NCERT Solutions Class 8 Maths Chapter 11

Q.1 A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?

Ans

$\begin{array}{l}\mathrm{Perimeter}\mathrm{of}\mathrm{square}=4\times \mathrm{side}\\ =4\times 60\mathrm{m}\\ =240\mathrm{m}\\ \mathrm{Perimeter}\mathrm{of}\mathrm{rectangle}=2(\mathrm{Length}+\mathrm{Breadth})\\ =2(80\mathrm{m}+\mathrm{Breadth})\\ =160\mathrm{m}+2\times \mathrm{Breadth}\end{array}$

$\begin{array}{l}\mathrm{Perimeter}\mathrm{of}\mathrm{the}\mathrm{square}=\mathrm{Perimeter}\mathrm{of}\mathrm{the}\mathrm{rectangle}\\ \Rightarrow 160\mathrm{m}+2\times \mathrm{Breadth}=240\mathrm{m}\\ \Rightarrow \mathrm{Breadth}\mathrm{of}\mathrm{the}\mathrm{rectangle}=\left(\frac{80}{2}\right)\mathrm{m}=40\mathrm{m}\\ \mathrm{Area}\mathrm{of}\mathrm{square}={\left(\mathrm{Side}\right)}^2={\left(60\mathrm{m}\right)}^2=3600{\mathrm{m}}^2\\ \mathrm{Area}\mathrm{of}\mathrm{rectangle}=\mathrm{Length}\times \mathrm{Breadth}\\ =(80\times 40){\mathrm{m}}^2\\ =3200{\mathrm{m}}^2\\ \\ \mathrm{Therefore},\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{square}\mathrm{field}\mathrm{is}\mathrm{larger}\mathrm{than}\\ \mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{rectangular}\mathrm{field}.\end{array}$

Q.2 Mrs. Kaushik has a square plot with the measurement as shown in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of ₹ 55 per m².

Ans

$\begin{array}{l}\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{square}\mathrm{plot}={\left(25\right)}^2{\mathrm{m}}^2=625{\mathrm{m}}^2\\ \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{house}=\left(15\mathrm{m}\right)\times \left(20\mathrm{m}\right)=300{\mathrm{m}}^2\\ \\ \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{remaining}\mathrm{portion}=\left(\begin{array}{l}\mathrm{Area}\mathrm{of}\\ \mathrm{square}\mathrm{plot}\end{array}\right)-\left(\begin{array}{l}\mathrm{Area}\mathrm{of}\mathrm{the}\\ \mathrm{house}\end{array}\right)\\ =625{\mathrm{m}}^2-300{\mathrm{m}}^2\\ =325{\mathrm{m}}^2\\ \\ \therefore \mathrm{Total}\mathrm{cost}\mathrm{of}\mathrm{developing}\mathrm{a}\mathrm{garden}\mathrm{around}\mathrm{the}\\ \mathrm{house}\mathrm{at}\mathrm{the}\mathrm{rate}\mathrm{of}\mathrm{Rs}55\mathrm{per}{\mathrm{m}}^2=₹(55\times 325)\\ =₹17,875\end{array}$

Q.3 The shape of a garden is rectangular in the middle and semi circular at the ends as shown in the diagram. Find the area and the perimeter of this garden [Length of rectangle is 20 – (3.5 + 3.5) metres].

Ans

$\begin{array}{l}\text{Length of the rectangle}=[\text{2}0-(\text{3}.\text{5}+\text{3}.\text{5})]\text{metres}=\text{13 m}\\ \text{Circumference of 1 semi}-\text{circular part}=\mathrm{\pi r}=\frac{22}{7}\times 3.5=11\mathrm{m}.\end{array}$

$\begin{array}{l}\therefore \mathrm{Perimeter}\mathrm{of}\mathrm{the}\mathrm{garden}=\mathrm{AB}+\mathrm{Length}\mathrm{of}\mathrm{both}\\ \mathrm{semi}–\mathrm{circular}\mathrm{regions}+\mathrm{CD}\\ =13\mathrm{m}+22\mathrm{m}+13\mathrm{m}\\ =48\mathrm{m}\\ \\ \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{garden}=\mathrm{Area}\mathrm{of}\mathrm{rectangle}+2\times \mathrm{Area}\mathrm{of}\mathrm{two}\\ \mathrm{semi}–\mathrm{circular}\mathrm{regions}\\ =[(13\times 7)+2\times \frac{1}{2}\times \frac{22}{7}\times {(3.5)}^2]{\mathrm{m}}^2\\ =(91+38.5){\mathrm{m}}^2\\ =129.5{\mathrm{m}}^2\\ \\ \therefore \mathrm{Area}\text{and perimeter}\mathrm{of}\mathrm{the}\mathrm{garden}\mathrm{are}129.5{\mathrm{m}}^2\mathrm{and}\text{\hspace{0.17em}48 m.}\end{array}$

Q.4 A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m²? (If required you can split the tiles in whatever way you want to fill up the corners).

Ans

$\begin{array}{l}\mathrm{Area}\mathrm{of}\mathrm{parallelogram}=\mathrm{Base}\times \mathrm{Height}\\ \mathrm{Hence},\mathrm{area}\mathrm{of}\mathrm{one}\mathrm{tile}=24\mathrm{cm}\times 10\mathrm{cm}=240{\mathrm{cm}}^2\\ \mathrm{Required}\mathrm{number}\mathrm{of}\mathrm{tiles}=\frac{\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{floor}}{\mathrm{Area}\mathrm{of}\mathrm{each}\mathrm{tile}}=\frac{1080{\mathrm{m}}^2}{240{\mathrm{cm}}^2}\\ \mathrm{Since}1\mathrm{m}=100\mathrm{cm}\\ \Rightarrow 1{\mathrm{m}}^2=10000{\mathrm{cm}}^2\\ \therefore 1080{\mathrm{m}}^2=10000\times 1080{\mathrm{cm}}^2\\ =10,800,000{\mathrm{cm}}^2\\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{number}\mathrm{of}\mathrm{tiles}=\frac{10,800,000{\mathrm{cm}}^2}{240{\mathrm{cm}}^2}\\ =45000\end{array}$

Q.5 An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression c = 2πr, where r is the radius of the circle.

Ans

$\begin{array}{l}\left(\text{a}\right)\text{Radius of semi-circular part}=\left\{\frac{2.8}{2}\right\}\text{cm =1.4 cm}\\ \text{Perimeter of the given figure}=\text{2}.\text{8 cm}+\mathrm{\pi r}\\ =\text{2}.\text{8 cm}+(\frac{22}{7}\times 1.4)\mathrm{cm}\\ =\text{2}.\text{8 cm}+4.4\text{}\mathrm{cm}\\ =7.2\text{}\mathrm{cm}\end{array}$

$\begin{array}{l}\left(\text{b}\right)\text{Radius of semi-circular part}=\left\{\frac{2.8}{2}\right\}\text{cm=1.4 cm}\\ \text{Perimeter of the given figure}=\text{1}.\text{5 cm}+\text{2}.\text{8 cm}+\text{1}.\text{5 cm}+\mathrm{\pi }(\text{1}.\text{4 cm})\\ =5.\text{8 cm}+\frac{22}{7}\times 1.4\mathrm{cm}\\ =5.\text{8 cm}+4.4\text{}\mathrm{cm}\\ =10.2\text{}\mathrm{cm}\end{array}$ \begin{array}{l}\end{array}

$\begin{array}{l}\text{(c})\text{Radius of semi-circular part}=\left\{\frac{2.8}{2}\right\}\text{cm=1.4 cm}\\ \text{Perimeter of the given figure}=\text{2 cm}+\mathrm{\pi r}+\text{2 cm}\\ =4\text{cm}+\frac{22}{7}\times 1.4\mathrm{cm}\\ =4\text{cm}+4.4\text{}\mathrm{cm}\\ =8.4\text{}\mathrm{cm}\end{array}$ \begin{array}{l}\end{array}

\begin{array}{l}\text{Thus},\text{the ant will have to take a longer round for the food}\\ \text{piece}\left(\text{b}\right),\text{because the perimeter of the figure given in}\\ \text{option}\left(\text{b}\right)\text{is the greatest among all}.\end{array} \begin{array}{l}\end{array}

Q.6 The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.

Ans

$\begin{array}{l}\text{Area of trapezium}=\frac{1}{2}\left(\text{Sum of parallel sides}\right)\times \left(\begin{array}{l}\text{Distances between}\\ \text{parallel sides}\end{array}\right)\\ =\left[\frac{1}{2}(1+1.2)(0.8)\right]{\mathrm{m}}^2\\ =0.88\text{}{\mathrm{m}}^2\end{array}$

Q.7 The area of a trapezium is 34 cm² and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.

Ans

$\begin{array}{l}\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that},\mathrm{area}\mathrm{of}\mathrm{trapezium}=34{\mathrm{cm}}^2\\ \mathrm{and}\mathrm{height}=4\mathrm{cm}\\ \mathrm{Let}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{one}\mathrm{parallel}\mathrm{side}\mathrm{be}\mathrm{a}.\\ \mathrm{We}\mathrm{know}\mathrm{that},\\ \mathrm{Area}\mathrm{of}\mathrm{trapezium}=\frac{1}{2}\left(\mathrm{Sum}\mathrm{of}\mathrm{parallel}\mathrm{sides}\right)\times \left(\begin{array}{l}\mathrm{Distances}\mathrm{between}\\ \mathrm{parallel}\mathrm{sides}\end{array}\right)\end{array}$

$\begin{array}{l}34{\mathrm{cm}}^2=\frac{1}{2}(10\text{}\mathrm{cm}+\mathrm{a})\times 4\mathrm{cm}\\ \Rightarrow 34{\mathrm{cm}}^2=2(10\text{}\mathrm{cm}+\mathrm{a})\\ \Rightarrow 17\mathrm{cm}=10\text{}\mathrm{cm}+\mathrm{a}\\ \Rightarrow \mathrm{a}=17\mathrm{cm}-10\text{}\mathrm{cm}=7\mathrm{cm}\\ \\ \text{Thus},\text{the length of the other parallel side is 7 cm}.\end{array}$

Q.8 Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

Ans

$\begin{array}{l}\mathrm{Length}\mathrm{of}\mathrm{the}\mathrm{fence}\mathrm{of}\text{a}\mathrm{trapezium}\mathrm{shaped}\mathrm{field}\mathrm{ABCD}=\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA}\\ \Rightarrow 120\mathrm{m}=\mathrm{AB}+48\mathrm{m}+17\mathrm{m}+40\mathrm{m}\\ \Rightarrow \mathrm{AB}=120\mathrm{m}-105\mathrm{m}\\ =15\mathrm{m}\end{array}$

$\begin{array}{l}\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{field}\mathrm{ABCD}=\frac{1}{2}(\mathrm{AD}+\mathrm{BC})\times \mathrm{AB}\\ =\left[\frac{1}{2}(40+48)\times 15\right]{\mathrm{m}}^2\\ =\left(\frac{1}{2}\times 88\times 15\right){\mathrm{m}}^2\\ =660\text{}{\mathrm{m}}^2\\ \\ \therefore \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{field}\mathrm{ABCD}=660\text{}{\mathrm{m}}^2\end{array}$

Q.9 The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.

Ans

$\begin{array}{l}\text{It is given that},\\ \text{Length of the diagonal},\mathrm{d}=\text{24 m}\\ \text{Length of the perpendiculars},{\mathrm{h}}_{\text{1}}\text{and}{\mathrm{h}}_{\text{2}},\text{from the opposite vertices}\\ \text{to the diagonal are 8 m and 13 m}\\ \text{Area of the quadrilateral =}\frac{1}{2}\mathrm{d}\text{}\left({\mathrm{h}}_1+{\mathrm{h}}_2\right)\\ =\frac{1}{2}\times 24\text{}\times \left(13\mathrm{m}+8\text{m}\right)\\ =\frac{1}{2}\times 24\text{}\times 21\\ =252{\mathrm{m}}^2\\ \\ \text{Thus},{\text{the area of the field is 252 m}}^{\text{2}}.\end{array}$

Q.10 The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

Ans

\begin{array}{l}\text{Area of rhombus}=\frac{1}{2}\left(\text{Product of its diagonals}\right)\\ \text{Therefore},\text{area of the given rhombus}=\frac{1}{2}\left(\text{7}\text{.5×12}\right)\\ ={\text{45 cm}}^{\text{2}}\end{array}

Q.11 Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.

Ans

$\begin{array}{l}\text{Let the other diagonal of the rhombus be}x.\\ \text{Since rhombus is also a parallelogram}\text{.}\\ \text{Therefore, area of the given rhombus}=\text{Base}\times \text{Height}\\ =\text{5 cm}\times \text{4.8 cm}\\ ={\text{24 cm}}^{\text{2}}\\ \text{Also, area of rhombus}=\frac{1}{2}\left(\text{Product of its diagonals}\right)\\ \Rightarrow 24{\text{cm}}^{\text{2}}=\frac{1}{2}(\text{8 cm}\times \mathrm{x})\\ \Rightarrow \mathrm{x}=\frac{24\times 2}{8}\text{cm}=6\text{cm}\\ \\ \therefore \text{The length of the other diagonal of the rhombus is 6 cm and}\\ {\text{area of the given rhombus is 24 cm}}^{\text{2}}.\end{array}$

Q.12 The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m² is ₹ 4.

Ans

$\begin{array}{l}\text{Area\hspace{0.33em}of\hspace{0.33em}rhombus\hspace{0.33em}=\hspace{0.33em}}\frac{\text{1}}{\text{2}}\left(\text{Product\hspace{0.33em}of\hspace{0.33em}its\hspace{0.33em}diagonals}\right)\\ \text{Area\hspace{0.33em}of\hspace{0.33em}each\hspace{0.33em}tile\hspace{0.33em}=\hspace{0.33em}}\left(\frac{\text{1}}{\text{2}}\text{×45×30}\right){\text{cm}}^{\text{2}}\\ {\text{= 675 cm}}^{\text{2}}\\ \text{Area\hspace{0.33em}of\hspace{0.33em}3000\hspace{0.33em}tiles\hspace{0.33em}=\hspace{0.33em}}\left(\text{675×3000}\right){\text{cm}}^{\text{2}}\\ {\text{\hspace{0.33em} \hspace{0.33em} \hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}2025000 cm}}^{\text{2}}\\ {\text{\hspace{0.33em} \hspace{0.33em} \hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}202.5 m}}^{\text{2}}\\ \therefore {\text{the\hspace{0.33em}total\hspace{0.33em}cost\hspace{0.33em}of\hspace{0.33em}polishing\hspace{0.33em}the\hspace{0.33em}floor,\hspace{0.33em}if\hspace{0.33em}the\hspace{0.33em}cost\hspace{0.33em}per\hspace{0.33em}m}}^{\text{2}}\text{\hspace{0.33em}is\hspace{0.33em}Rs\hspace{0.33em}4}\\ \begin{array}{l}=₹\left(\text{4}\times \text{2}0\text{2}.\text{5}\right)\\ =₹\text{81}0\end{array}\\ \text{Thus,\hspace{0.33em}the\hspace{0.33em}cost\hspace{0.33em}of\hspace{0.33em}polishing\hspace{0.33em}the\hspace{0.33em}floor\hspace{0.33em}is\hspace{0.33em}Rs\hspace{0.33em}810.}\end{array}$

Q.13 Mohan wants to buy a trapezium shaped field.

Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m² and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

Ans

\begin{array}{l}\text{Let the length of the field along the road be}l\text{m}.\\ \text{Hence},\text{the length of the field along the river will be 2}l\text{m}.\\ \text{Area of trapezium}=\frac{1}{2}\left(\text{Sum of parallel sides}\right)\left(\begin{array}{l}\text{Distance between}\\ \text{the parallel sides}\end{array}\right)\end{array}

$\begin{array}{l}\Rightarrow 10500{\text{m}}^{\text{2}}=\frac{1}{2}(\mathrm{l}+2\mathrm{l})\left(100\text{m}\right)\\ \Rightarrow 3\mathrm{l}=\left(\frac{2\times 10500}{100}\right)\text{m}\\ \Rightarrow 3\mathrm{l}=\text{210 m}\\ \Rightarrow \mathrm{l}=\text{70 m}\\ \\ \\ \text{Thus},\text{length of the field along the river}=(\text{2}\times \text{7}0)\text{m}=\text{14}0\text{m}\\ \end{array}$

Q.14 Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.

Ans

Side of regular octagon = 5 cm
Area of trapezium ABCH = Area of trapezium DEFG

$\begin{array}{l}\text{Area of trapezium ABCH\hspace{0.33em}=\hspace{0.33em}}\left[\frac{\text{1}}{\text{2}}\text{×4×}\left(\text{11+5}\right)\right]{\text{m}}^{\text{2}}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}}\left[\frac{\text{1}}{\text{2}}\text{×4×16}\right]{\text{m}}^{\text{2}}\\ {\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}\hspace{0.33em}32\hspace{0.17em}m}}^{\text{2}}\end{array}$

Area of rectangle HGDC = 11 × 5 = 55 m²
Area of octagon = Area of trapezium ABCH + Area of trapezium DEFG + Area of rectangle HGDC
= 32 m² + 32 m² + 55 m² = 119 m²
Therefore, the area of octagonal surface is 119 m²

Q.15 There is a pentagonal shaped park as shown in the figure.

For finding its area Jyoti and Kavita divided it in two different ways.

Find the area of this park using both ways. Can you suggest some other way of finding its area?

Ans

Jyoti way of calculating area:

\begin{array}{l}\text{Area of pentagon}=\text{2}\left(\text{Area of trapezium ABCF}\right)\\ =\left[2\times \frac{1}{2}\times (15+30)\left(\frac{15}{2}\right)\right]{\text{m}}^2\\ =337.5{\text{m}}^2\end{array}

Kavita’s way of calculating area:

\begin{array}{l}\text{Area of pentagon ABCDE}=\text{Area of ΔABE}+\text{Area of square BCDE}\\ =\left[\frac{1}{2}\times 15\times \left(30-15\right)+{\left(15\right)}^2\right]{\text{m}}^2\\ =\left[\frac{1}{2}\times 15\times 15+225\right]{\text{m}}^2\\ =337.5{\text{m}}^2\end{array}

No, there is no other way to find its area.

Q.16 Diagram of the adjacent picture frame has outer dimensions 24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is same.

Ans

It is given that the width of each section is same.

Therefore,

LM = BM = CN = NE = OF = OH = PK = PI

CH = CN + NO+ OH

28 = 2CN+20

2CN = 28 – 20

2CN = 8

CN = 4 cm

Hence, LM = BM = NE = OF = OH = PK = PI = 4 cm

$\begin{array}{l}\mathrm{Area}\mathrm{of}\mathrm{section}\mathrm{NDGO}=\mathrm{Area}\mathrm{of}\mathrm{section}\mathrm{AMPJ}\\ =\left[\frac{1}{2}(20+28)\left(4\right)\right]{\mathrm{cm}}^2\\ =96{\mathrm{cm}}^2\\ \therefore \mathrm{Area}\mathrm{of}\mathrm{section}\mathrm{NDGO}=\mathrm{Area}\mathrm{of}\mathrm{section}\mathrm{AMPJ}=96{\mathrm{cm}}^2\\ \\ \mathrm{Also},\mathrm{Area}\mathrm{of}\mathrm{section}\mathrm{AMND}=\mathrm{Area}\mathrm{of}\mathrm{section}\mathrm{PJGO}\\ =\left[\frac{1}{2}(16+24)\left(4\right)\right]\\ =80\mathrm{}{\mathrm{cm}}^2\\ \therefore \mathrm{Area}\mathrm{of}\mathrm{section}\mathrm{AMND}=\mathrm{Area}\mathrm{of}\mathrm{section}\mathrm{PJGO}=80\mathrm{}{\mathrm{cm}}^2\end{array}$

Q.17 There is a pentagonal shaped park as shown in the figure.

For finding its area Jyoti and Kavita divided it in two different ways.

Find the area of this park using both ways. Can you suggest some other way of finding its area?

Ans

Jyoti’s way of calculating area:

$\begin{array}{l}\text{Area of pentagon}=\text{2}\left(\text{Area of trapezium ABCF}\right)\\ =\left[2\times \frac{1}{2}\times (15+30)\left(\frac{15}{2}\right)\right]{\text{m}}^2\\ =337.5{\text{m}}^2\end{array}$

Kavita’s way of calculating area:

$\begin{array}{l}\text{Area of pentagon ABCDE}=\text{Area of ΔABE}+\text{Area of square BCDE}\\ =[\frac{1}{2}\times 15\times (30-15)+{\left(15\right)}^2]{\text{m}}^2\\ =[\frac{1}{2}\times 15\times 15+225]{\text{m}}^2\\ =337.5{\text{m}}^2\end{array}$

No, there is no other way to find its area.

Q.18 There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to

Ans

$\begin{array}{l}\text{We know that},\text{Total surface area of the cuboid}=\text{2}(\mathrm{lb}+\mathrm{bh}+\mathrm{hl})\\ \\ \text{Total surface area of cuboid}=\left[\text{2}\{\left(\text{6}0\right)\left(\text{4}0\right)+\left(\text{4}0\right)\left(\text{5}0\right)+\left(\text{5}0\right)\left(\text{6}0\right)\}\right]{\text{cm}}^{\text{2}}\\ =\left[\text{2}(\text{24}00+\text{2}000+\text{3}000)\right]{\text{cm}}^{\text{2}}\\ =\left[\text{2}(\text{24}00+\text{2}000+\text{3}000)\right](\text{2}\times \text{74}00){\text{cm}}^{\text{2}}\\ =\left[\text{2}(\text{24}00+\text{2}000+\text{3}000)\right]\text{148}00{\text{cm}}^{\text{2}}\\ \\ \text{Total surface area of the cube}=\text{6}{\left(\mathrm{l}\right)}^{\text{2}}\\ =\text{6}{\left(\text{5}0\text{cm}\right)}^{\text{2}}\\ =\text{15}000{\text{cm}}^{\text{2}}\\ \\ \text{Since the total surface area of cuboid is less than}\\ \text{the total surface area of cube,therefore, the cuboidal box will}\\ \text{require lesser amount of material}.\end{array}$

Q.19 A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?

Ans

$\begin{array}{l}\mathrm{A}\mathrm{suitcase}\mathrm{is}\mathrm{in}\mathrm{the}\mathrm{shape}\mathrm{of}\mathrm{a}\mathrm{cuboid}.\\ \therefore \mathrm{Total}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{a}\mathrm{cuboid}=2(\mathrm{lb}+\mathrm{bh}+\mathrm{hl})\\ \mathrm{Hence},\\ \mathrm{Total}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{suitcase}=2[\left(80\right)\left(48\right)+\left(48\right)\left(24\right)+\left(24\right)\left(80\right)]{\mathrm{cm}}^2\\ =2[3840+1152+1920]{\mathrm{cm}}^2\\ =13824{\mathrm{cm}}^2\\ \therefore \mathrm{Total}\mathrm{surface}\mathrm{area}\mathrm{of}100\mathrm{suitcases}=(13824\times 100){\mathrm{cm}}^2\\ =1382400{\mathrm{cm}}^2\\ \\ \mathrm{Required}\mathrm{tarpaulin}=\mathrm{Length}\times \mathrm{Breadth}\\ \Rightarrow 1382400{\mathrm{cm}}^2=\mathrm{Length}\times 96\mathrm{cm}\\ \mathrm{Length}=\left(\frac{1382400}{96}\right)\mathrm{cm}=14400\mathrm{cm}=144\mathrm{m}\\ \mathrm{Thus},144\mathrm{m}\mathrm{of}\mathrm{tarpaulin}\mathrm{is}\mathrm{required}\mathrm{to}\mathrm{cover}100\mathrm{suitcases}.\end{array}$

Q.20 Find the side of a cube whose surface area is 600 cm².

Ans

$\begin{array}{l}\mathrm{Given}\mathrm{that},\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{cube}=600{\mathrm{cm}}^2\\ \mathrm{Let}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{each}\mathrm{side}\mathrm{of}\mathrm{cube}\mathrm{be}\text{‘}\mathrm{x}‘.\\ \mathrm{Surface}\mathrm{area}\mathrm{of}\mathrm{cube}=6{\left(\mathrm{Side}\right)}^2\end{array}$

$\begin{array}{l}\Rightarrow 600{\mathrm{cm}}^2=6{\mathrm{x}}^2\\ \Rightarrow {\mathrm{x}}^2=100{\mathrm{cm}}^2\\ \Rightarrow \mathrm{x}=10\mathrm{cm}\\ \\ \therefore \mathrm{The}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{cube}\mathrm{is}10\mathrm{cm}.\end{array}$

Q.21 Rukhsar painted the outside of the cabinet of measure 1 m × 2 m × 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet?

Ans

$\begin{array}{l}\text{Length of the cabinet}=\text{2 m}\\ \text{Breadth of the cabinet}=\text{1 m}\\ \text{Height of the cabinet}=\text{1}.\text{5 m}\end{array}$

$\begin{array}{l}\text{Area of the cabinet that was painted}=\text{2}\mathrm{h}(\mathrm{l}+\mathrm{b})+\mathrm{lb}\\ =[\text{2}\times \text{1}.\text{5}\times (\text{2}+\text{1})+\left(\text{2}\right)\left(\text{1}\right)]{\text{m}}^{\text{2}}\\ =[\text{3}\left(\text{3}\right)+\text{2}]{\text{m}}^{\text{2}}\\ =(\text{9}+\text{2}){\text{m}}^{\text{2}}\\ ={\text{11 m}}^{\text{2}}\\ \\ \therefore {\text{Area of the cabinet that was painted is 11 m}}^{\text{2}}.\end{array}$