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NCERT Solutions for Class 7 Maths Chapter 11: Perimeter and Area - Exercise 11.1

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NCERT Solutions for Class 7 Maths Chapter 11 (EX 11.1)

Free PDF download of NCERT Solutions for Class 7 Maths Chapter 11 Exercise 11.1 and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.1 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails. Every NCERT Solution is provided to make the study simple and interesting on Vedantu. Subjects like Science, Maths, English will become easy to study if you have access to NCERT Solution for Class 7 Science , Maths solutions and solutions of other subjects.


Class:

NCERT Solutions for Class 7

Subject:

Class 7 Maths

Chapter Name:

Chapter 11 - Perimeter and Area

Exercise:

Exercise - 11.1

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.1

Exercise 11.1

1. The length and breadth of a rectangular piece of land are 500 m and 300 m respectively. Find: 

(i) Its area.

Ans: The area $A$ of a rectangle having its length as $l$ and breadth as $b$ is defined by $A = l \times b$.

Given, the length of the rectangular land is 500 m and breadth of the land is 300 m.

Hence, the value of $l$ and $b$ are 500 and 300 respectively.

Therefore, the area of the rectangular piece of land is as follows,

$\Rightarrow A = l \times b$

$\Rightarrow A = 500 \times 300$

$\Rightarrow A = 150000$

Therefore, the required area of the rectangular piece of land is $150000{\text{ }}{{\text{m}}^2}$.

(ii) The cost of the land, if $1{\text{ }}{{\text{m}}^2}$ of the land costs ₹10,000.

Ans: Given, the cost of $1{\text{ }}{{\text{m}}^2}$ of the land is ₹10,000.

The area of the land is $150000{\text{ }}{{\text{m}}^2}$.

Therefore the cost of the land is as follows.

\[\Rightarrow {\text{Cost}} = 10000 \times 150000\]

$\Rightarrow {\text{Cost}} = 1500000000$

Hence, the cost of the rectangular piece of land is ₹1,50,00,00,000.


2. Find the area of a square park whose perimeter is 320 m.

Ans: The perimeter $p$ of a square is defined by, $p = 4s$, where $s$ is the length of the sides of the square.

Given, perimeter of the square park is 320 m.

Therefore, $p = 320$.

In order to obtain the sides of the square, substitute $4s$ for $p$ in the equation $p = 320$ and solve it.

$\Rightarrow p = 320$

$\Rightarrow 4s = 320$

$\Rightarrow s = \frac{{320}}{4}$

$\Rightarrow s = 80$

Hence, the sides of the square park are 80 m.

The area $A$ of a square is denoted by, $A = s \times s$, where $s$ is the length of the sides of the square.

$\Rightarrow A = 80 \times 80$

$\Rightarrow A = 6400$

Hence, the required area of the square park is $6400{\text{ }}{{\text{m}}^2}$.


3. Find the breadth of a rectangular plot of land, if its area is $440{\text{ }}{{\text{m}}^2}$ and the length is 22 m. Also find its perimeter.

Ans: The area $A$ of a rectangle having its length as $l$ and breadth as $b$ is defined by $A = l \times b$.

Given, the length of the rectangular plot is 22 m and the area of the plot is $440{\text{ }}{{\text{m}}^2}$.

Hence, the value of $l$ and $A$ are 22 and 440 respectively.

In order to obtain the breadth of the plot, substitute 22 for $l$ and 440 for $A$ in the formula $A = l \times b$, and then solve for $b$.

\[\Rightarrow A = l \times b\]

\[\Rightarrow 440 = 22 \times b\]

\[\Rightarrow \frac{{440}}{{22}} = b\]

\[\Rightarrow 20 = b\]

\[\Rightarrow b = 20\]

Therefore, the breadth of the given plot is 20 m.

The perimeter $P$ of a rectangle with length $l$ and breadth $b$ is given by $P = 2\left( {l + b} \right)$.

$\Rightarrow P = 2\left( {22 + 20} \right)$

$\Rightarrow P = 2\left( {42} \right)$

$\Rightarrow P = 84$

Therefore, the perimeter of the given plot is 84 m.


4. The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also find the area.

Ans: The perimeter $P$ of a rectangle with length $l$ and breadth $b$ is given by $P = 2\left( {l + b} \right)$.

Given, the length of the rectangular sheet is 35cm and perimeter of the sheet is 100 cm.

Hence, the value of $l$ and $P$ are 35 and 100 respectively.

In order to obtain the breadth of the sheet, substitute 35 for $l$ and 100 for $P$ in the formula $P = 2\left( {l + b} \right)$, and then solve for $b$.

$\Rightarrow P = 2\left( {l + b} \right)$

$\Rightarrow 100 = 2\left( {35 + b} \right)$

$\Rightarrow \frac{{100}}{2} = \left( {35 + b} \right)$

$\Rightarrow 50 = \left( {35 + b} \right)$

$\Rightarrow 50 - 35 = b$

$\Rightarrow 15 = b$

$\Rightarrow b = 15$

Therefore, the obtained breadth of the given sheet is 15 cm.

The area $A$ of a rectangle with length $l$ and breadth $b$ is defined by $A = l \times b$.

$\Rightarrow A = l \times b$

$\Rightarrow A = 35 \times 15$

$\Rightarrow A = 525$

Hence, the required area of the rectangular sheet is ${\text{525 c}}{{\text{m}}^2}$.


5. The area of a square park is the same as of a rectangular park. If the side of the square park is 60 m and the length of the rectangular park is 90 m, find the breadth of the rectangular park.

Ans: The area $A$ of a square is denoted by, $A = s \times s$, where $s$ is the length of the sides of the square.

Given, sides of the square park $s = 60{\text{ m}}$.

$\Rightarrow A = 60 \times 60$

$\Rightarrow A = 3600$

Hence, the derived area of the square park is $3600{\text{ }}{{\text{m}}^2}$, which is equal to the area of the rectangular park.

The area $A$ of a rectangle with length $l$ and breadth $b$ is defined by $A = l \times b$.

It is given that the length of the rectangular park is 90 m.

Also, the area of the park is $3600{\text{ }}{{\text{m}}^2}$.

Hence, the value of $l$ and $A$ are 90 and 3600 respectively.

In order to obtain the breadth of the rectangular park, substitute 90 for $l$ and 3600 for $A$ in the formula $A = l \times b$, and then solve for $b$.

\[\Rightarrow A = l \times b\]

\[\Rightarrow 3600 = 90 \times b\]

\[\Rightarrow \frac{{3600}}{{90}} = b\]

\[\Rightarrow 40 = b\]

\[\Rightarrow b = 40\]

Therefore, the derived breadth of the given rectangular park is 40 m.


6. A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is rebent in the shape of a square, what will be the measure of each side? Also find which shape encloses more area?

Ans: The perimeter of a rectangle with length $l$ and breadth $b$ is given by $2\left( {l + b} \right)$.

Given, length and breadth of the rectangle are 40 cm and 22 cm respectively.

Therefore, $l = 40$ and $b = 22$.

The perimeter of a square is defined by, $4s$, where $s$ is the length of the sides of the square.

Since, the same wire is rebent into the shape of a square from the shape of a rectangle, the perimeter of the rectangle will be the same as the perimeter of the square.

Hence, equate the perimeter of both rectangle and square, in order to obtain the side of the square.

$\Rightarrow 4s = 2\left( {l + b} \right)$

$\Rightarrow 4s = 2\left( {40 + 22} \right)$

$\Rightarrow 4s = 2\left( {62} \right)$

$\Rightarrow 4s = 124$

$\Rightarrow s = \frac{{124}}{4}$

$\Rightarrow s = 31$

Therefore, the obtained sides of the square are 31 cm.

The area $A$ of a square is denoted by, $A = s \times s$, where $s$ is the length of the sides of the square.

The obtained sides of the square is $s = 31{\text{ cm}}$.

Therefore, the area of the square is as follows,

$\Rightarrow A = 31 \times 31$

$\Rightarrow A = 961$

Hence, the derived area of the square is $961{\text{ c}}{{\text{m}}^2}$.

The area $A$ of a rectangle with length $l$ and breadth $b$ is defined by $A = l \times b$.

Thus, the area of the rectangle is as follows,

$\Rightarrow A = 40 \times 22$

$\Rightarrow A = 880$

Hence, the derived area of the rectangle is ${\text{880 c}}{{\text{m}}^2}$.

Since $880 < 961$, the square shape encloses greater area.


7. The perimeter of a rectangle is 130 cm. If the breadth of the rectangle is 30 cm, find its length. Also, find the area of the rectangle. 

Ans: The perimeter $P$ of a rectangle with length $l$ and breadth $b$ is given by $P = 2\left( {l + b} \right)$.

Given, the breadth of the rectangle is 30cm and perimeter is 130 cm.

Hence, the value of $b$ and $P$ are 30 and 130 respectively.

In order to obtain the breadth of the sheet, substitute 30 for $b$ and 130 for $P$ in the formula $P = 2\left( {l + b} \right)$, and then solve for $l$.

$\Rightarrow P = 2\left( {l + b} \right)$

$\Rightarrow 130 = 2\left( {l + 30} \right)$

$\Rightarrow \frac{{130}}{2} = \left( {l + 30} \right)$

$\Rightarrow 65 = \left( {l + 30} \right)$

$\Rightarrow 65 - 30 = l$

$\Rightarrow 35 = l$

$\Rightarrow l = 35$

Therefore, the obtained length of the given rectangle is 35 cm.

The area $A$ of a rectangle having its length as $l$ and breadth as $b$ is defined by $\Rightarrow A = l \times b$.

$\Rightarrow A = l \times b$

$\Rightarrow A = 35 \times 30$

$\Rightarrow A = 1050$

Hence, the obtained area of the rectangle is ${\text{1050 c}}{{\text{m}}^2}$.


8. A door of length 2 m and breadth 1 m is fitted in a wall. The length of the wall is $4.5$ m and the breadth is \[3.6\] m. Find the cost of white washing the wall, if the rate of white washing the wall is ₹20per ${{\text{m}}^2}$.


Image Representing the Door


Ans: The area $A$ of a rectangle having its length as $l$ and breadth as $b$ is defined by $A = l \times b$.

Given length and breadth of the door are 2 m and 1 m respectively.

Therefore, $l = 2$ and $b = 1$.

Thus, the area of the door can be derived as follows.

$\Rightarrow A = l \times b$

$\Rightarrow A = 2 \times 1$

$\Rightarrow A = 2$

Therefore, the area of the door is ${\text{2 }}{{\text{m}}^2}$.

Given length of the wall is $4.5$ m and breadth is \[3.6\] m.

Therefore, $l = 4.5$ and $b = 3.6$.

Thus, the area of the wall can be derived as follows.

$\Rightarrow A = l \times b$

$\Rightarrow A = 4.5 \times 3.6$

$\Rightarrow A = 16.2$

Hence, the obtained area of the wall is ${\text{16}}{\text{.2 }}{{\text{m}}^2}$.

Therefore, the area of the wall excluding the door is $16.2 - 2 = 14.2{\text{ }}{{\text{m}}^2}$.

Given, the rate of white washing the wall excluding the door is ₹20 per ${{\text{m}}^2}$.

The required cost of white washing the wall is \[20 \times 14.2 = 284\].

Hence, the estimated cost is ₹284.


NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.1

Opting for the NCERT solutions for Ex 11.1 Class 7 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 11.1 Class 7 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 7 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 7 Maths Chapter 11 Exercise 11.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 7 Maths Chapter 11 Exercise 11.1, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

Do not delay any more. Download the NCERT solutions for Class 7 Maths Chapter 11 Exercise 11.1 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.