NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area (EX 11.4) Exercise 11.4

Q.1 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.2 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.3 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.4

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

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.5

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

Ans

\begin{array}{l}\text{(i)}\\ \text{Side (a) of square garden}=\text{30 m}\\ \text{Area of square garden including path}={\text{a}}^2\\ ={\left(\text{30 m}\right)}^2\\ ={\text{900 m}}^2\\ \text{From the figure, it can be observed that the side of the square}\\ \text{garden, when path is not included, is 28 m}\text{.}\\ \text{Area of the square garden not including the path}={\left(28\right)}^2\\ ={\text{784 m}}^2\\ \text{Area of path}\\ =\text{Area of the square garden including the path}\\ -\text{Area of square garden not including the path}\\ =900{\text{m}}^2-784{\text{m}}^2\\ =116{\text{m}}^2\\ \text{(ii)}\\ {\text{Cost of planting grass in 1 m}}^2\text{area of the garden}=\text{₹40}\\ {\text{Cost of planting grass in 784 m}}^2\text{area of the garden}=\text{₹40}\times \text{784}\\ =\text{₹31360}\end{array}

Q.6

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

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.7

$\begin{array}{l}\mathrm{Through}\mathrm{a}\mathrm{rectangular}\mathrm{field}\mathrm{of}\mathrm{length}90\mathrm{m}\mathrm{and}\mathrm{breadth}\\ 60\mathrm{m},\mathrm{two}\mathrm{roads}\mathrm{are}\mathrm{constructed}\mathrm{which}\mathrm{are}\mathrm{parallel}\mathrm{to}\mathrm{the}\\ \mathrm{sides}\mathrm{and}\mathrm{cut}\mathrm{each}\mathrm{other}\mathrm{at}\mathrm{right}\mathrm{angles}\mathrm{through}\mathrm{the}\mathrm{centre}\\ \mathrm{of}\mathrm{the}\mathrm{fields}.\mathrm{If}\mathrm{the}\mathrm{width}\mathrm{of}\mathrm{each}\mathrm{road}\mathrm{is}3\mathrm{m},\mathrm{find}\\ \left(\mathrm{i}\right)\mathrm{the}\mathrm{area}\mathrm{covered}\mathrm{by}\mathrm{the}\mathrm{roads}.\\ \left(\mathrm{ii}\right)\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{constructing}\mathrm{the}\mathrm{roads}\mathrm{at}\mathrm{the}\mathrm{rate}\mathrm{of}\\ ₹110\mathrm{per}{\mathrm{m}}^{\mathrm{2}}\mathrm{.}\end{array}$

Ans

\begin{array}{l}\text{Length (L) of the field}=\text{90 m}\\ \text{Breadth (B) of the field}=\text{60 m}\\ \text{Area of field}=\text{90}\times \text{60}\\ ={\text{5400 m}}^2\\ \text{Length of road PQRS}=\text{90 m}\\ \text{Length of road ABCD}=\text{60 m}\\ \text{Width of each road}=\text{3 m}\\ \text{Area of the roads}=\text{area(PQRS)+area(ABCD)}-\text{area(KLMN)}\\ =\left(90\times 3\right)+\left(60\times 3\right)-\left(3\times 3\right)\\ =270+180-100\\ =441{\text{m}}^2\\ {\text{Cost for constructing 1 m}}^2\text{road}=\text{₹110}\\ {\text{Cost for constructing 4411 m}}^2\text{road}=\text{₹110}\times 441\\ =₹48510\end{array}

Q.8 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.9 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.10 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.11

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

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}