RS Aggrawal 2020 2021 Solutions for Class 6 Maths Chapter 21 - Concept Of Perimeter And Area

Question 1:

Question 2:

We know: Perimeter of a rectangle = $2\times (\mathrm{Length}+\mathrm{Breadth})$

(i) Length = 16.8 cm
    Breadth = 6.2 cm
    Perimeter = $2\times (\mathrm{Length}+\mathrm{Breadth})$
                   = $2\times (16.8+6.2) =46 \mathrm{cm}$
(ii) Length = 2 m 25 cm
                  =(200+25) cm       (1 m = 100 cm )
                  = 225 cm  
    Breadth =1 m 50 cm
                  = (100+50) cm      (1 m = 100 cm )
                  = 150 cm    
    Perimeter = $2\times (\mathrm{Length}+\mathrm{Breadth})$
                     = $2\times (225+150) =750 \mathrm{cm}$
(iii) Length = 8 m 5 dm
                   = (80+5) dm   (1 m = 10 dm )
                   = 85 dm     
       Breadth = 6 m 8 dm
                     = (60+8) dm   (1 m = 10 dm )
                     = 68 dm  
    Perimeter = $2\times (\mathrm{Length}+\mathrm{Breadth})$
                   = $2\times (85+68) =306 \mathrm{dm}$

Question 3:

Length of the field = 62 m
Breadth of the field = 33 m
Perimeter of the field = 2(l + b) units
                                = 2(62 + 33) m =190 m
Cost of fencing per metre = Rs 16
Total cost of fencing = Rs (16$\times$190) = Rs 3040

Question 4:

Let the length of the rectangle be 5x m.
Breadth of the rectangle = 3x m
Perimeter of the rectangle = 2(l + b)
                                    = 2(5x + 3x) m
                                    = (16x) m
It is given that the perimeter of the field is 128 m.

$$\begin{gathered} \therefore 16x=128 \\ \Rightarrow x=\frac{128}{16}=8 \\ \therefore \mathrm{Length} =(5\times 8)=40\mathrm{m} \\ \mathrm{Breadth} =(3\times 8)=24\mathrm{m} \end{gathered}$$

Question 5:

Total cost of fencing = Rs 1980
Rate of fencing = Rs 18 per metre

Perimeter of the field = $\frac{\mathrm{Total} \mathrm{cost}}{\mathrm{Rate}}=\frac{\mathrm{Rs} 1980}{\mathrm{Rs} 18/\mathrm{m}}=\left(\frac{1980}{18}\right) \mathrm{m}=110 \mathrm{m}$

Let the length of the field be x metre.
Perimeter of the field = 2(x + 23) m

$$\begin{gathered} \therefore 2(x+23)=110 \\ \Rightarrow (x+23)=55 \\ x=(55-23)=32 \end{gathered}$$
Hence, the length of the field is 32 m.

Question 6:

Total cost of fencing = Rs 3300
Rate of fencing = Rs 25/m
Perimeter of the field = $\frac{\mathrm{Total} \mathrm{cost} }{\mathrm{Rate} \mathrm{of} \mathrm{fencing}}=\left(\frac{\mathrm{Rs} 3300}{\mathrm{Rs} 25/\mathrm{m}}\right)=\frac{3300}{25} \mathrm{m}=132 \mathrm{m}$

Let the length and the breadth of the rectangular field be 7x and 4x, respectively.
Perimeter of the field = 2(7x + 4x) = 22x

It is given that the perimeter of the field is 132 m.

$$\begin{gathered} \therefore 22\mathrm{x}=132 \\ \Rightarrow \mathrm{x}=\frac{132}{22}=6 \\ \therefore \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{field} =(7 \times 6) \mathrm{m}=42 \mathrm{m} \\ \mathrm{Breadth} \mathrm{of} \mathrm{the} \mathrm{field} =(4 \times 6) \mathrm{m}=24 \mathrm{m} \end{gathered}$$

Question 7:

(i) Side of the square = 3.8 cm
    Perimeter of the square = (4$\times$side)
                                    = (4$\times$3.8) = 15.2 cm

(ii) Side of the square = 4.6 cm
     Perimeter of the square = (4$\times$side)
                                     = (4$\times$4.6) = 18.4 cm

(iii) Side of the square = 2 m 5 dm
                                     = (20+5) dm   (1 m = 10 dm)
                                     = 25 dm
      Perimeter of the square = (4$\times$side)
                                       = (4$\times$25) = 100 dm

Question 8:

Total cost of fencing = Rs 4480
Rate of fencing = Rs 35/m
Perimeter of the field = $\frac{\mathrm{Total} \mathrm{cost}}{\mathrm{Rate}}=\frac{\mathrm{Rs} 4480}{\mathrm{Rs} 35/\mathrm{m}}=\frac{4480}{35} \mathrm{m}=128 \mathrm{m}$

Let the length of each side of the field be x metres.
Perimeter = (4x) metres
$$\begin{gathered} \therefore 4x=128 \\ \Rightarrow x=\frac{128}{4}=32 \end{gathered}$$

Hence, the length of each side of the field is 32 m.

Question 9:

Side of the square field = 21m
Perimeter of the square field = (4$\times$21) m
                                       = 84 m  

Let the length and the breadth of the rectangular field be 4x and 3x, respectively.
 Perimeter of the rectangular field = 2(4x + 3x) = 14x

 Perimeter of the rectangular field = Perimeter of the square field

$$\begin{gathered} \therefore 14x=84 \\ \Rightarrow x=\frac{84}{14}=6 \end{gathered}$$
$$\begin{gathered} \therefore \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{rectangular} \mathrm{field} =(4 \times 6) \mathrm{m}=24 \mathrm{m} \\ \mathrm{Breadth} \mathrm{of} \mathrm{the} \mathrm{rectangular} \mathrm{field} = (3 \times 6) \mathrm{m}=18 \mathrm{m} \end{gathered}$$

Question 10:

(i) Sides of the triangle are 7.8 cm, 6.5 cm and 5.9 cm. 
    Perimeter of the triangle = (First side + Second side + Third Side) cm
                                      = (7.8 + 6.5 + 5.9) cm
                                      = 20.2 cm

(ii) In an equilateral triangle, all sides are equal.
     Length of each side of the triangle = 9.4 cm
     $\therefore$ Perimeter of the triangle = (3 $\times$ Side) cm
                                            = (3 $\times$ 9.4) cm
                                            = 28.2 cm

(iii) Length of two equal sides = 8.5 cm
       Length of the third side = 7 cm
    $\therefore$ Perimeter of the triangle = {(2 $\times$ Equal sides) + Third side} cm
                                           = {(2 $\times$ 8.5) + 7} cm
                                           = 24 cm

Question 11:

(i) Length of each side of the given pentagon = 8 cm
   $\therefore$ Perimeter of the pentagon = (5$\times$8) cm
                                                  = 40 cm

(ii) Length of each side of the given octagon = 4.5 cm
   $\therefore$ Perimeter of the octagon = (8$\times$4.5) cm
                                                = 36 cm

(iii) Length of each side of the given decagon = 3.6 cm
   $\therefore$ Perimeter of the decagon = (10$\times$3.6) cm
                                                 = 36 cm

Question 1:

(i) Perimeter of the figure = Sum of all the sides
                                         =(27 + 35 + 35 + 45) cm
                                         = 142 cm
(ii) Perimeter of the figure = Sum of all the sides
                                         =(18 + 18 + 18 + 18) cm
                                         = 72 cm
(iii) Perimeter of the figure = Sum of all the sides
                                         =(8 + 16 + 4 + 12 + 12 + 16 + 4) cm
                                         = 72 cm

Question 2:

(i) Radius, r = 28 cm

$$\begin{gathered} \therefore \mathrm{Circumference} \mathrm{of} \mathrm{the} \mathrm{circle}, \mathrm{C}=2\mathrm{\pi }r \\ =(2\times \frac{22}{7}\times 28) \\ =176 \mathrm{cm} \\ \mathrm{Hence}, \mathrm{the} \mathrm{circumference} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{circle} \mathrm{is} 176 \mathrm{cm}. \end{gathered}$$

(ii) Radius, r = 10.5 cm
      $$\begin{gathered} \therefore \mathrm{Circumference} \mathrm{of} \mathrm{the} \mathrm{circle}, \mathrm{C}=2\mathrm{\pi }r \\ =(2\times \frac{22}{7}\times 10.5) \\ =66 \mathrm{cm} \\ \mathrm{Hence}, \mathrm{the} \mathrm{circumference} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{circle} \mathrm{is} 66 \mathrm{cm}. \end{gathered}$$

(iii) Radius, r = 3.5 m
     $$\begin{gathered} \therefore \mathrm{Circumference} \mathrm{of} \mathrm{the} \mathrm{circle}, \mathrm{C}=2\mathrm{\pi }r \\ =(2\times \frac{22}{7}\times 3.5) \\ =22 \mathrm{m} \\ \mathrm{Hence}, \mathrm{the} \mathrm{circumference} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{circle} \mathrm{is} 22 \mathrm{m}. \end{gathered}$$

Question 3:

(i)

$$\begin{gathered} \mathrm{Circumference}=2\mathrm{\pi }r \mathit{=}\mathrm{\pi }(2r) =\mathrm{\pi }\times \mathrm{Diameter} \mathrm{of} \mathrm{the} \mathrm{circle} (d) (\mathrm{Diameter}=2\times \mathrm{ra}di\mathrm{us}) \\ \Rightarrow \mathrm{Circumference}=\mathrm{Diameter}\times \mathrm{\pi } D\mathrm{iameter} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{circle} \mathrm{is} 14 \mathrm{cm}.\mathrm{Circumference} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{circle}=14\times \mathrm{\pi }\Rightarrow (14\times \frac{22 }{7})=44 \mathrm{cm}C\mathrm{ircumference} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{circle} \mathrm{is} 44 \mathrm{cm}. \end{gathered}$$

(ii)
$$\begin{gathered} \mathrm{Circumference}=2\mathrm{\pi }r \mathit{=}\mathrm{\pi }(2r\mathit{)}\mathit{}\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }=\mathrm{\pi }\times \mathrm{Diam}eter \mathrm{of} \mathrm{the} \mathrm{circle}(\mathrm{d}) (\mathrm{Diameter}=2\times \mathrm{Radius}) \\ \Rightarrow \mathrm{Circumference}=\mathrm{Diameter}\times \mathrm{\pi }\mathrm{Diameter} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{circle} \mathrm{is} 35 \mathrm{cm}.\Rightarrow \mathrm{Circumference} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{circle}=35\times \mathrm{\pi } \Rightarrow (35\times \frac{22 }{7})=110 \mathrm{cm}C\mathrm{ircumference} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{circle} \mathrm{is} 110 \mathrm{cm}. \end{gathered}$$

(iii)
$$\begin{gathered} \mathrm{Circumference}=2\mathrm{\pi }r \mathit{=}\mathrm{\pi }(2r) =\mathrm{\pi }\times \mathrm{Diameter} \mathrm{of} \mathrm{the} \mathrm{circle}(d) (\mathrm{Diameter}=2\times Radi\mathrm{us}) \\ \Rightarrow \mathrm{Circumference}=\mathrm{Diameter}\times \mathrm{\pi }\mathrm{Diameter} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{circle} \mathrm{is} 10.5 \mathrm{m}.\mathrm{Circumference} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{circle}=10.5\times \mathrm{\pi }\Rightarrow (10.5\times \frac{22 }{7})=33 \mathrm{m}\mathrm{Circumference} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{circle} \mathrm{is} 33 \mathrm{m}. \end{gathered}$$

Question 4:

Let the radius of the given circle be r cm.
Circumference of the circle = 176 cm
Circumference = $2\pi r$
$$\begin{gathered} \therefore 2\mathrm{\pi }r=176 \\ \Rightarrow r=\frac{176}{2\mathrm{\pi }} \\ \Rightarrow r=(\frac{176}{2}\times \frac{7}{22}) \\ \Rightarrow r=28 \\ \mathrm{The} \mathrm{radius} \mathrm{of} the \mathrm{given} \mathrm{circle} \mathrm{is} 28 \mathrm{cm}. \end{gathered}$$

Question 5:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{the} \mathrm{circle} \mathrm{be} r \mathrm{cm}. \\ \mathrm{Diameter}=2\times R\mathrm{adius}=2r \mathrm{cm} \\ \mathrm{Circumference} \mathrm{of} \mathrm{the} \mathrm{wheel}=264 \mathrm{cm} \\ \mathrm{Circumference} \mathrm{of} \mathrm{the} \mathrm{wheel}=2\mathrm{\pi r} \\ \therefore 2\pi r=264 \\ \Rightarrow 2r=\frac{264}{\mathrm{\pi }} \\ \Rightarrow 2r=(264\times \frac{7}{22}) \\ \Rightarrow 2r=84 \\ \mathrm{Diameter} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{wheel} is 84 \mathrm{cm}. \end{gathered}$$

Question 6:

Radius of the wheel =$\frac{\mathrm{Diameter} \mathrm{of} \mathrm{the} \mathrm{wheel}}{2}$
$\Rightarrow \mathrm{r}=\frac{77}{2}\mathrm{cm}$
Circumference of the wheel =$2\pi r$
$$\begin{gathered} =(2\times \frac{22}{7}\times \frac{77}{2}) \\ =242 \mathrm{cm} \end{gathered}$$

In 1 revolution the wheel covers a distance equal to its circumference.

$$\begin{gathered} \therefore \mathrm{Distance} \mathrm{covered} \mathrm{by} the w\mathrm{heel} \mathrm{in} 1 \mathrm{revolution}=242 \mathrm{cm} \\ \therefore \mathrm{Distance} \mathrm{covered} \mathrm{by} the w\mathrm{heel} \mathrm{in} 500 \mathrm{revolution}s=(500 \times 242) \mathrm{cm} \\ =121000 \mathrm{cm} \\ =1210 \mathrm{m} (100 \mathrm{cm}= 1\mathrm{m} ) \\ =1.21 \mathrm{km} (1000 \mathrm{m}=1 \mathrm{km} ) \end{gathered}$$

Question 1:

$$\begin{gathered} \mathrm{Radius} \mathrm{of} \mathrm{the} \mathrm{wheel}\left(r\right)=\frac{\mathrm{Diameter} \mathrm{of} the \mathrm{wheel}}{2} \\ r=\frac{70}{2}\mathrm{cm}=35 \mathrm{cm} \\ \mathrm{Circumference} \mathrm{of} \mathrm{the} \mathrm{wheel} = 2\mathrm{\pi }r =(2\times \frac{22}{7}\times 35) \\ =220 \mathrm{cm} \\ \mathrm{In} \mathrm{one} \mathrm{revolution}, the \mathrm{wheel} \mathrm{covers} the \mathrm{distance} \mathrm{equal} \mathrm{to} \mathrm{its} \mathrm{circumference}. \\ \therefore 220 \mathrm{cm} \mathrm{distance} =1 \mathrm{revolution} \\ \therefore 1 \mathrm{cm} \mathrm{distance} =\frac{1}{220} \mathrm{revolution} \\ \therefore 1\mathrm{km} (\mathrm{or} 100000 \mathrm{cm}) \mathrm{distance} =\frac{1\times 100000}{220} \mathrm{revolution} (\therefore 1 \mathrm{km}=100000 \mathrm{cm}) \\ \therefore 1.65 \mathrm{km} \mathrm{distance} = \frac{1.65\times 100000 }{220} \mathrm{revolutions} \\ = 750 \mathrm{revolutions} \\ Thus,t\mathrm{he} \mathrm{wheel} \mathrm{will} \mathrm{make} 750 \mathrm{revolutions} \mathrm{to} \mathrm{travel} 1.65 \mathrm{km}. \end{gathered}$$

Question 2:

The figure contains 12 complete squares.
    Area of 1 small square = 1 sq cm
      ∴ Area of the figure = Number of complete squares $\times$ Area of the square
                                         =$(12\times 1) \mathrm{sq} \mathrm{cm}$
                                         =12 sq cm

Question 3:

The figure contains 18 complete squares.
    Area of 1 small square = 1 sq cm
      ∴ Area of the figure = Number of complete squares $\times$ Area of the square
                                         =$(18\times 1) \mathrm{sq} \mathrm{cm}$
                                         =18 sq cm

Question 4:

The figure contains 14 complete squares and 1 half square.
    Area of 1 small square = 1 sq cm
      ∴ Area of the figure = Number of squares $\times$ Area of the square
                                         =$\left[(14 \times 1) + (1 \times \frac{1}{2})\right]\mathrm{sq} \mathrm{cm}$
                                         =$14\frac{1}{2}$ sq cm

Question 5:

The figure contains 6 complete squares and 4 half squares.
    Area of 1 small square = 1 sq cm
      ∴ Area of the figure = Number of squares $\times$ Area of the square
                                         =$\left[(6\times 1)+(4\times \frac{1}{2})\right] \mathrm{sq} \mathrm{cm}$
                                         =8 sq cm

Question 6:

The figure contains 9 complete squares and 6 half squares.
    Area of 1 small square = 1 sq cm
     ∴ Area of the figure = Number of squares $\times$ Area of the square
                                         =$\left[(9 \times 1) + (6 \times \frac{1}{2})\right] \mathrm{sq} \mathrm{cm}$
                                         =12 sq cm

Question 7:

The figure contains 16 complete squares.
    Area of 1 small square = 1 sq cm
     ∴ Area of the figure = Number of squares $\times$ Area of a square
                                         =$(16\times 1) \mathrm{sq} \mathrm{cm}$
                                         =16 sq cm

Question 8:

In the given figure, there are 4 complete squares, 8 more than half parts of squares and 4 less than half parts of squares.
We neglect the less than half parts and consider each more than half part of the square as a complete square.

                  ∴ Area = (4 + 8) sq cm
                            = 12 sq cm

Question 9:

In the given figure, there are 9 complete squares, 5 more than half parts of squares and 7 less than half parts of squares.
We neglect the less than half parts of squares and consider the more than half squares as complete squares.
∴ Area of the figure = (9 + 5) sq cm
                                      = 14 sq cm

Question 1:

The figure contains 14 complete squares and 4 half squares.
    Area of 1 small square = 1 sq cm
     Area of the figure = Number of squares $\times$ Area of one square
                                         =$\left[(14\times 1)+(4\times \frac{1}{2}) \right]\mathrm{sq} \mathrm{cm}$
                                         =16 sq cm

Question 2:

 (i) Length = 46 cm
Breadth = 25 cm
  Area of the rectangle = (Length $\times$Breadth) sq units
                              = (46$\times$25) cm² = 1150 cm²
 
(ii) Length = 9 m
Breadth = 6 m
  Area of the rectangle = (Length $\times$Breadth) sq units
                             = (9$\times$6) m² = 54 m²

 (iii) Length = 14.5 m
Breadth = 6.8 m
  Area of the rectangle = (Length $\times$Breadth) sq units
                             = ($\frac{145}{10}\times \frac{68}{10}$) m² = $\frac{9860}{100}$ m² =98.60 m²

 (iv) Length = 2 m 5 cm
                   = (200+5) cm   (1 m = 100 cm )
                   =205cm
       Breadth = 60 cm
       Area of the rectangle = (Length $\times$Breadth) sq units
                                    = (205$\times$60) cm² = 12300 cm²

 (v) Length = 3.5 km
Breadth = 2 km
  Area of the rectangle = (Length $\times$Breadth) sq units
                             = (3.5$\times$2) km² = $(\frac{35}{10}\times 2)$ km² =7 km² 

Question 3:

Side of the square plot = 14 m
Area of the square plot = (Side)² sq units
                                     = (14)² m²
                                     = 196  m²

Question 4:

Length of the table = 2 m 25 cm
                         = (2 + 0.25) m     (100 cm = 1 m)
                         = 2.25 m
Breadth of the table = 1 m 20 cm
                                 = (1 + 0.20) m    (100 cm = 1 m)
                                 =1.20 m
Area of the table = (Length × Breadth) sq units
                             = (2.25 × 1.20) m²
       
                              = $(\frac{225}{100}\times \frac{120}{100})$ m²
                              = 2.7 m²

Question 5:

Length of the carpet = 30 m 75 cm
                           =(30 + 0.75) cm        (100 cm = 1 m)
                           = 30.75 m
Breadth of the carpet = 80 cm
                            = 0.80 m                (100 cm = 1 m)

Area of carpet = ( Length $\times$ breadth ) sq units

                            $$\begin{gathered} =(30.75 \times 0.80) {\mathrm{m}}^2 \\ = (\frac{3075}{100}\times \frac{80}{100}) {\mathrm{m}}^2 \\ =24.6 {\mathrm{m}}^2 \end{gathered}$$

Cost of 1 m² carpet= Rs 150
Cost of 24.6 m² carpet = Rs (24.6$\times$150)
                                      =  Rs 3690

                      

Question 6:

Length of the sheet of paper = 3 m 24 cm = 324 cm
Breadth of the sheet of paper = 1 m 72 cm = 172 cm
Area of the sheet = (Length $\times$ Breadth)
                           $$\begin{gathered} =(324\times 172) {\mathrm{cm}}^2 \\ =55728 {\mathrm{cm}}^2 \end{gathered}$$

Length of the piece of paper required to make 1 envelope = 18 cm
Breadth of the piece of paper required to make 1 envelope = 12 cm
Area of the piece of paper required to make 1 envelope = (18$\times$12) cm²
                                                                                   = 216 cm²

$$\begin{gathered} \mathrm{No}. \mathrm{of} \mathrm{envelope}s \mathrm{that} \mathrm{can} \mathrm{be} \mathrm{made} =\frac{\mathrm{Area} \mathrm{of} the \mathrm{sheet}}{\mathrm{Area} \mathrm{of} the \mathrm{piece} \mathrm{of} \mathrm{paper} \mathrm{required} \mathrm{to} \mathrm{make} 1 \mathrm{envelope}} \\ \Rightarrow \mathrm{N}\mathrm{o}. \mathrm{o}\mathrm{f} \mathrm{e}\mathrm{nv}\mathrm{e}\mathrm{l}\mathrm{o}\mathrm{p}\mathrm{es} \mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t} \mathrm{c}\mathrm{a}\mathrm{n} \mathrm{b}\mathrm{e} \mathrm{m}\mathrm{a}\mathrm{d}\mathrm{e} =\frac{55728}{216}=258 \mathrm{envelopes} \end{gathered}$$

Question 7:

Length of the room = 12.5 m
Breadth of the room = 8 m
Area of the room = (Length$\times$Breadth)
                     $$\begin{gathered} =(12.5\times 8) {\mathrm{m}}^2 \\ = 100 {\mathrm{m}}^2 \end{gathered}$$
Side of the square carpet = 8 m
Area of the carpet = (Side)²
                              = 8² m²
                              = 64 m²

Area of the floor which is not carpeted = Area of the room − Area of the carpet
                                                               = (100 − 64) m²
                                                               = 36 m²

Question 8:

Length of the road = 150 m = 15000 cm
Breadth of the road = 9 m = 900 cm
Area of the road = (Length$\times$Breadth)
                     $$\begin{gathered} =15000\times 900 {\mathrm{cm}}^2 \\ =13500000 {\mathrm{cm}}^2 \end{gathered}$$
Length of the brick = 22.5 cm
Breadth of the brick = 7.5 cm
Area of one brick = (Length$\times$Breadth)
                             $$\begin{gathered} =(22.5\times 7.5) {\mathrm{cm}}^2 \\ =168.75 {\mathrm{cm}}^2 \end{gathered}$$

$\mathrm{Number} \mathrm{of} \mathrm{bricks} = \frac{\mathrm{Area} \mathrm{of} the \mathrm{road}}{\mathrm{Area} \mathrm{of} \mathrm{one} \mathrm{brick}}=\frac{13500000}{168.75}=80000 b\mathrm{ricks}$

Question 9:

Length of the room = 13 m
Breadth of the room = 9 m
Area of the room = (13$\times$9) m² = 117 m²

Let length of required carpet be x m.
Breadth of the carpet = 75 cm
                            = 0.75 m         (100 cm = 1 m)
Area of the carpet = (0.75$\times$x) m²
                       = 0.75x m²
For carpeting the room:
Area covered by the carpet = Area of the room
     $$\begin{gathered} \Rightarrow 0.75x=117 \\ \Rightarrow x=\frac{117}{0.75} \\ \Rightarrow x=117\times \frac{4}{3} \\ \Rightarrow x=156 \mathrm{m} \end{gathered}$$

So, the length of the carpet is 156 m.
Cost of 1 m carpet = Rs 65
Cost 156 m carpet = Rs (156$\times$65)
                              = Rs 10140

Question 10:

Let the length of the rectangular park be 5x.
∴ Breadth of the rectangular park = 3x
Perimeter of the rectangular field = 2(Length + Breadth)
                                                      =2(5x + 3x)
                                                      = 16x

It is given that the perimeter of rectangular park is 128 m.
$$\begin{gathered} \Rightarrow 16\mathrm{x}=128 \\ \Rightarrow \mathrm{x}=\frac{128}{16} \\ \Rightarrow \mathrm{x}=8 \\ \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{park}=(5\times 8) \mathrm{m} \\ =40 \mathrm{m} \\ \mathrm{Breadth} \mathrm{of} \mathrm{the} \mathrm{park}=(3\times 8) \mathrm{m} \\ =24 \mathrm{m} \end{gathered}$$

Area of the park = (Length $\times$ Breadth) sq units
                           
                          $$\begin{gathered} =(40\times 24) {\mathrm{m}}^2 \\ =960 {\mathrm{m}}^2 \end{gathered}$$

Question 11:

Side of the square plot = 64 m
Perimeter of the square plot = $(4\times S\mathrm{ide}) \mathrm{m} =(4\times 64) \mathrm{m}=256 \mathrm{m}$
Area of the square plot = (Side)²
= 64² m²
= 4096 m²

Let the breadth of the rectangular plot be x m.
Perimeter of the rectangular plot = 2(l+b)  m
= 2(70+x) m

Perimeter of the rectangular plot = Perimeter of the square plot   (Given)
$$\begin{gathered} \Rightarrow 2(70+x)=256 \\ \Rightarrow 140+2x=256 \\ \Rightarrow 2x=256-140 \\ \Rightarrow 2x=116 \\ \Rightarrow x=\frac{116}{2}=58 \end{gathered}$$
So, the breadth of the rectangular plot is 58 m.
Area of the rectangular plot = $(L\mathrm{ength} \times B\mathrm{readth})=(70 \times 58) {\mathrm{m}}^2=4060 {\mathrm{m}}^2$
Area of the square plot − Area of the rectangular plot
= (4096 − 4060)
= 36 m²
Area of the square plot is 36 m² greater than the rectangular plot.

Question 12:

Total cost of cultivating the field = Rs 71400
Rate of cultivating the field = Rs 35/m²


$\mathrm{Are}a \mathrm{of} \mathrm{the} \mathrm{field}=\frac{\mathrm{Total} \mathrm{cost} \mathrm{of} \mathrm{cultivating} \mathrm{the} \mathrm{field}}{\mathrm{Rate} \mathrm{of} \mathrm{cultivating}}=\frac{\mathrm{Rs} 71400}{\mathrm{Rs} 35/{\mathrm{m}}^2}=2040 {\mathrm{m}}^2$

Let the length of the field be x m.

$$\begin{gathered} \mathrm{Area} \mathrm{of} the \mathrm{field} = (L\mathrm{ength} \times W\mathrm{idth}) {\mathrm{m}}^2=(\mathrm{x} \times 40) {\mathrm{m}}^2 =40\mathrm{x} {\mathrm{m}}^2 \\ \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{field} is 2040 {\mathrm{m}}^2. \\ \Rightarrow 40\mathrm{x}=2040 \\ \Rightarrow \mathrm{x}=\frac{2040}{40}=51 \\ \therefore L\mathrm{ength} \mathrm{of} \mathrm{the} \mathrm{field} = 51 \mathrm{m} \end{gathered}$$

Perimeter of the field = 2(l+b)
= 2(51+40) m
= 182 m
Cost of fencing 1 m of the field = Rs 50
Cost of fencing 182 m of the field = Rs (182$\times$50)
= Rs 9100