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

Q.1

$\begin{array}{l}\mathrm{The}\mathrm{length}\mathrm{and}\mathrm{the}\mathrm{breadth}\mathrm{of}\mathrm{a}\mathrm{rectangular}\mathrm{piece}\mathrm{of}\mathrm{land}\\ \mathrm{are}500\mathrm{m}\mathrm{and}300\mathrm{m}\mathrm{respectively}.\mathrm{Find}\\ \left(\mathrm{i}\right)\mathrm{its}\mathrm{area}\\ \left(\mathrm{ii}\right)\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{the}\mathrm{land},\mathrm{if}1{\mathrm{m}}^{\mathrm{2}}\mathrm{of}\mathrm{the}\mathrm{land}\mathrm{costs}₹10,000.\end{array}$

Ans

\begin{array}{l}\text{(i)}\\ \text{Area}=\text{Length}\times \text{breadth}\\ =\text{500 m}\times \text{3000}\times \text{m}\\ ={\text{150000m}}^2\\ \text{(ii)}\\ {\text{1 m}}^2\text{land cost}=\text{₹ 10,000}\\ {\text{So, cost of 150000 m}}^2\text{land}=\text{₹ 10,000}\times \text{150000}\\ =\text{₹ 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 ‘}a\text{‘}\text{.}\\ \text{Since perimeter of a square}=\text{4}a\\ \text{So, we get}\\ 4a=3\text{20}\\ a=\frac{320}{4}\\ =80\text{m}\\ \text{Therefore, the area of the square park}=a^2\\ ={\left(80\text{m}\right)}^2\\ =6400{\text{m}}^2\end{array}

Q.3 Find the breadth of a rectangular plot of land, if its area is 440 m² 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{m}\times x\\ x=\frac{440{\text{m}}^2}{22\text{m}}\\ =20\text{m}\\ \text{Perimeter of the rectangular plot}=\text{2}\left(\text{length+breadth}\right)\\ =2\left(20\text{m}+22\text{m}\right)\\ =2\left(42\text{m}\right)\\ =84\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)\\ 100\text{cm}=2\left(35\text{cm}+x\right)\\ \frac{100}{2}\text{cm}=\left(35\text{cm+x}\right)\\ 50\text{cm}=35\text{cm+x}\\ x=50\text{cm}-35\text{cm}\\ =15\text{cm}\\ \text{Area of rectangular sheet}=\text{length}\times \text{breadth}\\ =\text{35 cm}\times \text{15 cm}\\ ={\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}\\ {\left(\text{60m}\right)}^2=90m\times x\\ 3600m^2=90m\times x\\ x=\frac{3600m^2}{90m}\\ =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{2}\left(\text{length+breadth}\right)=4\times \text{side}\\ \text{2}\left(40\text{m}+22\text{m}\right)=4\times \text{side}\\ \text{side}=\frac{124}{4}\\ =31\text{cm}\\ So,\\ \text{Area of rectangle}=\text{length}\times \text{breadth}\\ =\text{40cm}\times \text{22cm}\\ ={\text{880cm}}^2\\ \text{Area of square}={\left(\text{side}\right)}^2\\ ={\left(31\text{cm}\right)}^2\\ =961{\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)}\\ 130\text{cm}=\text{2(30 cm}+x\text{)}\\ \frac{130}{2}\text{cm}=\text{30 cm}+x\\ x=\text{65 cm}-\text{30 cm}\\ =\text{35 cm}\\ \text{Now,}\\ \text{area of rectangle}=\text{length}\times \text{breadth}\\ =\text{30 cm}\times \text{35 cm}\\ ={\text{1050 cm}}^2\end{array}

Q.8

$\begin{array}{l}\mathrm{A}\mathrm{door}\mathrm{of}\mathrm{length}2\mathrm{m}\mathrm{and}\mathrm{breadth}1\mathrm{m}\mathrm{is}\mathrm{fitted}\mathrm{in}\mathrm{a}\mathrm{wall}.\\ \mathrm{The}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{wall}\mathrm{is}4.5\mathrm{m}\mathrm{and}\mathrm{the}\mathrm{breadth}\mathrm{is}3.6\mathrm{m}.\\ \mathrm{Find}\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{white}\mathrm{washing}\mathrm{the}\mathrm{wall},\mathrm{if}\mathrm{the}\mathrm{rate}\mathrm{of}\\ \mathrm{white}\mathrm{washing}\mathrm{the}\mathrm{wall}\mathrm{is}₹20\mathrm{per}{\mathrm{m}}^{\mathrm{2}}\mathrm{.}\end{array}$

Ans

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