# NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area (EX 11.1) Exercise 11.1

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**NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area (EX 11.1) Exercise 11.1**

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

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**Exercise 11.1**

Here students can get the straightforward PDF solutions to the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.1 Chapter 11 Perimeter and Area. Students will review how to compute the length, width, perimeter, and area of a square and a rectangle in this activity from the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.1. By using the answers given here to the problems in NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.1 Perimeter and Area, students may learn more about these areas.

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The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.1 attempts to introduce the idea of area and perimeter through a variety of short tasks. Squares and rectangles’ perimeters and areas are discussed in this chapter. Finding the parameters for specified squares and rectangles is what the practise questions ask the students to do. In some circumstances, it may also be necessary to calculate the unknown dimensions of squares and rectangles from their area and perimeter. There are 8 questions in the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.1 Perimeter and Area, the majority of which are formula-based and simple to answer.

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

The area and the perimeter are two distinct metrics that cannot be compared, which the students must keep in mind. A perimeter is one-dimensional, having length and no breadth. There are two dimensions to the space: length and breadth. Below are the Chapter 11 Exercise 11.1 Perimeter and Area answers for Class 7 Mathematics.

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**Q.1 **

$\begin{array}{l}\mathrm{The}\mathrm{length}\mathrm{and}\mathrm{the}\mathrm{breadth}\mathrm{of}\mathrm{a}\mathrm{rectangular}\mathrm{piece}\mathrm{of}\mathrm{land}\\ \mathrm{are}500\mathrm{m}\mathrm{and}300\mathrm{m}\mathrm{respectively}.\mathrm{Find}\\ \left(\mathrm{i}\right)\mathrm{its}\mathrm{area}\\ \left(\mathrm{ii}\right)\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{the}\mathrm{land},\mathrm{if}1{\mathrm{m}}^{\mathrm{2}}\mathrm{of}\mathrm{the}\mathrm{land}\mathrm{costs}\u20b910,000.\end{array}$

**Ans**

\begin{array}{l}\text{(i)}\\ \text{Area}=\text{Length}\times \text{breadth}\\ \text{}\text{\hspace{0.17em}}=\text{500 m}\times \text{3000}\times \text{m}\\ \text{}\text{\hspace{0.17em}}={\text{150000m}}^{2}\\ \text{(ii)}\\ {\text{1 m}}^{2}\text{land cost}=\text{\u20b9 10,000}\\ {\text{So, cost of 150000 m}}^{2}\text{land}=\text{\u20b9 10,000}\times \text{150000}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{\u20b9 1,500,000,000}\end{array}

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

**Ans**

\begin{array}{l}\text{Let the side of the square park be \u2018}a\text{\u2018}\text{.}\\ \text{Since perimeter of a square}=\text{4}a\\ \text{So, we get}\\ 4a=3\text{20}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}a=\frac{320}{4}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=80\text{\hspace{0.17em}}\text{m}\\ \text{Therefore, the area of the square park}={a}^{2}\\ \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}}={\left(80\text{\hspace{0.17em}}\text{m}\right)}^{2}\\ \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}}=6400\text{\hspace{0.17em}}{\text{m}}^{2}\end{array}

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

**Ans**

\begin{array}{l}\text{Let the breadth of the rectangular plot be}x\text{.}\\ \text{Area of the rectangular plot}=\text{length}\times \text{breadth}\\ \text{SO, we get}\\ {\text{440m}}^{2}=22\text{\hspace{0.17em}}\text{m}\times x\\ \text{}x=\frac{440\text{\hspace{0.17em}}{\text{m}}^{2}}{22\text{\hspace{0.17em}}\text{m}}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=20\text{\hspace{0.17em}}\text{m}\\ \text{Perimeter of the rectangular plot}=\text{2}\left(\text{length+breadth}\right)\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\left(20\text{\hspace{0.17em}}\text{m}+22\text{\hspace{0.17em}}\text{m}\right)\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\left(42\text{\hspace{0.17em}}\text{m}\right)\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=84\text{\hspace{0.17em}}\text{m}\end{array}

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

**Ans**

\begin{array}{l}\text{Let the breadth of the rectangular sheet be}x\text{.}\\ \text{Perimeter of the rectangular sheet}=\text{2}\left(\text{length+breadth}\right)\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}100\text{cm}=2\left(35\text{\hspace{0.17em}}\text{cm}+x\right)\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\frac{100}{2}\text{cm}=\left(35\text{cm+x}\right)\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}50\text{\hspace{0.17em}}\text{cm}=35\text{cm+x}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=50\text{cm}-35\text{cm}\\ \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}}=15\text{cm}\\ \text{Area of rectangular sheet}=\text{length}\times \text{breadth}\\ \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{35 cm}\times \text{15 cm}\\ \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{525 cm}}^{2}\end{array}

**Q.5 **

$\begin{array}{l}\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{a}\mathrm{square}\mathrm{park}\mathrm{is}\mathrm{the}\mathrm{same}\mathrm{as}\mathrm{of}\mathrm{a}\mathrm{rectangular}\\ \mathrm{park}.\mathrm{If}\mathrm{the}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{square}\mathrm{park}\mathrm{is}60\mathrm{m}\mathrm{and}\mathrm{the}\mathrm{length}\\ \mathrm{of}\mathrm{the}\mathrm{rectangular}\mathrm{park}\mathrm{is}90\mathrm{m},\mathrm{find}\mathrm{the}\mathrm{breadth}\mathrm{of}\mathrm{the}\\ \mathrm{rectangular}\mathrm{park}.\end{array}$

**Ans**

\begin{array}{l}\text{Let the breadth of rectangular park be}x\text{.}\\ \text{Since, area of a square park is the same as of a rectangular}\\ \text{park}.\\ \text{Area of square}=\text{Area of rectangular park}\\ \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}}{\left(\text{60m}\right)}^{2}=90m\times x\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3600{m}^{2}=90m\times x\\ \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}}x=\frac{3600{m}^{2}}{90m}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=40m\end{array}

**Q.6 **

$\begin{array}{l}\mathrm{A}\mathrm{wire}\mathrm{is}\mathrm{in}\mathrm{the}\mathrm{shape}\mathrm{of}\mathrm{a}\mathrm{rectangle}.\mathrm{Its}\mathrm{length}\mathrm{is}40\mathrm{cm}\\ \mathrm{and}\mathrm{breadth}\mathrm{is}22\mathrm{cm}.\mathrm{If}\mathrm{the}\mathrm{same}\mathrm{wire}\mathrm{is}\mathrm{rebent}\mathrm{in}\mathrm{the}\\ \mathrm{shape}\mathrm{of}\mathrm{a}\mathrm{square},\mathrm{what}\mathrm{will}\mathrm{be}\mathrm{the}\mathrm{measure}\mathrm{of}\mathrm{each}\mathrm{side}.\\ \mathrm{Also}\mathrm{find}\mathrm{which}\mathrm{shape}\mathrm{encloses}\mathrm{more}\mathrm{area}?\end{array}$

**Ans**

\begin{array}{l}\text{Perimeter of a rectangle}=\text{Perimeter of square}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{2}\left(\text{length+breadth}\right)=4\times \text{side}\\ \text{}\text{}\text{\hspace{0.17em}}\text{2}\left(40\text{\hspace{0.17em}}\text{m}+22\text{\hspace{0.17em}}\text{m}\right)=4\times \text{side}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{side}=\frac{124}{4}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=31\text{\hspace{0.17em}}\text{cm}\\ So,\\ \text{Area of rectangle}=\text{length}\times \text{breadth}\\ \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{40cm}\times \text{22cm}\\ \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{880cm}}^{2}\\ \text{Area of square}={\left(\text{side}\right)}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left(31\text{\hspace{0.17em}}\text{cm}\right)}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=961\text{\hspace{0.17em}}{\text{cm}}^{2}\\ \text{Therefore, the square-shaped wire encloses more area}\text{.}\end{array}

**Q.7 **

$\begin{array}{l}\mathrm{The}\mathrm{perimeter}\mathrm{of}\mathrm{a}\mathrm{rectangle}\mathrm{is}130\mathrm{cm}.\mathrm{If}\mathrm{the}\mathrm{breadth}\\ \mathrm{of}\mathrm{the}\mathrm{rectangle}\mathrm{is}30\mathrm{cm},\mathrm{find}\mathrm{its}\mathrm{length}.\mathrm{Also}\mathrm{find}\mathrm{the}\\ \mathrm{area}\mathrm{of}\mathrm{the}\mathrm{rectangle}.\end{array}$

**Ans**

\begin{array}{l}\text{Let the length of the rectangle be}x\text{cm}\text{.}\\ \text{Perimeter of rectangle}=\text{2 (length}+\text{breadth)}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}130\text{\hspace{0.17em}}\text{cm}=\text{2(30 cm}+x\text{)}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{130}{2}\text{cm}=\text{30 cm}+x\\ \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}}x=\text{65 cm}-\text{30 cm}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{35 cm}\\ \text{Now,}\\ \text{area of rectangle}=\text{length}\times \text{breadth}\\ \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{30 cm}\times \text{35 cm}\\ \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{1050 cm}}^{2}\end{array}

**Q.8 **

$\begin{array}{l}\mathrm{A}\mathrm{door}\mathrm{of}\mathrm{length}2\mathrm{m}\mathrm{and}\mathrm{breadth}1\mathrm{m}\mathrm{is}\mathrm{fitted}\mathrm{in}\mathrm{a}\mathrm{wall}.\\ \mathrm{The}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{wall}\mathrm{is}4.5\mathrm{m}\mathrm{and}\mathrm{the}\mathrm{breadth}\mathrm{is}3.6\mathrm{m}.\\ \mathrm{Find}\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{white}\mathrm{washing}\mathrm{the}\mathrm{wall},\mathrm{if}\mathrm{the}\mathrm{rate}\mathrm{of}\\ \mathrm{white}\mathrm{washing}\mathrm{the}\mathrm{wall}\mathrm{is}\u20b920\mathrm{per}{\mathrm{m}}^{\mathrm{2}}\mathrm{.}\end{array}$

**Ans**

\begin{array}{l}\text{\hspace{0.17em}}\text{Area of the wall}=\text{4}\text{.5 cm}\times \text{3}\text{.6 cm}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{16}{\text{.2 m}}^{\text{2}}\\ \text{Area of the door}=\text{2}\times \text{1}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{2m}}^{\text{2}}\\ \text{Area to be white-washed}=\text{16}{\text{.2 m}}^{2}-{\text{2 m}}^{2}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{14}{\text{.2 m}}^{\text{2}}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{Cost of white-washing 1m}}^{\text{2}}\text{area}=\text{\u20b920}\\ \text{So, cost of white-washing 14}\text{.2m2 area}=\text{14}\text{.2}\times \text{20}\\ \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{\u20b9 284}\end{array}