# NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area (EX 11.4) Exercise 11.4

Class 7 lays the groundwork for higher levels. As a result, it is an essential class in a student’s life. Only class learning is insufficient in today’s fast-paced world of strong competition. Students should do self-study, acquire all ideas and theories, and understand how to apply the theories in everyday life. Extramarks guides students in achieving their objectives and learning more effectively. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4, developed by specialists, includes all of the Class 7 Maths Chapter 11 Exercise 11.4 Solution. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4 are designed in such a way that students can learn and grasp all of the ideas. The language is basic, grade-appropriate, and simple to follow.

Students must properly understand the subjects to perform well on their tests. They may learn rapidly if they comprehend the concepts properly. Understanding and comprehending concepts, as well as practising questions based on them, are important when it comes to Mathematics. Students may struggle to answer the questions if they do not have access to the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4. As a result, Extramarks provides chapter-by-chapter learning tools like NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4 in PDF format for students. Students may explore all of the subjects and answer questions to deepen their comprehension. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4 are written in simple language that students may grasp and retain.

The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4 available at the Extramarks website are created by highly skilled and experienced subject specialists. The NCERT Solutions for Class 7 Maths, Chapter 11, Exercise 11.4 are compiled to clear up any confusion that students may have while practising questions from the NCERT Mathematics textbook.Students will be able to perform better on their tests in this way. NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4 to every question in the Maths Class 7 Chapter 11 Exercise 11.4 written in simple language that any student can understand.

The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4 are essential, especially for the Class 7th Math Exercise 11.4. Students can be stress-free and prepare for the exam without worry by referring to the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4. Students can also register for Extramarks doubt sessions to get answers to their questions.

All students at Extramarks can access the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4. As a result, students must download and refer to these solutions frequently. They should also look at the other relevant study materials. Students do not have to worry about their study for Mathematics if they utilise the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4 as it will help them understand concepts quickly. Extramarks can provide them with all the solutions. Students should create and stick to a study programme. Every day, they must set aside time for Mathematics. Students must go through the chapters and practise the questions. Students should make full use of the materials on the Extramarks website, and they will be able to complete Class 7 Maths Chapter 11 Exercise 11.4 in no time.Also, reviewing everything before their examinations can help them a lot if they want to get good grades. If students want to cover Class 7 Math Chapter 11 Exercise 11.4, students should pay close attention to the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4. Every student can achieve good results in Class 7 Mathematics examination with consistent work and devotion.

**NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area (EX 11.4) Exercise 11.4 **

Extramarks provides the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4 with a PDF Download option. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4 provides a complete method for mastering all of the chapter’s themes. For CBSE students studying for examinations, the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4 are regarded as the finest option. This chapter contains several exercises. Extramarks provides the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4 in PDF format on its website. Students may study the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4 directly from the Extramarks website or mobile application, or they can download it as needed.

**NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.4**

NCERT Class 7 Mathematics, Chapter 11: Perimeter and Area- This chapter describes the Perimeter and Area of planar figures in depth. The first part of the section is dedicated to Squares and Rectangles. The Area of Parallelograms and Triangles is then calculated using formulae. Circles are described in the second half of the chapter, and students will learn about the Circumference and Area of Circles

The chapter concludes with unit conversion and application-based questions. One can learn how to convert or write one unit in terms of another. Furthermore, because students have learned about multiple formulas, their application becomes vital in situations where more than one formula might be used in the same subject. The NCERT Solutions for Class 7 Maths, Chapter 11, Exercise 11.4 assists students in quickly and easily learning a wide range of ideas.

There are lots of important formulas in NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4. These formulas will help while practising questions with help of the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4

The perimeter of a square

= 4 × side

Perimeter of a rectangle

= 2 × (length + breadth)

Area of a square

= side × side

Area of a rectangle

= length × breadth

Area of a parallelogram

= base × height

Area of a triangle

= ½ × base × height

Circumference of a circle

= 2πr

Area of a circle

= πr2

Extramarks’ NCERT Solutions for Class 7 Maths, Chapter 11, Exercise 11.4 provides answers to all questions as well as assistance in understanding concepts.For students who prefer self-study, the PDF of solutions is strongly suggested. If students still have doubts, Extramarks offers doubt sessions to assist them in resolving these doubts.

NCERT Solutions Class 7 Mathematics Chapters List

Chapter-1 Integers

The first chapter of NCERT Class 7 Mathematics has four tasks. It covers the fundamental notion of Integers as well as the methods for Adding, Subtracting, Multiplying, and Dividing Integers and their characteristics. This chapter discusses Positive and Negative Integer Multiplication as well as the Multiplication of two Negative Integers.

Chapter-2 Fractions And Decimals

This chapter lays the groundwork for Fractions and Decimals. It not only exposes students to the notion of Fractions and Decimals but also provides a thorough understanding of the entire topic. The chapter teaches students how to add, subtract, multiply, divide, and reciprocate Fractions and Decimals. It goes over other Multiplication concepts in depth, such as Multiplication of Fraction by a Fraction, Division of Fraction by a Fraction, Multiplication of Decimal Numbers, Multiplication of a Fraction by a Whole Number, Division of a Whole Number by a Fraction, Division of Fraction by a Fraction, Division of Fraction by a Fraction, and Division of Decimal Numbers.

Chapter-3 Data Handling

This chapter discusses data collection and organisation, as well as the usage of Bar Graphs for various reasons and the concepts of Mean, Median, and Mode. Arithmetic Mean and Mode of Large Data, as well as Chance and Probability, are taught to students.

Chapter-4 Simple Equations

Students study the definition of Equations and the numerous methods for solving Equations in this chapter. This chapter also teaches students how to create equations from solutions and how to apply simple equations to actual problems.

Chapter-5 Lines And Angles

This chapter covers a variety of Angle types, including Complementary Angles, Supplementary Angles, Adjacent Angles, Linear Pairs, and Vertically Opposite Angles. It discusses the concepts of Lines and Pairs Of Lines, such as Intersecting Lines, Transversals, Angles formed by a Transversal, and Transversal of Parallel Lines. Students will also learn how to check for Parallel Lines.

Chapter-6 The Triangle And Its Properties

This chapter exposes students to several sorts of Triangles, including the Equilateral Triangle, Isosceles Triangle, and Right-Angled Triangle. It also covers the Medians and Altitudes of a Triangle, Sums of the Lengths of Two Sides of a Triangle, the Angle Sum Property of a Triangle, Angle and its property, and the Pythagoras Property.

Chapter-7 Congruence Of Triangles

This chapter is a continuation of the preceding one. It is concerned with the Congruence of Numerous Triangles, as well as the Congruence of Plane Figures, Line Segments, and Angles. It goes over the requirements for Triangle Congruence.

Chapter-8 Comparing Quantities

This chapter teaches students how to compare amounts using Percentages, as well as how to convert Fraction Numbers into Percentages, Decimals to Percentages, Percentages to Fractions or Decimals, Ratios to Percentages, and Percentages to “How Many.” It also summarises Equivalent Ratios and teaches students how to represent an increase or decrease and Profit or Loss as Percentages and interpret Percentages.

Chapter-9 Rational Numbers

Students study the concept and importance of Rational Numbers, Positive and Negative Rational Numbers, and Rational Number Operations such as Addition, Subtraction, Multiplication, and Division. It also teaches students how to compare Rational Numbers as well as the Rational Numbers that exist between two Rational Numbers. It includes a visual depiction of Rational Numbers on a number line as well as instructions on how to recognise them.

Chapter-10 Practical Geometry

NCERT Grade 7 Mathematics, Chapter 10: Practical Geometry: This chapter teaches the fundamentals of drawing parallel lines and constructing triangles.The chapter discusses the creation of a line parallel to a given line that passes through a location that is not on the line. A well-labelled diagrammatic exposition of the topic is provided.

Chapter-11 Perimeter And Area

This chapter teach students how to find the Length, Width, and Area of Squares, Rectangles, the Circumference and Area of a Circle, the Area of a Parallelogram and Triangle, and their Applications, as well as Unit Conversion. It also shows Triangles as parts of Rectangles and generalises to other congruent Rectangles parts. To properly understand the material, students must refer to the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4.

Chapter-12 Algebraic Expression

In this chapter, students will study how expressions are constructed as well as expression terminology such as Like and Unlike terms.Students also learn how to discover the value of an expression, how to use formulas and rules in Algebraic Expressions, and how to add and subtract Algebraic Expressions. It is concerned with, as well as other Algebra concepts such as Monomials, Binomials, Trinomials, and Polynomials.

Chapter-13 Exponents And Power

This chapter introduces students to the concept of Exponents and the principles that govern them. It covers Multiplying and Dividing Powers with the Same Base, the Decimal Number System, Taking Power of a Power, and the Standard Form Expression of Large Numbers. This chapter contains Miscellaneous Examples Using Exponent Laws.

Chapter-14 Symmetry

The chapter will explain what Lines of Symmetry signify for a figure. It goes over the Lines of Symmetry for Regular Polygons, Line Symmetry, and Rotational Symmetry in great depth.

Chapter-15 Visualising Solid Shapes

This chapter provides an overview of Plane Figures and Solid Figures. It also assists students with visualising and understanding the faces, edges, and vertices of forms, as well as different ways to observe distinct areas of an item by cutting or slicing or shadow play. It shows a graphical depiction of various forms and nets for constructing 3-D shapes. Students must sketch solids on a level surface using Oblique Sketches, Isometric Sketches, and Visualising Solid Objects in this chapter. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4 for Class 7th Math 11.4 are chosen by experts to cover all of the important concepts.

**Exercise 11.4**

The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4 Perimeter and Area are given in simple PDF format on the Extramarks’ official website. Unit conversion and applications are included in the NCERT Solutions for Class 7 Maths, Chapter 11, Exercise 11.4.Students have probably seen applications like this in real life, such as in gardens or parks, where some space is left all around in the shape of a path or in between as cross paths.Students may improve their Mathematics scores by working through the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4.

Students are advised to use the solved examples inthis part to develop a better understandingof unit conversions and how they are used in computations. Furthermore, the solved examples in the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4, will give additional information and ideas for problem-solving approaches. A full understanding of ideas is required for studying; as a consequence, a review of prior subjects will always give more insight into deciding the best strategy to tackle any of the problems.

By studying the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4, students can improve their theoretical and practical knowledge. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4 includes every question from the Maths Class 7 Chapter 11 Exercise 11.4. In addition to the solutions, there are ideas and tips in these tools, that would make responding to inquiries easier. Students can acquire the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4 in PDF format from Extramarks. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4 covers all topics related to sound and those covered in the curriculum.

**NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.4**

When calculating the Area and Perimeter of figures, students will come across a variety of units. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4, shows how to convert units when it comes to the area since the area is expressed in the unit square, such as cm2, m2, and so on. Students should be comfortable converting length from one unit to another; similarly, they will become acquainted with converting area units. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4, has solutions, some of which include two figures overlapping and students being asked to calculate the Area and Perimeter of the overlapping or non-overlapping areas.

When dealing with unit conversion, students will notice that converting the area into a smaller unit, such as metres to centimetres or centimetres to millimetres, results in a larger number of units, such as 1000cm2 resulting in 100000mm2. There are 11 questions in Class 7 Math Chapter 11 Exercise 11.4 and the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4 contains accurate solutions for every question.

Mathematics is one of the most important courses for students of all grades. It is critical to raising the student’s overall grade. The Mathematics syllabus is extremely lengthy, and many students struggle to follow the procedures and ideas. Students are perplexed by the extensive syllabus. However, with the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4, each step of the solution is carefully described to assist students to obtain a comprehensive understanding of the issue they are confronted with. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4 enables students to grasp each concept.Students can use this strategy to answer any question from Class 7 Maths Chapter 11 Exercise 11.4.

Students find it difficult to determine the proper approach and apply methods in Mathematics. Mathematics self-study can be time-consuming and frustrating. Students must have a thorough comprehension of the ideas and formulas. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.4 assists in alleviating the load. Extramarks gives prepared solutions, and each chapter is meticulously separated into several portions. Step-by-step answers are offered for each subject, guaranteeing students have a better grasp and achieve higher grades.

**Q.1 ****A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectare.**

**Ans**

\begin{array}{l}\text{\hspace{0.17em}}\text{Length (L) of garden}=\text{90 m}\\ \text{Breadth (B) of garden}=\text{75 m}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Area of garden}=\text{L}\times \text{B}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{90 m}\times \text{75 m}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{6750 m}}^{2}\end{array}

\begin{array}{l}\text{From the figure, it can be observed that the new length}\\ \text{and breadth of the garden,when path is also included are}\\ \text{100 m and 85 m respectively}\\ \text{Area of the garden including path}=\text{100 m}\times \text{80 m}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{8000 m}}^{2}\\ \text{Area of the path}\\ =\text{Area of the garden including path}-\text{Area of garden}\\ =8000{\text{m}}^{2}-{\text{6750 m}}^{2}\\ =1750{\text{m}}^{2}\\ 1\text{hectare}={\text{10000 m}}^{2}\\ \text{Therefore,}\\ \text{area of the garden in hectare}=\frac{6750}{10000}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.675\text{hectare}\end{array}

**Q.2 ****A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65m. Find the area of the path.**

**Ans**

\begin{array}{l}\text{\hspace{0.17em}}\text{Length (L) of park}=\text{125 m}\\ \text{Breadth (B) of park}=\text{65 m}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Area of park}=\text{L}\times \text{B}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{125 m}\times \text{65 m}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{8125 m}}^{2}\end{array}

\begin{array}{l}\text{From the figure, it can be observed that the new length}\\ \text{and breadth of the park,when path is also included are}\\ \text{131 m and 71 m respectively}\\ \text{Area of the park including path}=\text{131 m}\times \text{71 m}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}={\text{9301 m}}^{2}\\ \text{Area of the path}\\ =\text{Area of the park including path}-\text{Area of park}\\ =9301{\text{m}}^{2}-{\text{8125 m}}^{2}\\ =1176{\text{m}}^{2}\end{array}

**Q.3 ****A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.**

**Ans**

\begin{array}{l}\text{\hspace{0.17em}}\text{Length (L) of cardboard}=\text{125 m}\\ \text{Breadth (B) of cardboard}=\text{65 m}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Area of cardboard including margin}=\text{L}\times \text{B}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{8 cm}\times \text{5 cm}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=40{\text{cm}}^{2}\end{array}

\begin{array}{l}\text{From the figure, it can be observed that the new length}\\ \text{and breadth of the cardboard,when margin is not included}\\ \text{are 5 cm and 2 cm respectively}\\ \text{Area of the cardboard not including margin}=\text{5 cm}\times \text{2 m}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{10 cm}}^{2}\\ \text{Area of the margin}\\ =\text{Area of the cardboard including margin}\\ -\text{Area of cardboard not including margin}\\ =40{\text{cm}}^{2}-{\text{10 cm}}^{2}\\ =30{\text{cm}}^{2}\end{array}

**Q.4 **

$\begin{array}{l}\mathrm{A}\mathrm{verandah}\mathrm{of}\mathrm{width}2.25\mathrm{m}\mathrm{is}\mathrm{constructed}\mathrm{all}\mathrm{along}\\ \mathrm{outside}\mathrm{a}\mathrm{room}\mathrm{which}\mathrm{is}5.5\mathrm{m\; long}\mathrm{and}4\mathrm{m}\mathrm{wide}.\mathrm{Find}:\\ \left(\mathrm{i}\right)\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{verandah}.\\ \left(\mathrm{ii}\right)\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{cementing}\mathrm{the}\mathrm{floor}\mathrm{of}\mathrm{the}\mathrm{verandah}\mathrm{at}\mathrm{the}\\ \mathrm{rate}\mathrm{of}\mathrm{Rs}200\mathrm{per}{\mathrm{m}}^{\mathrm{2}}\end{array}$

**Ans**

\begin{array}{l}\text{(i)}\\ \text{Length (L) of room}=\text{5}\text{.5 m}\\ \text{Breadth (B) of room}=\text{4 m}\\ \text{Area of room including margin}=\text{L}\times \text{B}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{5}\text{.5 m}\times \text{4 m}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=22{\text{m}}^{2}\end{array}

\begin{array}{l}\text{From the figure, it can be observed that the new length}\\ \text{and breadth of the room,when verandah is also included}\\ \text{are 10 m and 8}\text{.5 m respectively}\\ \text{Area of the verandah including margin}=\text{10 m}\times \text{8}\text{.5 m}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{85 m}}^{2}\\ \text{Area of the verandah}\\ =\text{Area of the room including verandah}-\text{Area of the room}\\ =85{\text{m}}^{2}-{\text{22 m}}^{2}\\ =63{\text{m}}^{2}\end{array} \begin{array}{l}\text{}\end{array}

\begin{array}{l}\text{(ii)}\\ {\text{Cost of cementing 1m}}^{\text{2}}\text{area of floor of verandah}=\u20b9200\\ {\text{Cost of cementing 63m}}^{\text{2}}\text{area of floor of verandah}=\u20b9200\times 63\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}=\u20b912600\end{array}

**Q.5 **

$\begin{array}{l}\mathrm{A}\mathrm{path}1\mathrm{m}\mathrm{wide}\mathrm{is}\mathrm{built}\mathrm{along}\mathrm{the}\mathrm{border}\mathrm{and}\mathrm{inside}\mathrm{a}\\ \mathrm{square}\mathrm{garden}\mathrm{of}\mathrm{side}30\mathrm{m}.\mathrm{Find}:\\ \left(\mathrm{i}\right)\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{path}\\ \left(\mathrm{ii}\right)\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{planting}\mathrm{grass}\mathrm{in}\mathrm{the}\mathrm{remaining}\mathrm{portion}\mathrm{of}\mathrm{the}\\ \mathrm{garden}\mathrm{at}\mathrm{the}\mathrm{rate}\mathrm{of}\u20b940\mathrm{per}{\mathrm{m}}^{\mathrm{2}}\mathrm{.}\end{array}$

**Ans**

\begin{array}{l}\text{(i)}\\ \text{Side (a) of square garden}=\text{30 m}\\ \text{Area of square garden including path}={\text{a}}^{2}\text{}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left(\text{30 m}\right)}^{2}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{900 m}}^{2}\\ \text{From the figure, it can be observed that the side of the square}\\ \text{garden, when path is not included, is 28 m}\text{.}\\ \text{Area of the square garden not including the path}={\left(28\right)}^{2}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}={\text{784 m}}^{2}\\ \text{Area of path}\\ =\text{Area of the square garden including the path}\\ -\text{Area of square garden not including the path}\\ =900{\text{m}}^{2}-784{\text{m}}^{2}\\ =116{\text{m}}^{2}\\ \text{(ii)}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{Cost of planting grass in 1 m}}^{2}\text{area of the garden}=\text{\u20b940}\\ {\text{Cost of planting grass in 784 m}}^{2}\text{area of the garden}=\text{\u20b940}\times \text{784}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{\u20b931360}\end{array}

**Q.6 **

$\begin{array}{l}\mathrm{Two}\mathrm{cross}\mathrm{roads},\mathrm{each}\mathrm{of}\mathrm{width}10\mathrm{m},\mathrm{cut}\mathrm{at}\mathrm{right}\mathrm{angles}\\ \mathrm{through}\mathrm{the}\mathrm{center}\mathrm{of}\mathrm{a}\mathrm{rectangular}\mathrm{park}\mathrm{of}\mathrm{length}700\mathrm{m}\\ \mathrm{and}\mathrm{breadth}300\mathrm{m}\mathrm{and}\mathrm{parallel}\mathrm{to}\mathrm{its}\mathrm{sides}.\mathrm{Find}\mathrm{the}\mathrm{area}\\ \mathrm{of}\mathrm{the}\mathrm{roads}.\mathrm{Also}\mathrm{find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{park}\mathrm{excluding}\mathrm{cross}\\ \mathrm{roads}.\mathrm{Give}\mathrm{the}\mathrm{answer}\mathrm{in}\mathrm{hectares}.\end{array}$

**Ans**

\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Length (L) of the park}=\text{700 m}\\ \text{Breadth (B) of the park}=\text{300 m}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Area of park}=\text{700}\times \text{300}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}={\text{210000 m}}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Length of road PQRS}=\text{700 m}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Length of road ABCD}=\text{300 m}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Width of each road}=\text{10 m}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Area of the roads}=\text{area(PQRS)+area(ABCD)}-\text{area(KLMN)}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(700\times 10\right)+\left(300\times 10\right)-\left(10\times 10\right)\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=7000+3000-100\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=9900{\text{m}}^{2}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.99\text{hectare}\end{array}

\begin{array}{l}\\ \text{Area of park excluding roads}=\text{210000-9900}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{200100 m}}^{2}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=20.01\text{hectare}\end{array}

**Q.7 **

$\begin{array}{l}\mathrm{Through}\mathrm{a}\mathrm{rectangular}\mathrm{field}\mathrm{of}\mathrm{length}90\mathrm{m}\mathrm{and}\mathrm{breadth}\\ 60\mathrm{m},\mathrm{two}\mathrm{roads}\mathrm{are}\mathrm{constructed}\mathrm{which}\mathrm{are}\mathrm{parallel}\mathrm{to}\mathrm{the}\\ \mathrm{sides}\mathrm{and}\mathrm{cut}\mathrm{each}\mathrm{other}\mathrm{at}\mathrm{right}\mathrm{angles}\mathrm{through}\mathrm{the}\mathrm{centre}\\ \mathrm{of}\mathrm{the}\mathrm{fields}.\mathrm{If}\mathrm{the}\mathrm{width}\mathrm{of}\mathrm{each}\mathrm{road}\mathrm{is}3\mathrm{m},\mathrm{find}\\ \left(\mathrm{i}\right)\mathrm{the}\mathrm{area}\mathrm{covered}\mathrm{by}\mathrm{the}\mathrm{roads}.\\ \left(\mathrm{ii}\right)\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{constructing}\mathrm{the}\mathrm{roads}\mathrm{at}\mathrm{the}\mathrm{rate}\mathrm{of}\\ \u20b9110\mathrm{per}{\mathrm{m}}^{\mathrm{2}}\mathrm{.}\end{array}$

**Ans**

\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Length (L) of the field}=\text{90 m}\\ \text{Breadth (B) of the field}=\text{60 m}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Area of field}=\text{90}\times \text{60}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}={\text{5400 m}}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Length of road PQRS}=\text{90 m}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Length of road ABCD}=\text{60 m}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Width of each road}=\text{3 m}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Area of the roads}=\text{area(PQRS)+area(ABCD)}-\text{area(KLMN)}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(90\times 3\right)+\left(60\times 3\right)-\left(3\times 3\right)\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=270+180-100\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=441{\text{m}}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{Cost for constructing 1 m}}^{2}\text{road}=\text{\u20b9110}\\ {\text{Cost for constructing 4411 m}}^{2}\text{road}=\text{\u20b9110}\times 441\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\u20b948510\end{array}

**Q.8 ****Pragya wrapped a cord around a circular pipe of radius 4 cm (adjoining figure) and cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown). Did she have any left? (π = 3.14)**

**Ans**

\begin{array}{l}\text{Perimeter of the square}=\text{4}\times \text{side of the square}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{4}\times \text{4}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{16 cm}\\ \text{Perimeter of circular pip}=\text{2}\pi \text{r}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{3}\times \text{3}\text{.14}\times \text{4}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{25}\text{.12 cm}\\ \text{Therefore,}\\ \text{length of chord left with Pragya}=\text{25}\text{.12 cm}-\text{16 cm}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}=\text{9}\text{.12 cm}\end{array}

**Q.9 ****The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find:
(i) the area of the whole land
(ii) the area of the flower bed
(iii) the area of the lawn excluding the area of the flower bed.**

**(iv) the circumference of the flower bed.**

**Ans**

\begin{array}{l}\text{(i) Area of whole land}=\text{Length}\times \text{Breadth}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{10}\times \text{5}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{50 m}}^{2}\\ \text{(ii) Area of flower bed}=\pi {\text{r}}^{2}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=3.14\times 2\times 2\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=12.56{\text{m}}^{2}\\ \text{(iii) Area of the lawn excluding the flower bed}\\ =\text{Area of whole land}-\text{Area of flower bed}\\ =\text{50-12}\text{.56}\\ =\text{37}{\text{.44 m}}^{2}\end{array}

**Q.10 ****In the following figures, find the area of the shaded portions:**

**Ans**

\begin{array}{l}\text{(i)}\\ \text{Area of EFDC}=\text{area(ABCD)}-\text{area(BCE)}\u2013\text{area(AFE)}\\ \text{}\text{}\text{}=\left(18\times 10\right)-\frac{1}{2}\left(10\times 8\right)-\frac{1}{2}\left(6\times 10\right)\\ \text{}\text{}\text{}=180-40-30\\ \text{}\text{}\text{}=110{\text{cm}}^{2}\\ \text{(ii)}\\ \text{area(QTU)}=\text{area(PQRS)}-\text{area(TSU)}-\text{area(RUQ)}-\text{area(PQT)}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(20\times 20\right)-\frac{1}{2}\left(10\times 10\right)-\frac{1}{2}\left(20\times 10\right)-\frac{1}{2}\left(20\times 10\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=400-50-100-100\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=150{\text{cm}}^{2}\end{array}

**Q.11 **

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{quadrilateral}\mathrm{ABCD}.\\ \mathrm{Here},\mathrm{AC}=22\mathrm{cm},\mathrm{BM}=3\mathrm{cm},\mathrm{DN}=3\mathrm{cm},\\ \mathrm{and}\mathrm{BM}\perp \mathrm{AC},\mathrm{DN}\perp \mathrm{AC}.\end{array}$

**Ans**

\begin{array}{l}\text{area(ABCD)}=\text{area(ABC)+area(ADC)}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}\left(3\times 22\right)\frac{1}{2}\left(3\times 22\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=33+33\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=66{\text{cm}}^{2}\end{array}