NCERT Solutions Class 7 Maths Chapter 11

Q.1 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.2 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.3 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.4

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

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.5

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

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.6

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

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.7 Find the area of each of the following parallelogram :

Ans

\begin{array}{l}\text{(a)}\\ \text{Area of Parallelogram}=\text{Base}\times \text{Height}\\ \text{Height}=\text{4 cm}\\ \text{Base}=\text{7 cm}\\ \text{So,}\\ \text{area of parallelogram}=\text{7}\times \text{4}\\ ={\text{28cm}}^2\\ \text{(b)}\\ \text{Area of Parallelogram}=\text{Base}\times \text{Height}\\ \text{Height}=\text{3 cm}\\ \text{Base}=\text{5 cm}\\ \text{So,}\\ \text{area of parallelogram}=\text{5}\times \text{3}\\ ={\text{15 cm}}^2\\ \text{(c)}\\ \text{Area of Parallelogram}=\text{Base}\times \text{Height}\\ \text{Height}=\text{3}\text{.5 cm}\\ \text{Base}=\text{2}\text{.5 cm}\\ \text{So,}\\ \text{area of parallelogram}=\text{2}\text{.5}\times \text{3}\text{.5}\\ =\text{8}{\text{.75 cm}}^2\end{array} \begin{array}{l}\text{(d)}\\ \text{Area of Parallelogram}=\text{Base}\times \text{Height}\\ \text{Height}=\text{4}\text{.8 cm}\\ \text{Base}=\text{5 cm}\\ \text{So,}\\ \text{area of parallelogram}=\text{5}\times \text{4}\text{.8}\\ ={\text{24 cm}}^2\\ \text{(e)}\\ \text{Area of Parallelogram}=\text{Base}\times \text{Height}\\ \text{Height}=\text{4}\text{.4 cm}\\ \text{Base}=\text{2 cm}\\ \text{So,}\\ \text{area of parallelogram}=\text{2}\times \text{4}\text{.4}\\ =\text{8}{\text{.8 cm}}^2\end{array}

Q.8 Find the area of each of the following triangles :

Ans

\begin{array}{l}\text{(a) Area of a triangle}=\frac{1}{2}\times \text{Base}\times \text{Height}\\ =\frac{1}{2}\times 4\times 3\\ =\frac{1}{2}\times 12=6{\text{cm}}^2\\ \text{(b) Area of a triangle}=\frac{1}{2}\times \text{Base}\times \text{Height}\\ =\frac{1}{2}\times 5\times 3.2\\ =\frac{1}{2}\times 16=8{\text{cm}}^2\\ \text{(c) Area of a triangle}=\frac{1}{2}\times \text{Base}\times \text{Height}\\ =\frac{1}{2}\times 3\times 4\\ =\frac{1}{2}\times 12=6{\text{cm}}^2\\ \text{(d) Area of a triangle}=\frac{1}{2}\times \text{Base}\times \text{Height}\\ =\frac{1}{2}\times 3\times 2\\ =\frac{1}{2}\times 6=3{\text{cm}}^2\end{array}

Q.9 Find the missing values :

Ans

\begin{array}{l}\text{Let the height be}h\text{and base be}b.\\ \text{(a) Area of parallelogram=Base}\times \text{Height}\\ \text{So,we get}\\ {\text{246 cm}}^{\text{2}}=\text{20 cm}\times \text{h}\\ \text{h=}\frac{246{\text{cm}}^2}{20\text{cm}}\\ =12.3\text{cm}\\ \text{(b) Area of parallelogram=Base}\times \text{Height}\\ \text{So,we get}\\ \text{154}{\text{.5 cm}}^{\text{2}}=\text{b}\times \text{15 cm}\\ \text{b=}\frac{154.5{\text{cm}}^2}{15\text{cm}}\\ =10.3\text{cm}\end{array} \begin{array}{l}\text{(c) Area of parallelogram=Base}\times \text{Height}\\ \text{So,we get}\\ \text{78}{\text{.72 cm}}^{\text{2}}=\text{b}\times \text{8}\text{.4 cm}\\ \text{b=}\frac{48.72{\text{cm}}^2}{8.4\text{cm}}\\ =5.8\text{cm}\\ \text{(d) Area of parallelogram=Base}\times \text{Height}\\ \text{So,we get}\\ \text{16}{\text{.38 cm}}^{\text{2}}=\text{15}\text{.6 cm}\times \text{h}\\ \text{h=}\frac{16.38{\text{cm}}^2}{15.6\text{cm}}\\ =1.05\text{cm}\\ \text{So, we get}\\ \begin{array}{cccc}\text{S}\text{.No}\text{.}& \text{Base}& \text{Height}& \text{Area of the Parallelogram}\\ \text{a}\text{.}& 20\text{cm}& 12.3\text{cm}& 246{\text{cm}}^2\\ \text{b}\text{.}& 10.3\text{cm}& 15\text{cm}& 154.5{\text{cm}}^2\\ \text{c}\text{.}& 5.8\text{cm}& 8.4\text{cm}& 78.72{\text{cm}}^2\\ \text{d}\text{.}& 15.6\text{cm}& 1.05\text{cm}& 16.38{\text{cm}}^2\end{array}\end{array}

Q.10 Find the missing values :

Ans

\begin{array}{l}\text{(a) Let the height be}h\text{and base be}b\text{.}\\ \text{Area of triangle}=\frac{1}{2}\times \text{base}\times \text{height}\\ {\text{87cm}}^2=\frac{1}{2}\times 15\text{cm}\times \text{h}\\ \text{h}=\frac{87{\text{cm}}^2\times 2}{15\text{cm}}\\ =11.6\text{cm}\\ \text{(b) Let the height be}h\text{and base be}b\text{.}\\ \text{Area of triangle}=\frac{1}{2}\times \text{base}\times \text{height}\\ {\text{1256 mm}}^2=\frac{1}{2}\times b\times \text{31}\text{.4 mm}\\ b=\frac{1256{\text{mm}}^2\times 2}{31.4\text{mm}}\\ =80\text{mm}\end{array} \begin{array}{l}\text{(c) Let the height be}h\text{and base be}b\text{.}\\ \text{Area of triangle}=\frac{1}{2}\times \text{base}\times \text{height}\\ \text{170}{\text{.5 cm}}^2=\frac{1}{2}\times 22\text{cm}\times \text{h}\\ \text{h}=\frac{170.5{\text{cm}}^2\times 2}{22\text{cm}}\\ =15.5\text{cm}\\ \text{So we get}\\ \begin{array}{ccc}\text{Base}& \text{Height}& \text{Area of Triangle}\\ 15\text{cm}& 11.6\text{cm}& 84{\text{cm}}^2\\ 80\text{mm}& 31.4\text{mm}& 1256{\text{mm}}^2\\ 22\text{cm}& 15.5\text{cm}& 170.5{\text{cm}}^2\end{array}\end{array}

Q.11

$\begin{array}{l}\mathrm{PQRS}\mathrm{is}\mathrm{a}\mathrm{parallelogram}.\hspace{0.33em}\mathrm{QM}\mathrm{is}\mathrm{the}\mathrm{height}\mathrm{from}\mathrm{Q}\mathrm{to}\mathrm{SR}\\ \mathrm{and}\mathrm{QN}\mathrm{is}\mathrm{the}\mathrm{height}\mathrm{from}\mathrm{Q}\mathrm{to}\mathrm{PS}.\mathrm{If}\mathrm{SR}=12\mathrm{cm}\mathrm{and}\\ \mathrm{QM}=\; 7.6\mathrm{cm}.\mathrm{Find}:\\ \left(\mathrm{a}\right)\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{parallegram}\mathrm{PQRS}\\ \left(\mathrm{b}\right)\mathrm{QN},\mathrm{if}\mathrm{PS}=\; 8\mathrm{cm}\end{array}$

Ans

\begin{array}{l}\text{(a)}\\ \text{Area of a parallelogram}=\text{Base}\times \text{Height}\\ =\text{SR}\times \text{QM}\\ =7.6\times 12\\ =91.2{\text{cm}}^2\\ \text{(b)}\\ \text{Area of a parallelogram}=\text{Base}\times \text{Height}\\ =\text{PS}\times \text{QN}\\ =91.2{\text{cm}}^2\\ \text{QN}\times 8=91.2\\ \text{QN}=\frac{91.2}{8}\\ =11.4\text{cm}\end{array}

Q.12

$\begin{array}{l}\mathrm{DL}\mathrm{and}\mathrm{BM}\mathrm{are}\mathrm{the}\mathrm{heights}\mathrm{on}\mathrm{sides}\mathrm{AB}\mathrm{and}\mathrm{AD}\mathrm{respectively}\\ \mathrm{of}\mathrm{parallelogram}\mathrm{ABCD}.\mathrm{If}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{parallelogram}\mathrm{is}\\ 1470{\mathrm{cm}}^{\mathrm{2}},\mathrm{AB}=\; 35\mathrm{cm}\mathrm{and}\mathrm{AD}=\; 49\mathrm{cm},\mathrm{find}\mathrm{the}\mathrm{length}\mathrm{of}\\ \mathrm{BMand}\mathrm{DL}.\end{array}$

Ans

\begin{array}{l}\text{Area of parallelogram}=\text{Base}\times \text{Height}\\ =\text{AB}\times \text{DL}\\ \text{1470}=\text{35}\times \text{DL}\\ \text{DL}=\frac{1470}{35}\\ =42\text{cm}\\ \text{Also,}\\ 1470=\text{AD}\times \text{BM}\\ \text{1470}=\text{49}\times \text{BM}\\ \text{BM}=\frac{1470}{49}\\ =30\text{cm}\end{array}

Q.13

$\begin{array}{l}△\mathrm{ABC}\mathrm{is}\mathrm{right}\mathrm{angled}\mathrm{at}\mathrm{A}.\mathrm{AD}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{BC}.\\ \mathrm{If}\mathrm{AB}=5\mathrm{cm},\hspace{0.33em}\mathrm{BC}=13\mathrm{cm}\mathrm{and}\mathrm{AC}=12\mathrm{cm},\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\\ △\mathrm{ABC}.\mathrm{Also}\mathrm{find}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{AD}.\end{array}$

Ans

\begin{array}{l}\text{Since, Area}=\frac{1}{2}\times \text{Base}\times \text{Height}\\ =\frac{1}{2}\times \text{5}\times \text{12}\\ ={\text{30 cm}}^2\\ \text{Also, area of triangle}=\frac{1}{2}\times \text{AD}\times \text{BC}\\ \text{30}=\frac{1}{2}\times \text{AD}\times \text{13}\\ \text{AD}=\frac{30\times 2}{12}\\ =4.6\text{cm}\end{array}

Q.14

$\begin{array}{l}△\mathrm{ABC}\mathrm{is}\mathrm{isosceles}\mathrm{with}\mathrm{AB}=\mathrm{AC}=7.5\mathrm{cm}\mathrm{and}\mathrm{BC}=9\mathrm{cm}.\\ \mathrm{The}\mathrm{height}\mathrm{AD}\mathrm{from}\mathrm{A}\mathrm{to}\mathrm{BC},\mathrm{is}6\mathrm{cm}.\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}△\mathrm{ABC}.\\ \mathrm{What}\mathrm{will}\mathrm{be}\mathrm{the}\mathrm{height}\mathrm{from}\mathrm{C}\mathrm{to}\mathrm{AB}\mathrm{i}.\mathrm{e}.,\mathrm{CE}?\end{array}$

Ans

\begin{array}{l}\text{Area of}\Delta \text{ABC}=\frac{1}{2}\times \text{Base}\times \text{Height}\\ =\frac{1}{2}\text{BC}\times \text{AD}=\frac{1}{2}\times \text{9}\times \text{6}={\text{27cm}}^{\text{2}}\\ \text{Also, area of}\Delta \text{ABC}=\frac{1}{2}\times \text{Base}\times \text{Height}\\ =\frac{1}{2}\text{AB}\times \text{CE}\\ 27=\frac{1}{2}\text{7}\text{.5}\times \text{CE}\\ \text{CE}=\frac{27\times 2}{7.5}=\text{7}\text{.2 cm}\end{array}

Q.15

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

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.16

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

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.17 If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take π =

$\frac{22}{7}$

)

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.18 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.19  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.20

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

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.21

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

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.22

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

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.23

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.24 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.25

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

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.26 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.27 How many times a wheel of radius 28 cm must rotate to go 352 m? (take π =

$\frac{22}{7}$

)

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.28 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}

Q.29 A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectare.

Ans

\begin{array}{l}\text{Length (L) of garden}=\text{90 m}\\ \text{Breadth (B) of garden}=\text{75 m}\\ \text{Area of garden}=\text{L}\times \text{B}\\ =\text{90 m}\times \text{75 m}\\ ={\text{6750 m}}^2\end{array} \begin{array}{l}\text{From the figure, it can be observed that the new length}\\ \text{and breadth of the garden,when path is also included are}\\ \text{100 m and 85 m respectively}\\ \text{Area of the garden including path}=\text{100 m}\times \text{80 m}\\ ={\text{8000 m}}^2\\ \text{Area of the path}\\ =\text{Area of the garden including path}-\text{Area of garden}\\ =8000{\text{m}}^2-{\text{6750 m}}^2\\ =1750{\text{m}}^2\\ 1\text{hectare}={\text{10000 m}}^2\\ \text{Therefore,}\\ \text{area of the garden in hectare}=\frac{6750}{10000}\\ =0.675\text{hectare}\end{array}

Q.30 A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65m. Find the area of the path.

Ans

\begin{array}{l}\text{Length (L) of park}=\text{125 m}\\ \text{Breadth (B) of park}=\text{65 m}\\ \text{Area of park}=\text{L}\times \text{B}\\ =\text{125 m}\times \text{65 m}\\ ={\text{8125 m}}^2\end{array} \begin{array}{l}\text{From the figure, it can be observed that the new length}\\ \text{and breadth of the park,when path is also included are}\\ \text{131 m and 71 m respectively}\\ \text{Area of the park including path}=\text{131 m}\times \text{71 m}\\ ={\text{9301 m}}^2\\ \text{Area of the path}\\ =\text{Area of the park including path}-\text{Area of park}\\ =9301{\text{m}}^2-{\text{8125 m}}^2\\ =1176{\text{m}}^2\end{array}

Q.31 A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.

Ans

\begin{array}{l}\text{Length (L) of cardboard}=\text{125 m}\\ \text{Breadth (B) of cardboard}=\text{65 m}\\ \text{Area of cardboard including margin}=\text{L}\times \text{B}\\ =\text{8 cm}\times \text{5 cm}\\ =40{\text{cm}}^2\end{array} \begin{array}{l}\text{From the figure, it can be observed that the new length}\\ \text{and breadth of the cardboard,when margin is not included}\\ \text{are 5 cm and 2 cm respectively}\\ \text{Area of the cardboard not including margin}=\text{5 cm}\times \text{2 m}\\ ={\text{10 cm}}^2\\ \text{Area of the margin}\\ =\text{Area of the cardboard including margin}\\ -\text{Area of cardboard not including margin}\\ =40{\text{cm}}^2-{\text{10 cm}}^2\\ =30{\text{cm}}^2\end{array}

Q.32

$\begin{array}{l}\mathrm{Two}\mathrm{cross}\mathrm{roads},\mathrm{each}\mathrm{of}\mathrm{width}10\mathrm{m},\mathrm{cut}\mathrm{at}\mathrm{right}\mathrm{angles}\\ \mathrm{through}\mathrm{the}\mathrm{centre}\mathrm{of}\mathrm{a}\mathrm{rectangular}\mathrm{park}\mathrm{of}\mathrm{length}700\mathrm{m}\\ \mathrm{and}\mathrm{breadth}300\mathrm{m}\mathrm{and}\mathrm{parallel}\mathrm{to}\mathrm{its}\mathrm{sides}.\mathrm{Find}\mathrm{the}\mathrm{area}\\ \mathrm{of}\mathrm{the}\mathrm{roads}.\mathrm{Also}\mathrm{find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{park}\mathrm{excluding}\mathrm{cross}\\ \mathrm{roads}.\mathrm{Give}\mathrm{the}\mathrm{answer}\mathrm{in}\mathrm{hectares}.\end{array}$

Ans

\begin{array}{l}\text{Length (L) of the park}=\text{700 m}\\ \text{Breadth (B) of the park}=\text{300 m}\\ \text{Area of park}=\text{700}\times \text{300}\\ ={\text{210000 m}}^2\\ \text{Length of road PQRS}=\text{700 m}\\ \text{Length of road ABCD}=\text{300 m}\\ \text{Width of each road}=\text{10 m}\\ \text{Area of the roads}=\text{area(PQRS)+area(ABCD)}-\text{area(KLMN)}\\ =\left(700\times 10\right)+\left(300\times 10\right)-\left(10\times 10\right)\\ =7000+3000-100\\ =9900{\text{m}}^2\\ =0.99\text{hectare}\end{array} \begin{array}{l}\\ \text{Area of park excluding roads}=\text{210000-9900}\\ ={\text{200100 m}}^2\\ =20.01\text{hectare}\end{array}

Q.33 Pragya wrapped a cord around a circular pipe of radius 4 cm (adjoining figure) and cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown). Did she have any left? (π = 3.14)

Ans

\begin{array}{l}\text{Perimeter of the square}=\text{4}\times \text{side of the square}\\ =\text{4}\times \text{4}\\ =\text{16 cm}\\ \text{Perimeter of circular pip}=\text{2}\pi \text{r}\\ =\text{3}\times \text{3}\text{.14}\times \text{4}\\ =\text{25}\text{.12 cm}\\ \text{Therefore,}\\ \text{length of chord left with Pragya}=\text{25}\text{.12 cm}-\text{16 cm}\\ =\text{9}\text{.12 cm}\end{array}

Q.34 The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find:
(i) the area of the whole land
(ii) the area of the flower bed
(iii) the area of the lawn excluding the area of the flower bed.
(iv) the circumference of the flower bed.

Ans

\begin{array}{l}\text{(i) Area of whole land}=\text{Length}\times \text{Breadth}\\ =\text{10}\times \text{5}\\ ={\text{50 m}}^2\\ \text{(ii) Area of flower bed}=\pi {\text{r}}^2\\ =3.14\times 2\times 2\\ =12.56{\text{m}}^2\\ \text{(iii) Area of the lawn excluding the flower bed}\\ =\text{Area of whole land}-\text{Area of flower bed}\\ =\text{50-12}\text{.56}\\ =\text{37}{\text{.44 m}}^2\end{array}

Q.35 In the following figures, find the area of the shaded portions:

Ans

\begin{array}{l}\text{(i)}\\ \text{Area of EFDC}=\text{area(ABCD)}-\text{area(BCE)}–\text{area(AFE)}\\ =\left(18\times 10\right)-\frac{1}{2}\left(10\times 8\right)-\frac{1}{2}\left(6\times 10\right)\\ =180-40-30\\ =110{\text{cm}}^2\\ \text{(ii)}\\ \text{area(QTU)}=\text{area(PQRS)}-\text{area(TSU)}-\text{area(RUQ)}-\text{area(PQT)}\\ =\left(20\times 20\right)-\frac{1}{2}\left(10\times 10\right)-\frac{1}{2}\left(20\times 10\right)-\frac{1}{2}\left(20\times 10\right)\\ =400-50-100-100\\ =150{\text{cm}}^2\end{array}

Q.36

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{quadrilateral}\mathrm{ABCD}.\\ \mathrm{Here},\mathrm{AC}=22\mathrm{cm},\mathrm{BM}=3\mathrm{cm},\mathrm{DN}=3\mathrm{cm},\\ \mathrm{and}\mathrm{BM}\perp \mathrm{AC},\mathrm{DN}\perp \mathrm{AC}.\end{array}$

Ans

\begin{array}{l}\text{area(ABCD)}=\text{area(ABC)+area(ADC)}\\ =\frac{1}{2}\left(3\times 22\right)\frac{1}{2}\left(3\times 22\right)\\ =33+33\\ =66{\text{cm}}^2\end{array}

Q.37

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

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.38

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

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}

Q.39

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

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.40

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

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.41

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

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.42

$\begin{array}{l}\mathrm{A}\mathrm{verandah}\mathrm{of}\mathrm{width}2.25\mathrm{m}\mathrm{is}\mathrm{constructed}\mathrm{all}\mathrm{along}\\ \mathrm{outside}\mathrm{a}\mathrm{room}\mathrm{which}\mathrm{is}5.5\mathrm{mlong}\mathrm{and}4\mathrm{m}\mathrm{wide}.\mathrm{Find}:\\ \left(\mathrm{i}\right)\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{verandah}.\\ \left(\mathrm{ii}\right)\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{cementing}\mathrm{the}\mathrm{floor}\mathrm{of}\mathrm{the}\mathrm{verandah}\mathrm{at}\mathrm{the}\\ \mathrm{rate}\mathrm{of}\mathrm{Rs}200\mathrm{per}{\mathrm{m}}^{\mathrm{2}}\end{array}$

Ans

\begin{array}{l}\text{(i)}\\ \text{Length (L) of room}=\text{5}\text{.5 m}\\ \text{Breadth (B) of room}=\text{4 m}\\ \text{Area of room including margin}=\text{L}\times \text{B}\\ =\text{5}\text{.5 m}\times \text{4 m}\\ =22{\text{m}}^2\end{array} \begin{array}{l}\text{From the figure, it can be observed that the new length}\\ \text{and breadth of the room,when verandah is also included}\\ \text{are 10 m and 8}\text{.5 m respectively}\\ \text{Area of the verandah including margin}=\text{10 m}\times \text{8}\text{.5 m}\\ ={\text{85 m}}^2\\ \text{Area of the verandah}\\ =\text{Area of the room including verandah}-\text{Area of the room}\\ =85{\text{m}}^2-{\text{22 m}}^2\\ =63{\text{m}}^2\end{array} \begin{array}{l}\end{array} \begin{array}{l}\text{(ii)}\\ {\text{Cost of cementing 1m}}^{\text{2}}\text{area of floor of verandah}=₹200\\ {\text{Cost of cementing 63m}}^{\text{2}}\text{area of floor of verandah}=₹200\times 63\\ =₹12600\end{array}

Q.43

$\begin{array}{l}\mathrm{A}\mathrm{path}1\mathrm{m}\mathrm{wide}\mathrm{is}\mathrm{built}\mathrm{along}\mathrm{the}\mathrm{border}\mathrm{and}\mathrm{inside}\mathrm{a}\\ \mathrm{square}\mathrm{garden}\mathrm{of}\mathrm{side}30\mathrm{m}.\mathrm{Find}:\\ \left(\mathrm{i}\right)\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{path}\\ \left(\mathrm{ii}\right)\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{planting}\mathrm{grass}\mathrm{in}\mathrm{the}\mathrm{remaining}\mathrm{portion}\mathrm{of}\mathrm{the}\\ \mathrm{garden}\mathrm{at}\mathrm{the}\mathrm{rate}\mathrm{of}₹40\mathrm{per}{\mathrm{m}}^{\mathrm{2}}\mathrm{.}\end{array}$