=
√[16s(s − a)(s − b)(s − c)]
= 4√[s(s − a)(s − b)(s − c)]
Therefore,
Area of △PQR = 4 × Area of △ABC.
Now check whether four copies fit.
Since every side of △PQR is twice the corresponding side of △ABC, joining the midpoints of the three sides of △PQR divides it into four triangles.
Each of these four triangles has sides
a, b, c.
Hence, each small triangle is congruent to △ABC.
Therefore, four copies of △ABC fit exactly into △PQR.
(iii) Triangles with sides tripled
Again, let the semiperimeter of △ABC be
a + b + c
2
= s
For △PQR, the sides are
3a, 3b, 3c.
Its semiperimeter is
3a + 3b + 3c
2
= 3s
By Heron’s formula,
Area of △PQR
=
√[(3s)(3s − 3a)(3s − 3b)(3s − 3c)]
Taking 3 from each of the four factors,
=
√[81s(s − a)(s − b)(s − c)]
= 9√[s(s − a)(s − b)(s − c)]
Therefore,
Area of △PQR = 9 × Area of △ABC.
Now check whether nine copies fit.
Divide each side of △PQR into three equal parts and draw lines parallel to the three sides.
This divides the large triangle into
9 congruent small triangles.
Each small triangle has sides
a, b, c.
Therefore, each small triangle is congruent to △ABC.
Hence, nine copies of △ABC fit exactly into △PQR.
Key Idea
When every linear dimension of a plane figure is multiplied by
a factor k, its area is multiplied by
k².
Thus,
scale factor 2 → area factor 4
and
scale factor 3 → area factor 9.
In this question, the larger rectangles and triangles can actually be divided into the required number of congruent copies of the smaller shape.
Final Answer
(i)
Area of PQRS = 4 × Area of ABCD, and
4 copies of ABCD fit exactly into PQRS.
(ii) Area of △PQR = 4 × Area of △ABC, and 4 copies of △ABC fit exactly into △PQR.
(iii) Area of △PQR = 9 × Area of △ABC, and 9 copies of △ABC fit exactly into △PQR.
End-of-Chapter Exercises • Question 16
Class 9 Maths Chapter 6 End-of-Chapter Exercises Question 16 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
Find the fraction of the shaded region in each of the following figures:
(i) What fraction of the triangle is shaded?
(ii) What fraction of the square is shaded?
(i) What fraction of the triangle is shaded?
(ii) What fraction of the square is shaded?
Fig. 6.43 and Fig. 6.44 — Question 16
Given
In Fig. 6.43, the left side of the triangle is divided
into two equal parts and the right side is divided into
three equal parts.
In Fig. 6.44, each side of the square is divided into two equal parts.
To Find:
The fraction of the whole figure that is shaded.
Solution Figure — (i)
Fig. 6.43 — Equal-area regions used in the solution
Solution Figure — (ii)
Parallel construction lines divide the square into 25 equal parts
Solution
(i) Fraction of the triangle shaded
Let the four regions be
a, b, c and d.
Since the marked point on the left side is the midpoint, the line through it gives equal bases on the same altitude.
Therefore,
Area a = Area b
Let
Area a = Area b = x
From the division shown on the other side, the median relationship gives
Area c = Area d = Area a + Area b
Therefore,
Area c = Area d = 2x
Hence the area of the whole triangle is
x + x + 2x + 2x = 6x
The shaded region consists of regions b and c.
Therefore,
Shaded area = x + 2x = 3x
Hence,
Fraction of triangle shaded
＝ 3x 6x
＝ 1 2
(ii) Fraction of the square shaded
Draw lines parallel to the sides of the square through
each vertex of the shaded region.
Since the four vertices of the shaded region lie at the corresponding intersection points, these parallel lines divide the whole square into
5 × 5 = 25 equal small squares.
The shaded region has the same total area as 5 of these equal small squares.
Therefore,
Shaded area = 5 equal parts
and
Whole square = 25 equal parts
Hence,
Fraction of square shaded
= 5 25
= 1 5
∴ (i) The fraction of the triangle that is shaded is
1
2
.
∴ (ii) The fraction of the square that is shaded is
1
5
.
End-of-Chapter Exercises • Question 17
Class 9 Maths Chapter 6 End-of-Chapter Exercises Question 17 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
What fraction of the rectangle is covered by the circles?
Given
In each rectangle, the circles have the same diameter as
the height of the rectangle.
Figure 6.45: The rectangle contains 3 equal circles.
Figure 6.46: The rectangle contains 4 equal circles.
To Find:
The fraction of each rectangle covered by the circles.
Figure 6.45 — Three Circles
Three equal circles, each of diameter d
Figure 6.46 — Four Circles
Four equal circles, each of diameter d
Solution
Step 1 — Let the diameter of each circle be d
Since each circle touches the top and bottom of the rectangle,
Height of rectangle = d
Also, the circles touch one another.
Therefore, each circle occupies a horizontal length equal to its diameter.
Step 2 — Figure 6.45: Three circles
Width of the rectangle
= 3d
Area of rectangle
= 3d × d
= 3d²
Radius of each circle
d 2
Area of one circle
= π × r²
= π × d² 4
Area of three circles
= 3 × πd² 4
= 3πd² 4
Step 3 — Fraction covered by the circles
Fraction covered
= Area of 3 circles Area of rectangle
= 3πd² 4 × 1 3d²
= π 4
Step 4 — Figure 6.46: Four circles
Width of the rectangle
= 4d
Area of rectangle
= 4d × d
= 4d²
Area of four circles
= 4 × πd² 4
= πd²
Therefore,
Fraction covered
= πd² 4d²
= π 4
Thus, in both figures the fraction covered by the circles is the same.
∴ The fraction of the rectangle covered by the circles is
π
4
in both figures.
End-of-Chapter Exercises • Question 18
Class 9 Maths Chapter 6 End-of-Chapter Exercises Question 18 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
Use the above to make a conjecture about the area occupied by
circles fitted into a rectangle in the manner shown. Test your
conjecture for particular cases: 10 circles; 20 circles; 50
circles. Then prove your conjecture!
Given
Equal circles are fitted inside a rectangle in the same manner
as in Question 17.
Let the diameter of each circle be d.
Let the number of circles be n.
Therefore, the height of the rectangle is d and its length is nd.
To Find:
The fraction of the rectangle occupied by the circles.
General Arrangement
n equal circles, each of diameter d, fitted inside a rectangle
Solution
Step 1 — Make the conjecture
Suppose there are n equal circles.
Let the diameter of each circle be d.
Then the height of the rectangle is d and its length is nd.
We therefore expect the fraction occupied by the circles to remain the same, whatever the value of n.
Conjecture:
The fraction of the rectangle occupied by the circles is
π 4
Step 2 — Test for 10 circles
For 10 circles, the rectangle has length
10d
Area of the rectangle
= 10d × d
= 10d2
Radius of each circle
d 2
Area of 10 circles
= 10 × π × d2 4
= 10πd2 4
Therefore, fraction occupied
10πd2 4 × 10d2
= π 4
Step 3 — Test for 20 circles
Area of 20 circles
= 20 × π × d2 4
Area of rectangle
= 20d2
Hence,
20πd2 4 × 20d2
= π 4
Step 4 — Test for 50 circles
Area of 50 circles
= 50 × π × d2 4
Area of rectangle
= 50d2
Therefore,
50πd2 4 × 50d2
= π 4
Step 5 — Prove the conjecture
Let there be n circles.
Diameter of each circle = d
Therefore,
Radius of each circle
d 2
Area of one circle
= πr2
= π × d2 4
Area of n circles
= n × π × d2 4
= nπd2 4
Since the rectangle has height d and length nd,
Area of rectangle
= nd × d
= nd2
Hence, fraction of rectangle occupied by circles
nπd2 4nd2
Cancelling n and d2,
= π 4
Thus, the fraction is independent of the number of circles.
Using the NCERT approximation
22 7 for π,
the fraction becomes
22 7 × 4
= 11 14
Therefore, about 11 out of every 14 equal parts of the rectangle are occupied by the circles.
∴ Conjecture proved:
If n equal circles of diameter d
are fitted side by side in a rectangle of height d,
then the fraction of the rectangle occupied by the circles is
π
4
=
11
14
.
using π =
22
7
.
End-of-Chapter Exercises • Question 19
Class 9 Maths Chapter 6 End-of-Chapter Exercises Question 19 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
The figure shows nine identical rectangles fitted together
to make a large rectangle whose area is 72 cm².
Find the perimeter of each small rectangle.
Given
There are 9 identical rectangles.
Let the length of each small rectangle be l cm and its breadth be b cm.
Area of the large rectangle = 72 cm².
To Find:
Perimeter of each small rectangle.
Figure — Arrangement of 9 Identical Rectangles
Four lengths and five breadths cover the same width.
Solution
Step 1 — Find the area of one small rectangle
There are 9 identical rectangles.
Therefore,
Area of one rectangle
=
72
9
= 8 cm²
Hence,
l × b = 8
Step 2 — Use the arrangement of the rectangles
From the figure, the total width of the large rectangle
can be written in two ways.
The top row contains 4 lengths.
The bottom row contains 5 breadths.
Therefore,
4l = 5b
Hence,
b =
4l
5
Step 3 — Find the length l
We know that
l × b = 8
Substituting
b =
4l
5
into the equation,
l ×
4l
5
= 8
Therefore,
4l²
5
= 8
Hence,
l² = 10
Since length is positive,
l = √10 cm
Step 4 — Find the breadth b
We have
b =
4l
5
Substituting
l = √10
,
b =
4√10
5
cm
Step 5 — Find the perimeter
Perimeter of a rectangle
= 2(l + b)
Substituting the values of l and b,
= 2
(
√10 +
4√10
5
)
Taking √10 common,
=
2√10
(
1 +
4
5
)
Therefore,
=
18√10
5
cm
Approximately,
≈ 11.38 cm
Final Answer
The perimeter of each small rectangle is
18√10
5
cm
≈ 11.38 cm
End-of-Chapter Exercises • Question 20
Class 9 Maths Chapter 6 End-of-Chapter Exercises Question 20 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
Show that the areas of the shaded blue triangle and the
shaded red triangle are equal.
Find a way of cutting up the blue triangle into some number
of pieces and rearranging the pieces to cover the red triangle.
Fig. 6.48 — Lines from a vertex to the points of trisection of the opposite side
Given
Let the large triangle be △ABC.
Let D and E be the points of trisection of BC.
Therefore,
BD = DE = EC = b
Let the perpendicular height of the large triangle be h.
The blue triangle is △ABD and the red triangle is △AEC.
To Prove:
Area of △ABD = Area of △AEC
To Show:
The blue triangle can be cut into pieces and rearranged to cover the red triangle.
Figure 1 — Equal Bases and Same Height
The blue and red triangles have equal bases and the same perpendicular height.
Figure 2 — Cut the Blue Triangle into Three Pieces
The blue triangle is divided into three pieces: P₁, P₂ and P₃.
Figure 3 — Rearrange the Three Pieces
Rearrangement of the Three Pieces
Three pieces P₁, P₂ and P₃ arranged together
By rotating the two upper pieces, the three pieces form a parallelogram.
Figure 4 — One More Cut and Slide
The extra triangular piece is cut and moved to change the slant while preserving its area.
Figure 5 — Final Arrangement Covers the Red Triangle
The finite cut-and-rearrange process converts the blue triangle into the red triangle.
Solution
Step 1 — Identify the equal bases
Since D and E are the
points of trisection of BC,
BD = DE = EC = b
Therefore, the base of the blue triangle △ABD and the base of the red triangle △AEC are equal.
Hence,
BD = EC = b
Step 2 — Both triangles have the same height
Both triangles have the same vertex A.
Their bases BD and EC lie on the same straight line BC.
Therefore, the perpendicular distance from A to the line BC is the same for both triangles.
Let this common height be h.
Step 3 — Compare their areas
We know,
Area of a triangle = 1 2 × base × height
For the blue triangle △ABD,
Area of △ABD = 1 2 × BD × h
Since BD = b,
Area of △ABD = 1 2 × b × h
For the red triangle △AEC,
Area of △AEC = 1 2 × EC × h
Since EC = b,
Area of △AEC = 1 2 × b × h
Therefore,
Area of △ABD = Area of △AEC
The shaded blue triangle and the shaded red triangle
have equal areas because they have
equal bases + the same perpendicular height.
equal bases + the same perpendicular height.
Step 4 — Cut the blue triangle into pieces
Draw a line through the midpoints of the two sides of
the blue triangle, parallel to its base.
This cuts off a smaller triangle from the top and leaves a trapezium below.
Now draw the altitude of the smaller top triangle.
Thus, the blue triangle is divided into three pieces:
P₁, P₂ and P₃.
Step 5 — Rearrange the three pieces
Rotate the two small upper pieces through
180°
about the appropriate endpoints of the middle cut.
They fit against the lower trapezium and form a parallelogram.
Its base is the same as the base of the original blue triangle, namely b, and its height is h upon 2.
No piece is stretched or shrunk.
Only rotation and rearrangement are used.
Step 6 — Change the slant
The red triangle has the same base length and the same
height as the blue triangle.
The parallelogram obtained from the blue triangle can therefore be changed into the corresponding parallelogram for the red triangle.
Make one more straight cut through the parallelogram.
Move the small triangular piece to the opposite side.
This changes only the position of the piece; its shape and area remain unchanged.
The resulting arrangement has exactly the slant required for the red triangle.
Step 7 — Cover the red triangle
The pieces obtained from the original blue triangle can
now be arranged to occupy the complete red triangle.
There is no gap and no overlap.
Since every operation was only a cut, rotation or translation, the total area remains unchanged.
Key Idea
Equal area does not mean that the two triangles must have
the same shape.
Here the blue and red triangles have equal bases and the same perpendicular height, so their areas are equal.
The cutting-and-rearranging construction shows how one region can be transformed into the other without changing its total area.
∴
Area of the shaded blue triangle = Area of the shaded red triangle.
The blue triangle can be cut into a finite number of pieces and rearranged to cover the red triangle completely.
The blue triangle can be cut into a finite number of pieces and rearranged to cover the red triangle completely.
End-of-Chapter Exercises • Question 21
Class 9 Maths Chapter 6 End-of-Chapter Exercises Question 21 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
The figure shows a quarter circle in a square. Its centre is at
one vertex, and it passes through two adjacent vertices. There are
two semicircles on two adjacent sides as diameters. They create
the shaded regions A and B. Show that A and B have equal areas.
Given
Let the side of the square be
a.
The quarter circle has radius a.
Each semicircle has diameter a.
Therefore, the radius of each semicircle is a 2 .
To Prove:
Area of region A = Area of region B.
Solution
Step 1 — Find the area of the quarter circle
The side of the square is a.
Therefore, the radius of the quarter circle is a.
Area of the quarter circle
=
πa²
4
Step 2 — Find the total area of the two semicircles
Each semicircle has diameter a.
Therefore, its radius is
a
2
Area of one semicircle
=
1
2
× π ×
a
2
2
=
πa²
8
Therefore, area of two semicircles
=
πa²
8
+
πa²
8
=
πa²
4
Hence,
Total area of the two semicircles
= Area of the quarter circle
Step 3 — Compare regions A and B
The two semicircles overlap in region A.
Therefore, the area covered by the two semicircles is
= Total area of two semicircles − Area of A
= Area of quarter circle − Area of A
On the other hand, region B is the part of the
quarter circle which is not covered by the two semicircles.
= Area of quarter circle − Area of A
Therefore, the same covered area can also be written as
Area of quarter circle − Area of B
Step 4 — Show that A and B have equal areas
Since both expressions represent the same covered area,
Area of quarter circle − Area of A
=
Area of quarter circle − Area of B
Subtracting the area of the quarter circle from both sides,
=
Area of quarter circle − Area of B
− Area of A = − Area of B
Therefore,
Area of A
=
Area of B
∴ Area of region A = Area of region B.
Hence, the two shaded regions have equal areas.
Hence, the two shaded regions have equal areas.
End-of-Chapter Exercises • Question 22
Class 9 Maths Chapter 6 End-of-Chapter Exercises Question 22 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
In Fig. 6.50, four semicircles have been drawn within the
given square whose side is 2 units. The centres of these
semicircles are the midpoints of the sides. They create a
4-petalled flower (shown in blue). Find the perimeter and
the area of this flower.
Given
Side of the square = 2 units.
The diameter of each semicircle is equal to the side of the square.
Therefore,
Diameter = 2 units
Hence, radius of each semicircle is
2
2
= 1 unit
To Find:1. Perimeter of the flower
2. Area of the flower
Figure 6.50 — Four-Semicircle Flower
Four semicircles form a 4-petalled flower
Solution
Step 1 — Find the radius of each semicircle
The diameter of each semicircle is equal to the side
of the square.
Therefore,
Radius
=
2
2
= 1 unit
Step 2 — Find the perimeter of the flower
Look carefully at the flower.
Its boundary consists of 8 quarter-circle arcs.
Each quarter-circle has radius 1 unit.
Length of one quarter-circle arc
=
1
4
× 2π × 1
Therefore,
=
π
2
units
There are 8 such arcs.
Hence,
Perimeter
=
8 ×
π
2
= 4π units
Using π = 22/7,
Perimeter
=
4 ×
22
7
Perimeter =
88
7
units
Step 3 — Find the area of the four semicircles
Area of one semicircle
=
1
2
× π × 1²
=
π
2
square units
Therefore, area of four semicircles
= 4 ×
π
2
= 2π square units
Step 4 — Find the area of the flower
The four semicircles together cover the whole square.
But the four petals are the regions where two adjacent semicircles overlap. These overlapping regions are counted twice when we add the areas of all four semicircles.
Therefore,
Area of flower
= Area of four semicircles − Area of square
Area of the square
= 2 × 2
= 4 square units
Hence,
Area of flower
= 2π − 4
Using π = 22/7,
= 2 ×
22
7
− 4
=
44
7
−
4
=
44
7
−
28
7
=
16
7
square units
Therefore,
Perimeter =
88
7
units
Area of flower = 16 7 square units
Area of flower = 16 7 square units
∴
Perimeter of the flower =
88
7
units
and Area of the flower = 16 7 square units.
and Area of the flower = 16 7 square units.
End-of-Chapter Exercises • Question 23
Class 9 Maths Chapter 6 End-of-Chapter Exercises Question 23 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
In Fig. 6.51 we see two concentric circles with a common
centre O. A chord BC of the larger circle is drawn,
touching the smaller circle at A. The length of BC is
l. Show that the area of the green region enclosed
between the two circles is
1
4
πl2.
Given
Two concentric circles have common centre
O.
BC is a chord of the larger circle and touches the smaller circle at A.
Length of chord BC = l.
Let the radius of the larger circle be R and the radius of the smaller circle be r.
To Show:
Area of the green region
=
1
4
πl2
Solution
Step 1 — Use the tangent property
Since BC touches the smaller circle at
A, BC is a tangent to the smaller circle.
The radius drawn to the point of contact of a circle is perpendicular to the tangent.
OA ⟂ BC
Therefore, triangle OAB is a
right-angled triangle at A.
Step 2 — Find the half of the chord
In a circle, the perpendicular from the centre to a chord
bisects the chord.
Hence,
AB = AC
=
BC
2
Since BC = l,
AB =
l
2
Step 3 — Apply Pythagoras theorem
In right-angled triangle OAB,
OB is the radius of the larger circle, so OB = R.
OA is the radius of the smaller circle, so OA = r.
Also, AB = l 2 .
By Pythagoras theorem,
OB2 = OA2 + AB2
Substituting the values,
R2
=
r2
+
l2
4
Therefore,
R2 − r2
=
l2
4
Step 4 — Find the area of the green region
The green region is the area between the two concentric
circles.
Therefore,
Area of green region
=
πR2 − πr2
Taking π common,
= π(R2 − r2)
From Step 3,
R2 − r2
=
l2
4
Hence,
Area of green region
=
π ×
l2
4
Therefore,
Area of green region
=
1
4
πl2
∴
Area of the green region =
1
4
πl2.
Hence proved.
End-of-Chapter Exercises • Question 24
Class 9 Maths Chapter 6 End-of-Chapter Exercises Question 24 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
In Fig. 6.52, semicircles have been drawn on all the sides
of a right-angled triangle as shown. Show that
Area (A) + Area (B) = Area (C).
Given
A right-angled triangle has semicircles drawn on
all its three sides.
Let the two perpendicular sides be 2a and 2b, and the hypotenuse be 2c.
C represents the area of the right-angled triangle.
To Show:
Area (A) + Area (B) = Area (C)
Solution
Step 1 — Apply Pythagoras theorem
The two perpendicular sides are 2a and
2b, and the hypotenuse is 2c.
Since the triangle is right-angled,
(2a)2 + (2b)2 = (2c)2
Dividing by 4,
a2 + b2 = c2
Step 2 — Compare the areas of the semicircles
Area of a semicircle is
1
2
× π × r2
For the two smaller semicircles,
Area of two smaller semicircles
=
1
2
πa2
+
1
2
πb2
Taking
π
and
1/2
common,
=
1
2
π(a2 + b2)
Using
a2 + b2 = c2,
=
1
2
πc2
But
1
2
πc2
is the area of the largest semicircle.
Therefore,
Area of two smaller semicircles
=
Area of largest semicircle
Step 3 — Find Area (A) + Area (B)
From the figure,
Area (A) + Area (B)
= Area (C) + Area of two smaller semicircles − Area of largest semicircle
But from Step 2,
= Area (C) + Area of two smaller semicircles − Area of largest semicircle
Area of two smaller semicircles
=
Area of largest semicircle
So these two areas cancel.
Therefore,
Area (A) + Area (B)
=
Area (C)
∴
Area (A) + Area (B) = Area (C)
Hence proved.
End-of-Chapter Exercises • Question 25
Class 9 Maths Chapter 6 End-of-Chapter Exercises Question 25 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
Fig. 6.53 shows two circles passing through each other’s
centres. Find the area of the region enclosed by the two
circles in terms of the common radius r.
Given
The two circles are congruent and have common radius
r.
Each circle passes through the centre of the other circle.
A and B are the centres, while C and D are the points where the circles intersect.
To Find:
Area of the common region
Solution
Step 1 — Identify the equilateral triangles
Since each circle passes through the centre of the other
circle,
AB = AC = BC = r
Therefore, triangle ABC is equilateral.
Similarly,
AB = AD = BD = r
Therefore, triangle ABD is also equilateral.
Hence,
∠CAB = ∠DAB = 60°
Step 2 — Find the area of the upper half
The upper half of the common region is made from
two equal 60° sectors and the equilateral triangle
ABC.
Therefore,
Area of upper half
=
Area of two 60° sectors
−
Area of △ABC
Area of one 60° sector is
60
360
× πr2
=
π
6
r2
So, area of two such sectors is
2 ×
π
6
r2
=
π
3
r2
Area of an equilateral triangle of side
r is
√3
4
r2
Therefore,
Area of upper half
=
π
3
r2
−
√3
4
r2
Step 3 — Find the complete common region
The lower half is equal to the upper half.
Therefore, the complete common region has twice the area
found above.
Area of common region
=
2 ×
[
π
3
r2
−
√3
4
r2
]
Therefore,
=
2π
3
r2
−
√3
2
r2
Taking r2 common,
Area of common region
=
2π
3
−
√3
2
times r2
∴
Area of the region enclosed by the two circles
=
2π
3
−
√3
2
r2
square units.
End-of-Chapter Exercises • Question 25
Method 2 — Common Chord Method
Finding the common region using the common chord CD
✦ Idea of Method 2
Here CD is the common chord of the two circles.
We use the perpendicular from the centre to a chord,
the Pythagoras theorem, and the area of a sector and triangle.
Construction
CD is the common chord and M is its midpoint.
✍️ Solution
Given
Two congruent circles have common radius
r and pass through each other’s centres.
Their common chord is CD.
Step 1: Find the angle of the sector
Since each circle passes through the centre of the other circle,
AB = AC = BC = r
Therefore, ΔABC is an equilateral triangle.
∠CAB = 60°
Similarly, ΔABD is also equilateral.
∠DAB = 60°
Hence,
∠CAD = 60° + 60° = 120°
So, each circular segment of the common region is formed
by a 120° sector and the triangle ACD.
Step 2: Find the length of the common chord CD
In equilateral ΔABC, the perpendicular from C to AB
divides AB into two equal parts.
Therefore, the perpendicular height is obtained using
Pythagoras theorem:
height² = r² −
r²
4
=
3r²
4
Hence,
height =
√3 r
2
The same height is obtained for ΔABD on the other side of AB.
Therefore,
CD = √3 r
Step 3: Find the area of ΔACD
Let M be the midpoint of the common chord CD.
Since the perpendicular from the centre of a circle
to a chord bisects the chord,
CM = DM =
√3 r
2
Also, AM ⟂ CD.
In right ΔAMC,
AC² = AM² + CM²
Therefore,
r² = AM² +
3r²
4
Hence,
AM² =
r²
4
So,
AM = r/2
Therefore,
Area of ΔACD
=
1
2
× √3r ×
r
2
Area of ΔACD =
√3
4
r²
Step 4: Area of one circular segment
Area of the 120° sector:
=
120
360
× πr²
=
πr²
3
Therefore,
Area of one segment
=
πr²
3
−
√3r²
4
=
(
π
3
−
√3
4
)r²
Step 5: Area of the common region
The common region consists of two equal circular segments.
Therefore,
Area of common region
=
2
×
(
π
3
−
√3
4
)r²
Hence,
Area =
(
2π
3
−
√3
2
)r²
✅ Final Answer
Area of the region enclosed by the two circles
=
2π
3
r²
−
√3
2
r²
=
(
2π
3
−
√3
2
)r²
square units
End-of-Chapter Exercises • Question 26
Areas of Three Triangles Inside a Rectangle
Proving the area relation using simple area formulas
Question 26
In Fig. 6.54, we see three triangles within a rectangle.
The areas of the triangles are A, B, C, as marked.
Show that the area of the rectangle is
2(A+C)(B+C)
C
.
Fig. 6.54
Three triangles of areas A, B and C inside the rectangle.
Geometry Used in the Solution
Width = x + y and height = h + k.
✍️ Solution
Let
Let the horizontal lengths be x and
y, and the vertical lengths be
h and k.
Width of rectangle = x + y
Height of rectangle = h + k
Step 1: Write the areas of A, B and C
For triangle A,
A =
1
2
× x × h
Therefore,
A =
xh
2
For triangle C,
C =
1
2
× y × h
Therefore,
C =
yh
2
For triangle B,
B =
1
2
× y × k
Therefore,
B =
yk
2
Step 2: Find A + C
A + C =
xh + yh
2
Hence,
A + C =
(x + y)h
2
Step 3: Find B + C
B + C =
yk + yh
2
Hence,
B + C =
y(h + k)
2
Step 4: Substitute in the given expression
Consider
2(A + C)(B + C)
C
Substituting the values,
=
2 ×
(x + y)h
2
×
y(h + k)
2
÷
yh
2
Cancelling common factors,
= (x + y)(h + k)
Step 5: Find the area of the rectangle
Width of the rectangle = x + y
Height of the rectangle = h + k
Therefore,
Area of rectangle
=
(x + y)(h + k)
But from Step 4,
(x + y)(h + k)
=
2(A + C)(B + C)
C
✅ Final Answer
Area of rectangle
=
2(A + C)(B + C)
C
💡 Key Idea
Use
Area = ½ × base × height
for the three triangles. After adding the appropriate pairs of
triangle areas, the required expression simplifies directly to
width × height, which is the area of the rectangle.
End-of-Chapter Exercises • Question 26
Class 9 Maths Chapter 6 End-of-Chapter Exercises Question 26 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
In Fig. 6.54, we see three triangles within a rectangle.
The areas of the triangles are A, B, C,
as marked. Show that the area of the rectangle is
2(A+C)(B+C)
C
.
Given
The areas of the three triangles are
A, B and C.
We need to show that the area of the rectangle is
2(A+C)(B+C)
C
Solution
Step 1 — Express the three triangle areas
Let the horizontal parts be x and
y, and the vertical parts be
h and k.
For triangle A,
A
=
1
2
× x × h
Therefore,
A =
xh
2
For triangle C,
C
=
1
2
× y × h
Therefore,
C =
yh
2
For triangle B,
B
=
1
2
× y × k
Therefore,
B =
yk
2
Step 2 — Find A + C
A + C
=
xh + yh
2
Taking h common,
A + C
=
(x+y)h
2
Step 3 — Find B + C
B + C
=
yk + yh
2
Taking y common,
B + C
=
y(h+k)
2
Step 4 — Substitute in the required expression
2(A+C)(B+C)
C
Substituting the values,
=
2 ×
(x+y)h
2
×
y(h+k)
2
÷
yh
2
On cancelling common factors,
= (x+y)(h+k)
Step 5 — Identify the area of the rectangle
The width of the rectangle is
x + y
and its height is
h + k.
Therefore,
Area of rectangle
=
(x+y)(h+k)
But from Step 4,
(x+y)(h+k)
=
2(A+C)(B+C)
C
Hence proved.
∴
Area of the rectangle
=
2(A+C)(B+C)
C
square units.
Thus, the required result is proved.
💡 Key Idea
Use
Area = ½ × base × height
for the three triangles. After adding the appropriate
areas and substituting them in the required expression,
the common factors cancel and the remaining product is
exactly the area of the rectangle.
End-of-Chapter Exercises • Question 27
Class 9 Maths Chapter 6 End-of-Chapter Exercises Question 27 – Solution
Ganita Manjari 2026–27 | Measuring Space: Perimeter and Area
In the figure we see two shaded regions formed by a quarter
circle, a semicircle, and a triangle.
Show that the areas of the two shaded regions are equal.
Given
OA = OB = OC = r, where
O is the centre of the semicircle
AC.
AB is the diameter of the smaller semicircle, whose centre is D.
F is the common unshaded region.
To Prove:
Area of the two shaded regions are equal
Solution
Step 1 — Find AB
In right triangle AOB,
∠AOB = 90°
By Pythagoras theorem,
AB2
=
OA2 + OB2
Therefore,
AB2
=
r2 + r2
=
2r2
Hence,
AB = r√2
Step 2 — Area of the semicircle on AB
Since AB is the diameter of the smaller
semicircle,
Radius
=
AB
2
=
r√2
2
Therefore,
Area of semicircle
=
1
2
× π
×
(
r√2
2
)2
Hence,
Area of semicircle on AB
=
πr2
4
Step 3 — Area of the quarter circle AOB
The quarter circle AOB has radius
r.
Therefore,
Area of quarter circle AOB
=
1
4
πr2
Thus,
Area of semicircle on AB
=
Area of quarter circle AOB
Step 4 — Compare the shaded regions
Let the common unshaded region be F.
For the semicircle on AB,
Area of semicircle
=
Area of left shaded region
+
Area of F
For the quarter circle AOB,
Area of quarter circle
=
Area of △AOB
+
Area of F
But the areas of the semicircle and quarter circle are equal.
Therefore, subtracting the same area F from
both sides,
Area of left shaded region
=
Area of △AOB
Step 5 — Conclusion
Hence, the areas of the two shaded regions are equal.
Hence proved.
∴
The areas of the two shaded regions are equal.
The semicircle on AB and the quarter circle AOB have equal
areas, and the same unshaded region F is removed from both.
📚 Continue Learning
Congratulations! You have completed the Class 9 Maths Chapter 6 End of Chapter Exercises Solutions. You have now worked through all 27 questions from the End of Chapter Exercises of Chapter 6 – Measuring Space: Perimeter and Area. Continue your learning journey by revisiting the chapter exercises, reviewing the important formulas, or exploring other Class 9 Maths Chapter 6 Solutions.
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❓ Frequently Asked Questions
Find quick answers to common questions about Class 9 Maths Chapter 6 End of Chapter Exercises Solutions. These FAQs cover the important concepts and question types included in the 27 End of Chapter questions, including perimeter and area, triangles, Heron’s formula, circles, trapeziums, kites, rectangles, area ratios and shaded regions. Use them for homework, revision, self-study and CBSE examination preparation.
What is included in the Class 9 Maths Chapter 6 End of Chapter Exercises?
The Class 9 Maths Chapter 6 End of Chapter Exercises contain 27 questions covering the important concepts of Chapter 6 – Measuring Space: Perimeter and Area. The questions include problems based on areas and perimeters of geometric figures, triangles, circles, trapeziums, kites, rectangles, shaded regions and relationships between areas.
How many questions are included in the End of Chapter Exercises?
The End of Chapter exercise set contains 27 questions. The questions provide a complete revision of the major concepts studied in Class 9 Maths Chapter 6 – Measuring Space: Perimeter and Area. The question navigator on this page allows students to jump directly to any question from Q1 to Q27.
What concepts are covered in Chapter 6 End of Chapter Exercises?
The questions revise important concepts from the chapter, including perimeter and area of triangles, Heron’s formula, area and circumference of circles, areas of trapeziums and kites, rectangles and other plane figures, area ratios and shaded regions. Several questions also require students to combine more than one geometrical concept to find the required area.
When is Heron’s formula used in Chapter 6?
Heron’s formula is useful when the three sides of a triangle are known and its area needs to be calculated. If the sides are a, b and c, first find the semi-perimeter s = (a + b + c) / 2, and then use Area = √[s(s − a)(s − b)(s − c)]. Students should check that the given lengths form a valid triangle before applying the formula.
How do we find the area of a circle in Chapter 6?
The area of a circle is calculated using Area = πr², where r is the radius of the circle. If the diameter is given, first use r = d/2 and then substitute the radius in the area formula. The value of π should be taken according to the instruction given in the individual question.
How do we calculate the area of a trapezium?
For a trapezium, the area is calculated using Area = ½ × (sum of parallel sides) × height. If the parallel sides are a and b and the perpendicular height is h, then Area = ½(a + b)h. The height must be the perpendicular distance between the two parallel sides.
How is the area of a kite calculated?
The area of a kite can be calculated using its diagonals: Area = ½ × d₁ × d₂, where d₁ and d₂ are the lengths of its two diagonals. When solving a question, make sure the diagonal lengths are identified correctly from the given figure.
How are shaded region questions solved?
For a shaded region, first identify the complete figure and the unshaded part. Depending on the diagram, the required area can often be found by Area of shaded region = Area of larger figure − Area of unshaded region. For composite figures, calculate each required part separately and combine the areas carefully. The diagram should always be examined before choosing the formula.
How are area ratios used in Chapter 6 questions?
Area ratios are useful when two or more figures are related through their dimensions. First express the required areas using common dimensions or known relationships. Then simplify the ratio carefully. Students should distinguish between a ratio of lengths and a ratio of areas, because areas depend on the square of corresponding linear dimensions.
Are these Class 9 Maths Chapter 6 End of Chapter Exercises Solutions based on Ganita Manjari 2026–27?
Yes. These Class 9 Maths Chapter 6 End of Chapter Exercises Solutions are prepared according to the NCERT Ganita Manjari (2026–27) Chapter 6 exercise material provided for this page. The solutions are arranged question-wise from Q1 to Q27 in a clear, student-friendly format to support classwork, homework, self-study, revision and CBSE examination preparation.
👨🏫 Reviewed & Prepared By
Rakesh Kumar Singh
Mathematics Educator • Founder of
Maths Gurukulam & Newton Study Point
18+ Years of Mathematics Teaching Experience • Teaching Since 2006
These Class 9 Maths Chapter 6 End of Chapter Exercises Solutions are carefully prepared according to the latest NCERT Ganita Manjari (2026–27) and the CBSE curriculum. The solutions cover all 27 questions from the End of Chapter Exercises and follow a clear, step-by-step approach to help students understand and solve problems based on perimeter and area, triangles, Heron’s formula, circles, trapeziums, kites, rectangles, area ratios, shaded regions and related geometry concepts.
📘 NCERT Ganita Manjari (2026)
🎯 CBSE Aligned
📝 Step-by-Step Solutions
💡 Concept-Based Learning
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