# NCERT Solutions Class 6 Maths Chapter 10

## NCERT Solutions for Class 6 Maths Chapter 10 – Mensuration

Students wanting to score well in Maths must solve all the sums given in their textbook exercises to get the required practice. For this, students need a reliable guide. Extramarks offers NCERT Solutions for Class 6 Maths Chapter 10 to help students with their Maths preparation. These are clear and concise solutions to all the questions given in their textbook exercises and prepared by subject matter experts with substantial experience in their respective fields. Students can use these whenever they are stuck on an exercise problem. They can also refer to these solutions to understand the approach of solving different kinds of questions. Access these solutions from the download links below and take a step closer to acing your Maths exam.

## NCERT Solutions for Class 6 Maths Chapter 10 – Mensuration Other Related Chapters

### NCERT Solutions Class 6 Maths Chapter 10 Mensuration

The art of measuring is known as mensuration. It is concerned with measurement, specifically the derivation and application of algebraic formulae to calculate the areas, volumes, and various parameters of geometric figures. The chapter also has numerous questions with practical applications, making it critical for students to answer them.

Extramarks provides accurate and reliable NCERT Solutions for Class 6 Maths Chapter 10 prepared by subject matter experts. Students can use NCERT Solutions for Class 6 Maths Chapter 10 as a guide to prepare answers for questions on Mensuration. Every NCERT Solution is offered to make studying simple and enjoyable. Students can access these solutions in both online and offline modes at any time and anywhere.

### Important Topics Under NCERT Solutions for Class 6 Maths Chapter 10

Mensuration is covered in Class 6 Maths Chapter 10. This chapter is further divided into three subtopics.  The following lists out the topics addressed in this chapter.

• Introduction
• Perimeter
1. Perimeter of Rectangle
2. Perimeter of Regular Figures
• Area
1. Area of Rectangle
2. Area of Square

Extramarks recommends that students read through each of these sections thoroughly before attempting and practising the NCERT Solutions for this chapter on Mensuration.

### NCERT Solutions for Class 6 Maths Chapter 10 – Mensuration

Students will study the concepts of perimeter and area in NCERT Class 6 Chapter 10 Mensuration. This chapter covers a lot of ground when it comes to Perimeter and Area. The perimeters of a rectangle, square, and equilateral triangle will be discussed. They will also learn how to calculate the area of a rectangle and a square.

The following are the subdivisions of the chapter.

#### 10.1 Introduction

Students are reminded of the concept of plain shapes that they learned in third grade. They will then learn about the boundaries, perimeters of the plane figures as well as how to compare two or more figures in measuring terms.

#### 10.2 Perimeter

This section of Chapter 10 of Maths Class 6 discusses perimeter. Perimeter can be described as the distance covered along the boundary forming a closed figure when you go around the figure once. It also provides a thorough explanation by displaying figures and covering them with a wire or thread.

#### 10.2.1 Perimeter of Rectangle

Because students understand that opposing sides of a rectangle are equal, they can simply calculate the perimeter of a rectangle by measuring both sides, one of which is called the length and the other is called the breadth. The formula to calculate the perimeter of a rectangle is:

#### 10.2.2 Perimeter of Regular figures

In this section, students will learn to determine the perimeter of regular shapes that they may encounter in their environment. It may be a square, rectangle, triangle, or other shape. Students are taught about different formulas for finding the perimeter of each shape. If you wish to find the circumference of a square, for example, it has four equal sides. The square’s perimeter will then be,

Perimeter = 4 times the side length

Similarly, if students want to calculate the perimeter of a triangle, particularly an equilateral triangle with all three sides equal, the perimeter of an equilateral triangle is calculated as follows:

Perimeter = 3 times the length of a side.

#### 10.3 Area

This section covers the calculations to determine the area of various figures. Students can learn what the area of a shape is and how to calculate it for various forms here. The area of a closed figure is the amount of surface it covers.

#### 10.3.1 Area of Rectangle

In this section, students will learn to find the area of a rectangle by analysing the length and breadth numbers. The area of the rectangle, like the perimeter, can be computed using the formula:

Area of Rectangle = length x breadth.

#### 10.3.2 Area of Square

In this section, students will learn to find the area of a square. A graph paper or plain paper can be used to calculate the area of a square. They can calculate the area of a square by utilising the formula below.

Area of a square = length of the side x length of the side.

### NCERT Solutions for Class 6 Maths Chapter 10 Exercises are given in the table below:

 Chapter 10 – Mensuration Exercises Exercise 10.1 17 Questions & Solutions Exercise 10.2 1 Questions & Solutions Exercise 10.3 12 Questions & Solutions

### Key Features of NCERT Solutions Class – 6 Maths Chapter – 10

For students, NCERT Solutions is a wonderful opportunity.

• Our NCERT Solutions for Class 6 Maths Chapter 10 are well-structured and follow the guidelines of CBSE.
• Subject matter experts write these answers which make these NCERT solutions credible and precise.
• Students can access these solutions as per their convenience from Extramarks.

Q.1 Find the perimeter of each of the following figures:

Ans-

(a) Perimeter of figure = 4 cm + 2 cm + 1 cm
+ 5 cm
= 12 cm
(b) Perimeter of figure = 23 cm + 35 cm + 35 cm
+ 40 cm
= 133 cm
(c) Perimeter of figure = 15 cm + 15 cm + 15 cm
+ 15 cm
= 60 cm
(d) Perimeter of figure = 4 cm + 4 cm + 4 cm
+ 4 cm + 4 cm
= 20 cm

(e) Perimeter of figure = 2.5 cm + 0.5 cm + 4 cm
+ 1 cm + 4 cm + 0.5 cm
+ 2.5 cm
= 15 cm
(f) Perimeter of figure = 4 cm + 1 cm + 3 cm
+ 2 cm+ 3 cm+ 4 cm+ 1 cm + 3 cm
+ 2 cm + 3 cm + 4 cm + 1 cm + 3 cm
+ 2 cm + 3 cm + 4 cm + 1 cm + 3 cm
+ 2 cm + 3 cm
= 52 cm

Q.2 The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?

Ans-

Since, lid of rectangular box is like a rectangle. So,

Length of tape = 2(length + Breadth)
= 2(40 + 10)
= 100 cm
= 1 m

Q.3 A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?

Ans-

Length of table-top = 2 m 25 cm

= 2.25 m
Breadth of table-top = 1 m 50 cm

= 1.5 m
Perimeter of table-top = 2(Length + Breadth)

= 2(2.25 + 1.50)
= 2(3.75)
= 7.50 m

Q.4 What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?

Ans-

Length of photograph = 32 cm
Breadth of photograph = 21 cm
Perimeter of photograph = 2(Length + Breadth)
= 2(32 + 21)
= 106 cm

Q.5 A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?

Ans-

Length of rectangular piece of land = 0.7 km
Breadth of rectangular piece of land = 0.5 km
Perimeter of rectangular piece of land = 2(0.7 + 0.5)
= 2(1.2) km
= 2.4 km

Length of required wire = 4(2.4) km
= 9.6 km
[Since, Length of required wire = 4(Perimeter of land)]

Q.6 Find the perimeter of each of the following shapes:
(a) A triangle of sides 3 cm, 4 cm and 5 cm.
(b) An equilateral triangle of side 9 cm.
(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.

Ans-

(a) Perimeter of triangle = 3 cm + 4 cm + 5 cm
= 12 cm
(b) Perimeter of equilateral triangle
= 9 cm + 9 cm + 9 cm
= 27 cm
(c) Perimeter of isosceles triangle
= 8 cm + 8 cm + 6 cm
= 22 cm

Q.7 Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.

Ans-

Perimeter of triangle = 10 cm + 14 cm + 15 cm
= 39 cm

Q.8 Find the perimeter of a regular hexagon with each side measuring 8 m.

Ans-

Perimeter of regular hexagon = 6(Length of side)
= 6(8 cm)
= 48 cm

Q.9 Find the side of the square whose perimeter is 20 m.

Ans-

$\begin{array}{l}\mathrm{Perimeter}\text{of square}=\text{4}×\text{side}\\ \text{\hspace{0.17em}\hspace{0.17em}Side of square}=\frac{\mathrm{Perimeter}}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}Side of square}=\frac{20\text{\hspace{0.17em}}\mathrm{m}}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5\text{\hspace{0.17em}}\mathrm{m}\end{array}$

Q.10 The perimeter of a regular pentagon is 100 cm. How long is its each side?

Ans-

$\begin{array}{l}\mathrm{Perimeter}\text{of regular pentagon}=\text{5}×\text{side}\\ \text{\hspace{0.17em}Side of regular pentagon}=\frac{\mathrm{Perimeter}}{5}\\ \text{\hspace{0.17em}\hspace{0.17em}Side of regular pentagon}=\frac{100\text{\hspace{0.17em}}}{4}\mathrm{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=25\text{\hspace{0.17em}}\mathrm{cm}\end{array}$

Q.11 A piece of string is 30 cm long. What will be the length of each side if the string is used to form :
(a) a square?
(b) an equilateral triangle?
(c) a regular hexagon?

Ans-

$\begin{array}{l}\text{Length of a piece of string}=\text{3}0\text{m}\mathrm{}\\ \left(\text{a}\right)\mathrm{Perimeter}\text{of square}=\text{Length of a piece of string}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} 4}×\mathrm{side}\text{of square}=\text{30 m}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}side of square}=\frac{30}{4}\text{\hspace{0.17em}}\mathrm{m}\\ =7.5\text{\hspace{0.17em}}\mathrm{m}\\ \left(\text{b}\right)\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{Perimeter}\text{of an equilateral triangle}\\ =\text{Length of a piece of string}\\ \text{3}×\mathrm{side}\text{of equilateral triangle}\\ =\text{30 m}\\ \text{Side of equilateral triangle}\\ =\frac{30}{3}\text{\hspace{0.17em}}\mathrm{m}\\ =10\text{\hspace{0.17em}}\mathrm{m}\\ \left(\text{c}\right)\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{Perimeter}\text{of a regular hexagon}\\ =\text{Length of a piece of string}\\ \text{6}×\mathrm{side}\text{of regular hexagon}\\ =\text{30 m}\\ \text{Side of regular hexagon}\\ =\frac{30}{6}\text{\hspace{0.17em}}\mathrm{m}\\ =5\text{\hspace{0.17em}}\mathrm{m}\end{array}$

Q.12 Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?

Ans-

Let third side of triangle be x cm.
Perimeter of triangle = 12 cm + 14 cm + x cm
36 cm = 26 cm + x
x = 36 cm – 26 cm
= 10 cm
Thus, third side of triangle is 10 cm.

Q.13 Sweety runs around a square park of side 75m. Bulbul runs around a rectangular park with length 60 m and breadth 45 m. Who covers less distance?

Ans-

Side of square park = 75 m
Perimeter of square park = 4 (75 m)
= 300 m
Distance covered by Sweety =300 m
Length of park = 60 m
Breadth of park = 45 m
Perimeter of rectangular park = 2 (60 m + 45 m)
= 210 m
Distance covered by Bulbul = 210 m
Thus, Bulbul covers less distance.

Q.14 What is the perimeter of each of the following figures? What do you infer from the answers?

Ans-

(a) Perimeter of square = 4(25 cm)
= 100 cm
(b) Perimeter of rectangle = 2(30 cm + 20 cm)
= 2(50 cm)
= 100 cm
(c) Perimeter of rectangle = 2(40 cm + 10 cm)
= 2(50 cm)
= 100 cm
(d) Perimeter of triangle = 30 cm + 30 cm+ 40 cm
= 100 cm

Q.15

Ans-

$\begin{array}{l}\mathrm{Length}\text{of each side of square paving}=\frac{1}{2}\text{\hspace{0.17em}}\mathrm{m}\\ \mathrm{Number}\text{of square paving}=\text{9}\\ \text{When square paving laid in the form of square,}\\ \text{Number of square paving in each side of square form}\\ \text{3}\\ \text{So, the length of each side of new square}\\ \text{\hspace{0.17em}\hspace{0.17em}}=3×\left(\mathrm{side}\mathrm{of}\mathrm{square}\mathrm{paving}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=3×\frac{1}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{3}{2}\\ \left(\mathrm{a}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Perimeter}\text{of new square}=4×\mathrm{side}\\ \text{\hspace{0.17em}\hspace{0.17em}}=4×\frac{3}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}}=6\text{\hspace{0.17em}}\mathrm{m}\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Perimeter}\text{of fig}\left(\mathrm{ii}\right)=4×\left(\mathrm{side}\mathrm{having}\mathrm{single}\mathrm{paving}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}}8×\left(\text{side having double}\mathrm{paving}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=4×\frac{1}{2}+8×\frac{1}{2}×2\\ \text{\hspace{0.17em}\hspace{0.17em}}=2+8\\ \text{\hspace{0.17em}\hspace{0.17em}}=10\text{\hspace{0.17em}}\mathrm{m}\\ \left(\mathrm{c}\right)\text{Cross figure i.e., fig}\left(\mathrm{ii}\right)\text{has greater perimeter.}\\ \left(\mathrm{d}\right)\text{Yes, if we put all paving in a straight line. It forms}\\ \text{a rectangle whose dimensions are:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Length}=9×\frac{1}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}}=4.5\text{\hspace{0.17em}}\mathrm{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Breadth}=\frac{1}{2}\text{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\text{0.5\hspace{0.17em}cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Perimeter}\text{of rectangle}=2\left(\mathrm{L}+\mathrm{B}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=2\left(4.5+0.5\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=10\text{​\hspace{0.17em}}\mathrm{cm}\end{array}$

Q.16 Find the areas of the following figures by counting square:

Ans-

(a) Number of squares in given figure = 9
So, area of given figure = 9 Sq. unit

(b) Number of squares in given figure = 5
So, area of given figure = 5 Sq. unit

(c) Number of squares in given figure = 2 + 4(1/2)

= 4
So, area of given figure = 4 Sq. unit

(d) Number of squares in given figure = 8
So, area of given figure = 8 Sq. unit

(e) Number of squares in given figure = 10
So, area of given figure = 10 Sq. unit

(f) Number of squares in given figure = 2 + 4(1/2)

= 4
So, area of given figure = 4 Sq. unit

(g) Number of squares in given figure = 4 + 4(1/2)

= 6
So, area of given figure = 6 Sq. unit

(h) Number of squares in given figure = 5
So, area of given figure = 5 Sq. unit

(i) Number of squares in given figure = 9
So, area of given figure = 9 Sq. unit

(j) Number of squares in given figure = 2 + 4(1/2)

= 4
so, area of given figure = 4 Sq. unit

(k) Number of squares in given figure = 4 + 2(1/2)

= 5
So, area of given figure = 5 Sq. unit

(l)

 Covered area Number Area estimate (sq units) Fully filled squares 2 2 Half filled squares 2 2(1/2) = 1 More than filled squares 5 5 Less than filled squares 2 0

So, the area of given figure = 2+1+5
= 8 square cm.

(m)

 Covered area Number Area estimate (sq units) Fully filled squares 7 2 Half filled squares 2 2(1/2) = 1 More than filled squares 6 6 Less than filled squares 8 0

So, the area of given figure = 7+1+6
=14 square cm.

(n)

 Covered area Number Area estimate (sq units) Fully filled squares 9 9 Half filled squares 4 4(1/2) = 2 More than filled squares 7 7 Less than filled squares 8 0

So, the area of given figure = 9+2+7
=18 square cm.

Q.17 Find the areas of the rectangles whose sides are:
(a) 3 cm and 4 cm (b) 12 m and 21 m
(c) 2 km and 3 km (d) 2 m and 70 cm

Ans-

$\begin{array}{l}\because \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Area}\text{of rectangle}=\mathrm{Length}×\mathrm{Breadth}\\ \left(\mathrm{a}\right)\mathrm{Area}\text{of rectangle}=3\text{\hspace{0.17em}}\mathrm{cm}×4\text{\hspace{0.17em}}\mathrm{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=12{\mathrm{cm}}^{2}\\ \left(\mathrm{b}\right)\mathrm{Area}\text{of rectangle}=12\text{\hspace{0.17em}}\mathrm{cm}×21\text{\hspace{0.17em}}\mathrm{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=252\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \left(\mathrm{c}\right)\mathrm{Area}\text{of rectangle}=2\text{\hspace{0.17em}}\mathrm{km}×3\text{\hspace{0.17em}}\mathrm{km}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6\text{\hspace{0.17em}}{\mathrm{km}}^{2}\\ \left(\mathrm{d}\right)\mathrm{Length}\text{of rectangle}=2\text{\hspace{0.17em}}\mathrm{m}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=200\text{\hspace{0.17em}}\mathrm{cm}\\ \mathrm{Area}\text{of rectangle}=200\text{\hspace{0.17em}}\mathrm{cm}×70\text{\hspace{0.17em}}\mathrm{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=14000\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\end{array}$

Q.18 Find the areas of the squares whose sides are:

(a) 10 cm (b) 14 cm (c) 5 m

Ans-

Area of square = (side of square)2
(a) Area of square = (10)2

= 100 cm2
(b) Area of square = (14)2
= 196 cm2
(c) Area of square = (5)2

= 25 cm2

Q.19 The length and breadth of three rectangles are as given below:
(a) 9 m and 6 m
(b) 17 m and 3 m
(c) 4 m and 14 m
Which one has the largest area and which one has the smallest?

Ans-

$\begin{array}{l}\text{\hspace{0.17em}}\mathrm{Area}\text{of rectangle}=\mathrm{Length}×\mathrm{Breadth}\\ \left(\mathrm{a}\right)\mathrm{Area}\text{of rectangle}=9\text{\hspace{0.17em}}\mathrm{m}×6\text{\hspace{0.17em}}\mathrm{m}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=54\text{\hspace{0.17em}}{\mathrm{m}}^{2}\\ \left(\mathrm{b}\right)\mathrm{Area}\text{of rectangle}=17\text{\hspace{0.17em}}\mathrm{m}×3\text{\hspace{0.17em}}\mathrm{m}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=51\text{\hspace{0.17em}}{\mathrm{m}}^{2}\\ \left(\mathrm{c}\right)\mathrm{Area}\text{of rectangle}=4\text{\hspace{0.17em}}\mathrm{m}×14\text{\hspace{0.17em}}\mathrm{m}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=56\text{\hspace{0.17em}}{\mathrm{m}}^{2}\\ \text{Rectangle}\left(\text{c}\right)\text{has the largest area and rectangle}\left(\mathrm{b}\right)\text{has the}\\ \text{smallest area.}\end{array}$

Q.20 The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.

Ans-

$\begin{array}{c}\text{Area of rectangular garden}=\text{3}00\text{sq m}\\ \text{Length of rectangular garden}=50\text{\hspace{0.17em}}\mathrm{m}\\ \mathrm{Breadth}\text{of rectangular garden}=\frac{\mathrm{Area}\text{of garden}}{\mathrm{Length}\text{of garden}}\\ =\frac{300}{50}\\ =6\text{\hspace{0.17em}}\mathrm{m}\end{array}$

Q.21 A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?

Ans-

Length of table top = 2 m

Breadth of table top = 1 m 50 cm

= (1 + 0.50) m

= 1.5 m

Area of table-top = Length x Breadth

= 2 m x 1.5 m

= 3 sq m

Q.22 A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet is needed to cover the floor of the room?

Ans-

Length of room = 4 m

Breadth of room = 3 m 50 cm

= (3 + 0.50) m

= 3.5 m

Area of floor = Length x Breadth

= 4 m x 3.5 m

= 14 sq m
Therefore, 14 sq m carpet is required to cover the floor of the room.

Q.23 A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted.

Ans-

$\begin{array}{c}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Area}\text{of rectangular floor}=5\mathrm{m}×4\mathrm{m}\\ =20\text{\hspace{0.17em}}{\mathrm{m}}^{2}\\ \text{\hspace{0.17em}}\mathrm{Length}\text{of side of square carpet}=3\text{\hspace{0.17em}}\mathrm{m}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Area}\text{of square carpet}={\left(3\text{\hspace{0.17em}}\mathrm{m}\right)}^{2}\\ =9\text{\hspace{0.17em}}{\mathrm{m}}^{2}\\ \mathrm{Area}\text{of floor that is not carpeted}=20-9\\ =11{\mathrm{m}}^{2}\end{array}$

Q.24 Five square flower beds each of sides 1 m are dug on a piece of land 5 m long and 4 m wide. What is the area of the remaining part of the land?

Area of rectangular land = 5m x 4m

= 20 m2
Area of 1 flower bed = (1m)2
= 1m2
Area of 5 flower beds = 5m2
Area of remaining part of land= 20 – 5
= 15 m2

Q.25 By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).

Ans-

(a)Area of given figure = 3 × 3 + 1 × 2 +3 × 3
+ 2 × 4
= 9 + 2 + 9 + 8

= 28 cm2
(b) Area of given figure = 3 × 1 + 3 × 1 +3 × 1
= 3 + 3 + 3

= 9 cm2

Q.26 Split the following shapes into rectangles and find their areas. (The measures are given in centimeters)

Ans-

(a) Area of given figure = 12 x 2 + 8 x 2

= 24 + 16

= 40 cm2
(b) Area of given figure = 7 x 7 + 7 x 21 +7 x 7
= 49 + 147 + 49

= 245 cm2

(c) Area of given figure = 1 x 5 + 4 x 1
= 5 + 4

= 9 cm2

Q.27 How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively:

(a) 100 cm and 144 cm

(b) 70 cm and 36 cm.

Ans-

$\begin{array}{l}\left(\mathrm{a}\right)\text{Number of bricks in wall}=\frac{\mathrm{Area}\text{of rectangular region}}{\mathrm{Area}\text{of surface of brick}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{100×144}{12×5}\\ \text{\hspace{0.17em}\hspace{0.17em}}=20×12\\ \text{\hspace{0.17em}\hspace{0.17em}}=240\\ \left(\mathrm{b}\right)\text{Number of bricks in wall}=\frac{\mathrm{Area}\text{of rectangular region}}{\mathrm{Area}\text{of surface of brick}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{70×36}{12×5}\\ \text{\hspace{0.17em}\hspace{0.17em}}=14×3\\ \text{\hspace{0.17em}\hspace{0.17em}}=42\end{array}$

Q.28 Find the cost of fencing a square park of side 250 m at the rate of ₹20 per metre.

Ans-

Side of square park = 250 m
Rate of fencing the park = ₹20 per metre
Perimeter of square park = 4(250 m)
= 1000 m
Cost of fencing the square park

= ₹20 x 1000 m
= ₹20,000
Thus, the cost of fencing the square park is ₹20,000

Q.29 Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of ₹12 per metre.

Ans-

Length of a rectangular park = 175 m
Breadth of a rectangular park = 125 m
Rate of fencing per metre = ₹12
Length of fencing of park = 2(L + B)
= 2(175+ 125) m
= 2(300) m
= 600 m
Cost of fencing the park = ₹12(600 m)
= ₹7200

Q.30 What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of ₹8 per hundred sq m.?

Area of rectangular plot = 500 m × 200 m

= 1, 00, 000 sq m
Rate of tilling per sq m = ₹8 per hundred
= ₹(8/100)
Cost of tilling the plot = ₹(8/100) × 1,00,000

= ₹8000

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### 1. What is the difference between a minor and major arc?

The shortest arc connecting two points on a circle is called a minor arc. The centre angle of a minor arc is less than 180°. A major arc is the longer arc that connects two circle endpoints. These two are essential components of a circle and must be mastered in order to tackle many circle-related difficulties.

### 2. What do you mean by Mensuration?

Mensuration is a discipline of Maths concerned with the computation of geometric figures and their properties, such as area, length, volume, lateral surface area, and surface area. It covers all of the fundamental equations and features of a wide range of geometric forms and figures, as well as the basics of calculation.

### 3. Explain the perimeter of irregular shapes as per the Chapter 10 of NCERT Solutions for Class 6 Maths.

Irregular forms are those that do not have the same number of sides and angles on all sides. The perimeter of an irregular shape is equal to the whole length it covers. The perimeter of the diagram is equal to the sum of all sides.

### 4. What is a circle's tangent?

A tangent to a circle is a straight line that joins the circle at any or only one point (tangency), and the tangent is at 90 degrees with regard to the radius of the circle at this point. Tangent lines to circles are the subject of many theorems and are employed in countless geometrical structures and proofs.

### 5. Why should I consult the NCERT Solutions Class 6 Maths Mensuration Chapter 10?

The NCERT Solutions Class 6 Maths Chapter 10 Mensuration includes well-explained answers as well as logical explanations for each question. These NCERT solutions also include visuals to help students understand concepts like area and perimeter. Moreover, the answers come from a credible source as subject matter experts prepare them.