NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area (EX 11.3) Exercise 11.3

Q.1

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{circumference}\mathrm{of}\mathrm{the}\mathrm{circles}\mathrm{with}\mathrm{the}\mathrm{following}\\ \mathrm{radius}:\left(\mathrm{Take}\mathrm{\pi }\mathrm{=}\frac{\mathrm{22}}{\mathrm{7}}\right)\\ \left(\mathrm{a}\right)14\mathrm{cm}\left(\mathrm{b}\right)28\mathrm{mm}\left(\mathrm{c}\right)21\mathrm{cm}\end{array}$

Ans

\begin{array}{l}\text{(a)}\\ \text{Circumference of a circle is}=\text{2}\pi \text{r}\\ =\text{2}\times \frac{22}{7}\times 14\\ =2\times 22\times 2\\ =88\text{cm}\\ \text{(b)}\\ \text{Circumference of a circle is}=\text{2}\pi \text{r}\\ =\text{2}\times \frac{22}{7}\times 28\\ =2\times 22\times 4\\ =176\text{mm}\end{array}

\begin{array}{l}\text{(c)}\\ \text{Circumference of a circle is}=\text{2}\pi \text{r}\\ =\text{2}\times \frac{22}{7}\times 21\\ =2\times 22\times 3\\ =132\text{cm}\end{array}

Q.2

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{circles},\mathrm{given}\mathrm{that}:\\ \left(\mathrm{a}\right)\mathrm{radius}=\; 14\mathrm{mm}\left(\mathrm{Take}\mathrm{\pi }\mathrm{=}\frac{\mathrm{22}}{\mathrm{7}}\right)\\ \left(\mathrm{b}\right)\mathrm{diameter}=\; 49\mathrm{m}\left(\mathrm{c}\right)\mathrm{radius}=\; 5\mathrm{cm}\end{array}$

Ans

\begin{array}{l}\text{(a)}\\ \text{Area of a circle}=\pi {\text{r}}^2\\ =\frac{22}{7}{\left(14\right)}^2\\ =616{\text{mm}}^2\end{array}

\begin{array}{l}\text{(b)}\\ \text{Radius}=\frac{\text{Diameter}}{2}=\frac{49}{2}=24.5\text{mm}\\ \text{Area of a circle}=\pi {\text{r}}^2\\ =\frac{22}{7}{\left(24.5\right)}^2\\ =1886.5{\text{m}}^2\\ \text{(c)}\\ \text{Area of a circle}=\pi {\text{r}}^2\\ =\frac{22}{7}{\left(5\right)}^2\\ =\frac{550}{7}{\text{cm}}^2\end{array}

Q.3 If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take π =22/7)

Ans

\begin{array}{l}\text{Since, circumference}=\text{2}\pi \text{r}\\ \text{So, we get}\end{array}

\begin{array}{l}\text{154}=\text{2}\times \frac{22}{7}\times r\\ r=\frac{154\times 7}{2\times 22}\\ =\frac{49}{2}\text{m}=4.5\text{m}\\ \text{Now,}\\ \text{Area}=\pi {\text{r}}^2\\ =\frac{22}{7}\times \frac{49}{2}\times \frac{49}{2}\\ =1886.5{\text{m}}^2\end{array}

Q.4

$\begin{array}{l}\mathrm{A}\mathrm{gardener}\mathrm{wants}\mathrm{to}\mathrm{fence}\mathrm{a}\mathrm{circular}\mathrm{garden}\mathrm{of}\mathrm{diameter}\\ 21\mathrm{m}.\mathrm{Find}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{rope}\mathrm{he}\mathrm{needs}\mathrm{to}\mathrm{purchase},\mathrm{if}\mathrm{he}\\ \mathrm{makes}2\mathrm{rounds}\mathrm{of}\mathrm{fence}.\mathrm{Also}\mathrm{find}\mathrm{the}\mathrm{costs}\mathrm{of}\mathrm{the}\mathrm{rope},\mathrm{if}\mathrm{it}\\ \mathrm{cost}₹\; 4\mathrm{per}\mathrm{meter}.\left(\mathrm{Take\; \pi }\mathrm{=}\frac{\mathrm{22}}{\mathrm{7}}\right)\end{array}$

Ans

\begin{array}{l}\text{Given,}d=\text{21 m}\\ \text{So,}r=\frac{21}{2}\text{m}\\ \text{Circumference}=\text{2}\pi \text{r}\\ =\text{2}\times \frac{22}{7}\times \frac{21}{2}\\ =66\text{m}\\ \text{Length of the rope required for fencing}=\text{2}\times \text{66 m}\\ =\text{132 m}\\ \text{Cost of 1 m rope}=\text{₹4}\\ \text{So, cost of 132 m rope}=\text{₹ 4}\times \text{132}\\ =\text{₹528}\end{array}

Q.5 From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)

Ans

\begin{array}{l}\text{Outer radius of circular sheet}=\text{4 cm}\\ \text{Inner radius of circular sheet}=\text{3 cm}\end{array}

\begin{array}{l}\text{Remaining area}=-\left(3.14\times 3\times 3\right)\\ =50.24-28.26\\ =21.98{\text{cm}}^2\end{array}

Q.6

$\begin{array}{l}\mathrm{Saima}\mathrm{wants}\mathrm{to}\mathrm{put}\mathrm{a}\mathrm{lace}\mathrm{on}\mathrm{the}\mathrm{edge}\mathrm{of}\mathrm{a}\mathrm{circular}\mathrm{table}\\ \mathrm{cover}\mathrm{of}\mathrm{diameter}1.5\mathrm{m}.\mathrm{Find}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{lace}\\ \mathrm{required}\mathrm{and}\mathrm{also}\mathrm{find}\mathrm{its}\mathrm{cost}\mathrm{if}\mathrm{one}\mathrm{meter}\mathrm{of}\mathrm{the}\mathrm{lace}\\ \mathrm{costs}₹\; 15.\left(\mathrm{Take\pi }\hspace{0.33em}=\; 3.14\right)\end{array}$

Ans

\begin{array}{l}\text{Circumference}=\text{2}\pi \text{r}\\ =\text{2}\times \text{3}\text{.14}\times \frac{d}{2}\\ =2\times 3.14\frac{1.5}{2}\\ =4.71\text{m}\\ \text{Cost of 1 m of lace}=\text{₹ 15}\\ \text{Therefore cost of 4}\text{.71 m of lace}=\text{₹15}\times \text{4}\text{.71}\\ =\text{₹70}\text{.65}\end{array}

Q.7 Find the perimeter of the adjoining figure, which is a semicircle including its diameter.

Ans

\begin{array}{l}\text{Radius}=\text{5 cm}\\ \text{Length of curved part}=\pi \text{r}\\ =\frac{22}{7}\times 5\\ =15.71\text{cm}\\ \text{Total permieter}=\text{Length of the curved part+Length of diameter}\\ =\text{15}\text{.71+10}\\ =\text{25}\text{.71 cm}\end{array}

Q.8

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{polishing}\mathrm{a}\mathrm{circular}\mathrm{table}–\mathrm{top}\mathrm{of}\mathrm{diameter}\\ 1.6\mathrm{m},\mathrm{if}\mathrm{the}\mathrm{rate}\mathrm{of}\mathrm{polishing}\mathrm{is}\mathrm{Rs}15/{\mathrm{m}}^{\mathrm{2}}\mathrm{.}\left(\mathrm{Take\pi }\mathrm{=\; 3.14}\right)\end{array}$

Ans

\begin{array}{l}\text{Diameter}=\text{1}\text{.6 m}\\ \text{Radius}=\frac{1.6}{2}=0.8\text{m}\\ \text{Area}=\text{3}\text{.14}\times \text{0}\text{.8}\times \text{0}\text{.8}\\ =\text{2}{\text{.0096 m}}^2\\ \text{Cost for polishing 1 m2 area}=\text{₹15}\\ \text{Cost for polishing 2}\text{.0096 m2 area}=\text{₹15}\times \text{2}\text{.0096}\\ =\text{₹30}\text{.14}\\ \text{So, it will cost ₹30}\text{.14 for polishing such circular table}\text{.}\end{array}

Q.9

$\begin{array}{l}\mathrm{Shazli}\mathrm{took}\mathrm{a}\mathrm{wire}\mathrm{of}\mathrm{length}44\mathrm{cm}\mathrm{and}\mathrm{bent}\mathrm{it}\mathrm{into}\mathrm{the}\\ \mathrm{shape}\mathrm{of}\mathrm{a}\mathrm{circle}.\mathrm{Find}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{that}\mathrm{circle}.\mathrm{Also}\mathrm{find}\\ \mathrm{its}\mathrm{area}.\mathrm{If}\mathrm{the}\mathrm{same}\mathrm{wire}\mathrm{is}\mathrm{bent}\mathrm{into}\mathrm{the}\mathrm{shape}\mathrm{of}\mathrm{a}\\ \mathrm{square},\mathrm{what}\mathrm{will}\mathrm{be}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{each}\mathrm{of}\mathrm{its}\mathrm{sides}?\\ \mathrm{Which}\mathrm{figure}\mathrm{encloses}\mathrm{more}.\end{array}$

Ans

\begin{array}{l}\text{Circumference}=\text{2}\pi \text{r}=\text{44 cm}\\ \text{2}\times \frac{22}{7}\times r=44\\ r=7\text{cm}\end{array}

\begin{array}{l}\text{Area}=\pi {\text{r}}^2\\ =\frac{22}{7}\times 7\times 7\\ =154{\text{cm}}^2\\ \text{If the wire is bent into a square, then the length of each}\\ \text{side would be}=\frac{44}{4}=11\text{cm}\\ \text{Area of square}={\left(11\right)}^2\\ =121{\text{cm}}^2\\ \text{Therefore, circle encloses more area}\text{.}\end{array}

Q.10

$\begin{array}{l}\mathrm{From}\mathrm{a}\mathrm{circular}\mathrm{card}\mathrm{sheet}\mathrm{of}\mathrm{radius}14\mathrm{cm},\mathrm{two}\mathrm{circles}\mathrm{of}\\ \mathrm{radius}3.5\mathrm{cm}\mathrm{and}\mathrm{a}\mathrm{rectangle}\mathrm{of}\mathrm{length}3\mathrm{cm}\mathrm{and}\mathrm{breadth}\\ 1\mathrm{cm}\mathrm{are}\mathrm{removed}.\; (\mathrm{as}\mathrm{shown}\mathrm{in}\mathrm{the}\mathrm{adjoining\; figure}).\\ \mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{remaining}\mathrm{sheet}.\left(\mathrm{Take\pi }\mathrm{=}\frac{\mathrm{22}}{\mathrm{7}}\right)\end{array}$

Ans

\begin{array}{l}\text{Area of bigger cirle}=\frac{22}{7}\times 14\times 14\\ =616{\text{cm}}^2\\ \text{Area of 2 small circles}=\text{2}\times \pi {\text{r}}^2\\ =2\times \frac{22}{7}\times 3.5\times 3.5\\ =77{\text{cm}}^2\\ \text{Area of rectangle}=\text{Length}\times \text{Breadth}\\ =\text{3}\times \text{1}={\text{3 cm}}^2\\ \text{Remaining area of sheet}={\text{616 cm}}^2-{\text{77 cm}}^2-{\text{3 cm}}^2\\ ={\text{536 cm}}^2\end{array}

Q.11

$\begin{array}{l}\mathrm{A}\mathrm{circle}\mathrm{of}\mathrm{radius}2\mathrm{cm}\mathrm{is}\mathrm{cut}\mathrm{out}\mathrm{from}\mathrm{a}\mathrm{square}\mathrm{piece}\mathrm{of}\mathrm{an}\\ \mathrm{aluminium}\mathrm{sheet}\mathrm{of}\mathrm{side}6\mathrm{cm}.\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{left}\\ \mathrm{over}\mathrm{aluminium}\mathrm{sheet}?\left(\mathrm{Take}\mathrm{\pi }=3.14\right)\end{array}$

Ans

\begin{array}{l}\text{Area of square-shaped sheet}={\left(\text{side}\right)}^2\\ ={\left(6{\text{cm}}^2\right)}^2=36{\text{cm}}^2\\ \text{Area of circle}=\text{3}\text{.14}\times 2\times 2\\ =12.56{\text{cm}}^2\\ \text{Remaining area}={\text{36 cm}}^{\text{2}}-\text{12}{\text{.56 cm}}^{\text{2}}\\ =\text{23}{\text{.44 cm}}^{\text{2}}\end{array}

Q.12 The circumference of a circle is 31.4 cm. Find the radius and the area of the circle ? (Take π = 3.14)

Ans

\begin{array}{l}\text{Circumference}=\text{2}\pi r=31.4\text{cm}\\ \text{2}\times \text{3}\text{.14}\times \text{r}=\text{31}\text{.4}\\ \text{r}=\frac{31.4}{2\times 3.14}\\ =5\text{cm}\\ \text{So, Area}=\text{3}\text{.14}\times \text{5}\times \text{5}\\ =\text{78}{\text{.50 cm}}^2\end{array}

Q.13 A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (π = 3.14)

Ans

\begin{array}{l}\text{Radius of the flower bed}=\frac{66}{2}=33\text{m}\\ \text{So, radius of flower bed and path together}=\text{33+4}\\ =\text{37m}\\ \text{Area of flower baed and path together}=\text{3}\text{.14}\times 37\times 37\\ =4298.66{\text{m}}^2\\ \text{Area of flower bed}=\text{3}\text{.14}\times \text{33}\times \text{33}\\ =\text{3419}{\text{.46 m}}^2\\ \text{Area of path}=\text{4298}{\text{.66 m}}^{\text{2}}-\text{3419}{\text{.46m}}^{\text{2}}\\ =\text{879}{\text{.20m}}^2\end{array}

Q.14

$\begin{array}{l}\mathrm{A}\mathrm{circular}\mathrm{flower}\mathrm{garden}\mathrm{has}\mathrm{an}\mathrm{area}\mathrm{of}314{\mathrm{m}}^{\mathrm{2}}.\mathrm{A}\mathrm{sprinkle\; r}\\ \mathrm{at}\mathrm{the}\mathrm{centre}\mathrm{of}\mathrm{the}\mathrm{garden}\mathrm{can}\mathrm{cover}\mathrm{an}\mathrm{area}\mathrm{that}\mathrm{has}\mathrm{a}\\ \mathrm{radius}\mathrm{of}12\mathrm{m}.\mathrm{Will}\mathrm{the}\mathrm{sprinkler}\mathrm{water}\mathrm{the}\mathrm{entire}\mathrm{garden}?\\ \left(\mathrm{Take\pi }=3.14\right)\end{array}$

Ans

\begin{array}{l}\text{Area}=\pi {\text{r}}^2=314{\text{m}}^2\\ 3.14\times r^2=314\\ r^2=100\\ r=10\text{m}\\ \text{Yes, the sprinkle will water the whole garden}\text{.}\end{array}

Q.15 Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14)

Ans

\begin{array}{l}\text{Radius of outer circle}=\text{19 m}\\ \text{Circumference}=\text{2}\pi \text{r}\\ =\text{2}\times \text{3}\text{.14}\times \text{19}\\ =\text{119}\text{.32 m}\\ \text{Radius of the inner circle}=\text{19}-\text{10}=\text{9 m}\\ \text{Circumference}=\text{2}\pi \text{r}\\ =\text{2}\times \text{3}\text{.14}\times \text{9}\\ =\text{56}\text{.52 m}\end{array}

Q.16 How many times a wheel of radius 28 cm must rotate to go 352 m? (take π=22/7)

Ans

\text{Radius, r}=\text{28 cm}

\begin{array}{l}\text{Circumference}=\text{2}\pi \text{r}\\ =\text{2}\times \frac{22}{7}\times 28\\ =176\text{cm}\\ \text{Number of rotatinos}=\frac{\text{Total distance to be covered}}{\text{Circumference of the wheel}}\\ =\frac{352\text{m}}{176\text{cm}}\\ =\frac{35200}{176}\\ =200\\ \text{Therefore, it will rotate 200 times}\text{.}\end{array}

Q.17 The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour. (Take π = 3.14)

Ans

\begin{array}{l}\text{Since, Distance travelled by the tip of minute hand}\\ =\text{Circumference of the clock}\\ =\text{2}\pi \text{r}\\ =\text{2}\times \text{3}\text{.14}\times 1\text{5}\\ =\text{94}\text{.2 cm}\\ \text{Therfore, minute hand move 94}\text{.2 cm in 1 hour}\text{.}\end{array}