Rs Aggarwal for Class 6 Math Chapter 1 - Number System

Answer:

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 = $2\times (\mathrm{Length}+\mathrm{Breadth})$
                   = $2\times (85+68) =306 \mathrm{dm}$

Answer:

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

Answer:

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.

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

Answer:

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.

Answer:

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.

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

Answer:

(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

Answer:

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.

Answer:

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

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

Answer:

(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

Answer:

(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

Answer:

(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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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.

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

 (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 

Answer:

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

Answer:

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

Answer:

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

                      

Answer:

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

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

Answer:

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

Answer:

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)
                             $$\begin{gathered} =(22.5\times 7.5) {\mathrm{cm}}^2 \\ =168.75 {\mathrm{cm}}^2 \end{gathered}$$

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

Answer:

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

Answer:

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
                           
                          $$\begin{gathered} =(40\times 24) {\mathrm{m}}^2 \\ =960 {\mathrm{m}}^2 \end{gathered}$$

Answer:

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.

Answer:

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

Answer:

Let the width of the rectangle be x cm.
Length of the rectangle = 36 cm
Area of the rectangle = ($\mathrm{Length} \times \mathrm{Width}$) = ($36 \times \mathrm{x}$) cm²
It is given that the area of the rectangle is 540 cm².

$$\begin{gathered} \Rightarrow 36 \times x= 540 \\ \Rightarrow x=\frac{540}{36} \\ \Rightarrow x=15 \\ \therefore W\mathrm{idth} \mathrm{of} \mathrm{the} \mathrm{rectangle} =15 \mathrm{cm} \end{gathered}$$

Perimeter of the rectangle = 2(Length + Width) cm
= 2(36 + 15) cm
= 102 cm

Answer:

Length of the wall = 4 m = 400 cm
Breadth of the wall = 3 m = 300 cm
Area of the wall = (400×300) cm² = 120000 cm²

Length of the tile = 12 cm
Breadth of the tile = 10 cm
Area of one tile = (12×10) cm² = (120) cm²

$\mathrm{Number} \mathrm{of} \mathrm{tiles} \mathrm{required} \mathrm{to} \mathrm{cover} \mathrm{the} \mathrm{wall}=\frac{\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{wall}}{\mathrm{Area} \mathrm{of} \mathrm{one} \mathrm{tile}}=\frac{120000}{120}=1000 \mathrm{tiles}$
Cost of 1 tile = Rs 22.50
Cost of 1000 tiles = (1000 × 22.50) = Rs 22500

Thus, the total cost of the tiles is Rs 22500.

Answer:

Let the length of the rectangle be x cm.
Breadth of the rectangle is 25 cm.
Area of the rectangle = (Length × Breadth) cm²
                                   = (x×25) cm²
                                   =25x cm²

It is given that the area of the rectangle is 600 cm².
$$\begin{gathered} \Rightarrow 25x=600 \\ \Rightarrow x=\frac{600}{25}=24 \end{gathered}$$
So, the length of the rectangle is 24 cm.
Perimeter of the rectangle = 2(Length + Breadth) units
                                           = 2(25 + 24) cm
                                           = 98 cm

Answer:

Area of the square =$\left\{\frac{1}{2}\times (D\mathrm{iagonal}{)}^2\right\} \mathrm{sq} \mathrm{units}$
$$\begin{gathered} = \left\{\frac{1}{2}\times (5\sqrt{2}{)}^2\right\} {\mathrm{cm}}^2 \\ =\left\{\frac{1}{2}\times (5{)}^2\times (\sqrt{2}{)}^2\right\} {\mathrm{cm}}^2 \\ =\left\{\frac{1}{2}\times 25\times 2\right\} {\mathrm{cm}}^2 \\ =(\frac{1}{2}\times 50) {\mathrm{cm}}^2= 25 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

(i) Area of rectangle ABDC = Length $\times$ Breadth
                                         = AB$\times$AC                    (AC = AE − CE)
                                         = $\left(1\times 8\right){\mathrm{m}}^2$
                                         = 8 m²
   Area of rectangle CEFG = Length $\times$ Breadth
                                         = CG$\times$GF               (CG = GD + CD)           
                                         = $\left(9\times 2\right){\mathrm{m}}^2$
                                         = 18 m²
   Area of the complete figure = Area of rectangle ABDC + Area of rectangle CEFG
                                                 = (8 + 18) m²
                                                 =  26 m²


(ii) Area of rectangle AEDC = Length $\times$ Breadth
                                         = ED $\times$ CD              
                                         = $\left(12\times 2\right){\mathrm{m}}^2$
                                         = 24 cm²
   Area of rectangle FJIH = Length $\times$ Breadth
                                         = HI $\times$ IJ                   
                                         = $\left(1\times 9\right){\mathrm{m}}^2$
                                         = 9 m²
Area of rectangle ABGF = Length $\times$ Breadth
                                         = AB $\times$ AF                                  {(AB = FJ − GJ) and AF = EH − (EA + FH)}                 
                                         = $\left(7\times 1.5\right){\mathrm{m}}^2$
                                         = 10.5 m²

   Area of the complete figure = Area of rectangle AEDC + Area of rectangle FJIH + Area of rectangle ABGF
                                         = (24 + 9 + 10.5) m²
                                         = 43.5 m²



(iii) Area of the shaded portion = Area of the complete figure − Area of the unshaded figure
                                           = Area of rectangle ABCD − Area of rectangle GBFE
                                           =(CD$\times$AD) − (GB$\times$BF)
                                           =$\left\{(12\times 9)-(7.5\times 10)\right\}{\mathrm{m}}^2$                                   (BF = BC − FC)
                                           =(108 − 75) m²

                                          =33 m²

Answer:

(i) Area of square BCDE= (Side)²
                                        = (CD)²
                                        = (3)² cm²        
                                        = 9 cm²
      Area of rectangle ABFK = $\mathrm{Length} \times \mathrm{Breadth}$
                                             = AK$\times$AB             [(AB = AC − BC) and (AK = AL + LK)
                                             = (5$\times$1) cm²  
                                             = 5 cm²

     Area of rectangle MLKG = $\mathrm{Length} \times \mathrm{Breadth}$
                                             = ML $\times$ MG
                                             = (2 $\times$ 3) cm²
                                             = 6 cm²
     Area of rectangle JHGF= $\mathrm{Length} \times \mathrm{Breadth}$
                                             = JH$\times$HG
                                             = (2$\times$4) cm²
                                             = 8 cm²
       Area of the figure = Area of rectangle ABFK + Area of rectangle MLKG + Area of rectangle JHGF + Area of square BCDE
                               = (9 + 5 + 6 + 8) cm²
                               = 28 cm²
                   
(ii) Area of rectangle CEFG= $\mathrm{Length} \times \mathrm{Breadth}$
                                             = EF$\times$CE
                                             = (1$\times$5) cm²          (CE = EA − AC)
                                             = 5 cm²
      Area of rectangle ABDC = $\mathrm{Length} \times \mathrm{Breadth}$
                                             = AB$\times$BD
                                             = (1$\times$2) cm²  
                                             = 2 cm²

     Area of rectangle HIJG = $\mathrm{Length} \times \mathrm{Breadth}$
                                             = HI $\times$ IJ
                                             = (1$\times$2) cm²
                                             = 2 cm²
       Area of the figure = Area of rectangle CEFG + Area of rectangle HIJG + Area of rectangle ABDC
                               = (5+2+2) cm²
                               = 9 cm²       
                      
(iii) In the figure, there are 5 squares, each of whose sides are 6 cm in length.
     Area of the figure = 5 $\times$ Area of square
                                   = 5$\times$(side)²
                                   = 5$\times$(6)² cm²
                                   = 180 cm²

Answer:

(b) 28 cm

Let the length and the breadth of the rectangle be 7x cm and 5x cm, respectively.
It is given that the perimeter of the rectangle is 96 cm.
Perimeter of the rectangle = 2(7x+5x) cm

$$\begin{gathered} \Rightarrow 2(7\mathrm{x}+5x)=96 \\ =2\left(12x\right)=96 \\ =24x=96 \\ \Rightarrow x=\frac{96}{24}=4 \\ \therefore \mathrm{Length} =(7\times 4)\mathrm{cm}=28 \mathrm{cm} \end{gathered}$$

Answer:

(d) 126 cm
Let length of the rectangle be L cm.
Area of the rectangle = 650 cm²
Area of the rectangle = ($\mathrm{L}\times 13$) cm²
$$\begin{gathered} \Rightarrow (\mathrm{L}\times 13)=650 \\ \Rightarrow \mathrm{L}=\frac{650}{13}=50 \\ \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{rectangle} \mathrm{is} 50 \mathrm{cm} \end{gathered}$$

Perimeter of the rectangle = 2(Length + Breadth) cm = 2(50+13) cm = 126 cm

Answer:

(b) Rs 2340
Perimeter of the rectangular field = 2(Length + Breadth)
                                                  = 2(34 + 18) m = 104 m
Cost of fencing 1 metre = Rs 22.50
Cost of fencing 104 m = Rs (22.50$\times$104) = Rs 2340

Answer:

(b) 16 m
Total cost of fencing = Rs 2400
Rate of fencing = Rs 30/m
Perimeter of the rectangular field = $\frac{\mathrm{Total} \mathrm{cost}}{\mathrm{Rate}}=\frac{\mathrm{Rs} 2400}{\mathrm{Rs} 30/\mathrm{m}}=80 \mathrm{m}$
Let the breadth of the rectangular field be x m.
Perimeter of the rectangular field = 2(24 + x) m
$$\begin{gathered} \Rightarrow 2(24+x)=80 \\ \Rightarrow 48+2x=80 \\ \Rightarrow 2x=(80-48) \\ \Rightarrow 2x=32 \\ \Rightarrow x=\frac{32}{2}=16 \\ \mathrm{So}, \mathrm{the} \mathrm{breadth} \mathrm{of} \mathrm{the} \mathrm{rectangular} \mathrm{field} \mathrm{is} 16 \mathrm{m}. \end{gathered}$$

Answer:

(c) 17 m
Let the length and the breadth of the rectangle be L m and B m, respectively.

Area of the rectangular carpet = ($L\times B$) m²
$\Rightarrow LB=120 ... \left(\mathrm{i}\right)$
Perimeter of the rectangular carpet = $2(L+B)$
$$\begin{gathered} \Rightarrow 2(L+B)=46 \\ \Rightarrow (L+B)=\frac{46}{2} \\ \Rightarrow (L+B)=23 ...\left(\mathrm{ii}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Diagonal} \mathrm{of} \mathrm{the} rectangle\; = \sqrt{L^2+B^2} \mathrm{m} \\ =\sqrt{(L+B{)}^2-2LB} \mathrm{m} \\ =\sqrt{(23{)}^2-240} \mathrm{m} (\mathrm{from} \mathrm{equatio}n\mathrm{s} (\mathrm{i}) \mathrm{and} (\mathrm{ii}\left)\right) \\ =\sqrt{529-240} \mathrm{m} \\ =\sqrt{289} \mathrm{m} \\ =17 \mathrm{m} \end{gathered}$$

Answer:

(a) 48 cm
Let the width  and the length of the rectangle be x cm and 3x cm, respectively.

Applying Pythagoras theorem:

$$\begin{gathered} (\mathrm{Diagonal}{)}^2=(\mathrm{Length}{)}^2+(\mathrm{Width}{)}^2 \\ \Rightarrow (6\sqrt{10}{)}^2=(3\mathrm{x}{)}^2+(\mathrm{x}{)}^2 \\ \Rightarrow 360=9{\mathrm{x}}^2+{\mathrm{x}}^2 \\ \Rightarrow 360=10{\mathrm{x}}^2 \\ \Rightarrow {\mathrm{x}}^2=\frac{360}{10} \\ \Rightarrow {\mathrm{x}}^2=36 \\ \Rightarrow \mathrm{x}=\pm 6 \\ \mathrm{Since} \mathrm{the} \mathrm{width} \mathrm{cannot} \mathrm{be} \mathrm{negative}, \mathrm{we} \mathrm{will} \mathrm{neglect} -6. \end{gathered}$$

So, width of the rectangle is 6 cm.
Length of the rectangle = $\left(3\times 6\right)=18 \mathrm{cm}$
Perimeter of the rectangle = 2(Length + Breadth) = 2(18 + 6) = 48 cm

Answer:

(b) 2 : 1
Let the breadth of the plot be b cm.

Let the length of the plot be x cm.
Perimeter of the plot = 3x cm

Perimeter of the plot =2(Length + Breadth)= 2(x + b) cm
$$\begin{gathered} \Rightarrow 2(x+b)=3x \\ 2x+2b=3x \\ \Rightarrow 2b=3x-2x \\ \Rightarrow 2b=x \\ \Rightarrow b=\frac{x}{2} \\ \therefore \mathrm{Ratio} o\mathrm{f} \mathrm{the} \mathrm{length} \mathrm{and} \mathrm{the} \mathrm{breadth} \mathrm{of} \mathrm{the} \mathrm{plot} =\frac{\mathrm{x}}{\left({\displaystyle \frac{\mathrm{x}}{2}}\right)}=\frac{\mathrm{x}}{\mathrm{x}} \times 2 = \frac{2}{1} \\ \therefore \mathrm{Ratio} \mathrm{of} \mathrm{the} \mathrm{length} \mathrm{and} \mathrm{the} \mathrm{breadth} \mathrm{of} \mathrm{the} \mathrm{plot} =2:1 \end{gathered}$$

Answer:

(b) 200 cm²
Area of the square = $\left\{\frac{1}{2}\times (\mathrm{Diagonal}{)}^{2 }\right\} \mathrm{sq} \mathrm{u}\mathrm{nits}$
$$\begin{gathered} =\left\{\frac{1}{2}\times (20{)}^2 \right\} {\mathrm{cm}}^2 \\ =\left\{\frac{1}{2}\times \left(20\right)\times \left(20\right)\right\} {\mathrm{cm}}^2 \\ =(20\times 10) {\mathrm{cm}}^2 \\ =200 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

(c) 20 m
Let one side of the square field be x m.
Total cost of fencing a square field = Rs 2000
Rate of fencing the field = Rs 25/m

$\mathrm{Perimeter} \mathrm{of} \mathrm{the} \mathrm{square} \mathrm{field}=\frac{\mathrm{Total} \mathrm{cost} \mathrm{of} \mathrm{fencing} \mathrm{the} \mathrm{field}}{\mathrm{Rate} \mathrm{of} \mathrm{fencing} \mathrm{the} \mathrm{field}}=\frac{\mathrm{Rs} 2000}{\mathrm{Rs} 25/\mathrm{m}}=\frac{2000}{25} \mathrm{m}=80 \mathrm{m}$

Perimeter of the square field = ($4\times \mathrm{side}$) = 4x m
$$\begin{gathered} \Rightarrow 4x=80 \\ \Rightarrow x=\frac{80}{4} \\ \Rightarrow x=20 \\ \mathrm{Each} \mathrm{side} \mathrm{of} \mathrm{the} \mathrm{field} \mathrm{is} 20 \mathrm{m}. \end{gathered}$$

Answer:

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

Answer:

(a) 28 cm
$$\begin{gathered} \mathrm{Circumference} \mathrm{of} \mathrm{the} \mathrm{circle} \mathrm{is} 88 \mathrm{cm}. \\ \mathrm{Let} \mathrm{the} \mathrm{radius} \mathrm{be} r \mathrm{cm}. \\ \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{circumference} \mathrm{of} \mathrm{the} \mathrm{circle} \mathrm{is} \left(2\mathrm{\pi r}\right) \mathrm{cm}. \\ \Rightarrow 2\pi r=88 \\ \Rightarrow 2\times \frac{22}{7}\times r=88 \\ \Rightarrow r=\frac{1}{2}\times \frac{7}{22}\times 88 \\ \Rightarrow r=14 \\ \therefore \mathrm{Radius}=14 \mathrm{cm} \\ \mathrm{Diameter}=(2\times \mathrm{Radius})=(2 \times 14) \mathrm{cm} =28 \mathrm{cm} \end{gathered}$$

Answer:

(b) 110 m
$$\begin{gathered} \mathrm{Radius} \mathrm{of} \mathrm{the} \mathrm{wheel}=\frac{\mathrm{Diameter}}{2}=\frac{70}{2}=35 \mathrm{cm} \\ \mathrm{Circumference} \mathrm{of} \mathrm{the} \mathrm{wheel} =2\mathrm{\pi r}=(2\times \frac{22}{7}\times 35) \mathrm{cm} =220 \mathrm{cm} \\ \mathrm{The} \mathrm{distance} \mathrm{covered} \mathrm{by} \mathrm{the} \mathrm{wheel} \mathrm{in} \mathrm{one} \mathrm{revolution} \mathrm{is} \mathrm{equal} \mathrm{to} \mathrm{its} \mathrm{circumference}. \\ \mathrm{Distance} \mathrm{covered} \mathrm{by} \mathrm{the} \mathrm{wheel} \mathrm{in} 1 \mathrm{revolution} = 220 \mathrm{cm} \\ \therefore \mathrm{Distance} \mathrm{covered} \mathrm{by} \mathrm{the} \mathrm{wheel} \mathrm{in} 50 \mathrm{revolution} = (50 \times 220)\mathrm{cm}=11000 \mathrm{cm}=110 \mathrm{m} \end{gathered}$$

Answer:

(d) 80000
Length of the road = 150 m = 15000 cm
Breadth of the road = 9 m = 900 cm
Area of the road = (Length × Breadth)
= (15000 × 900) cm²
= 13500000 cm²

Length of the brick = 22.5 cm
Breadth of the brick = 7.5 cm
Area of one brick = (Length × Breadth)
= ( 22.5 × 7.5 ) cm²
= 168.75 cm²

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{bricks} = \frac{\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{road}}{\mathrm{Area} \mathrm{of} \mathrm{one} \mathrm{brick}} \\ =\frac{13500000 {\mathrm{cm}}^2}{168.75 {\mathrm{cm}}^2}=80000 \mathrm{bricks} \end{gathered}$$

Answer:

(b) 24.3 m²

Length of the room = 5 m 40 cm = 5.40 m
Breadth of the room = 4 m 50 cm = 4.50 m

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{room} = (\mathrm{Length} \times \mathrm{Breadth})=(5.40 \times 4.50) {\mathrm{m}}^2 \\ =\left(\frac{540}{100}\times \frac{450}{100}\right){\mathrm{m}}^2 \\ =\left(\frac{27}{5}\times \frac{9}{2}\right){\mathrm{m}}^2 \\ =\frac{243}{10}{\mathrm{m}}^2=24.3 {\mathrm{m}}^2 \end{gathered}$$

Answer:

(d) 16

Length of the sheet of paper = 72 cm
Breadth of the sheet of paper = 48 cm
Area of the sheet = (Length × Breadth)
⇒ ( 72 × 48 ) cm²  = 3456 cm²

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 × 12) cm²
= 216 cm²

$$\begin{gathered} \mathrm{No}. \mathrm{of} \mathrm{envelopes} \mathrm{that} \mathrm{can} \mathrm{be} \mathrm{made} = \frac{\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{sheet}}{\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{piece} \mathrm{of} \mathrm{paper} \mathrm{required} \mathrm{to} \mathrm{make} 1 \mathrm{envelope}} \\ \Rightarrow \mathrm{No}. \mathrm{of} \mathrm{envelopes} \mathrm{th}a\mathrm{t} \mathrm{can} \mathrm{be} \mathrm{made} =\frac{3456}{216}=16 \mathrm{envelopes} \end{gathered}$$

Answer:

(i) Sides of the triangle are 5.4 cm, 4.6 cm and 6.8 cm. 
    Perimeter of the triangle = (First side + Second side + Third Side)
                                   = (5.4 + 4.6 + 6.8) cm = 16.8 cm

(ii) Length of each side of the given hexagon = 8 cm
   ∴ Perimeter of the hexagon = (6 × 8) cm = 48 cm

(iii) Length of the two equal sides = 6 cm
     Length of the third side = 4.5 cm
    ∴ Perimeter of the triangle = {(2 × equal sides) + third side} cm = (2 × 6) + 4.5 = 16.5 cm

Answer:

Let the length of the rectangle be x m.
Breadth of the rectangle = 75 m
Perimeter of the rectangle = 2(Length + Breadth)
                                  = 2(x + 75) m = (2x + 150) m
It is given that the perimeter of the field is 360 m.
$$\begin{gathered} \Rightarrow 2x+150=360 \\ \Rightarrow 2x=360-150 \\ \Rightarrow 2x=210 \\ \Rightarrow x=\frac{210}{2}=105 \end{gathered}$$
So, the length of the rectangle is 105 m.

Answer:

Let the length of the rectangle be 5x m.
Breadth of the rectangle = 4x m
Perimeter of the rectangle = 2(Length + Breadth)
                                  = 2(5x + 4x) m = 18x m
It is given that the perimeter of the field is 108 m.
∴ 18 x = 108
⇒ x = $\frac{108}{18}=6$
∴ Length of the field = ( 5 × 6 )m = 30 m
Breadth of the field = ( 4 × 6 )m = 24 m

Answer:

Let one side of the square be x cm.
Perimeter of the square = $(4\times \mathrm{side})=(4\times x) \mathrm{cm} =4x \mathrm{cm}$
It is given that the perimeter of the square is 84 cm.
$$\begin{gathered} \Rightarrow 4\mathrm{x}=84 \\ \Rightarrow x=\frac{84}{4}=21 \\ \mathrm{Thus}, \mathrm{one} \mathrm{side} \mathrm{of} \mathrm{the} \mathrm{square} \mathrm{is} 21 \mathrm{cm}. \\ \mathrm{Area} \mathrm{of} the \mathrm{square} = (S\mathrm{ide}{)}^2=(21{)}^2 {\mathrm{cm}}^2= 441{\mathrm{cm}}^2 \end{gathered}$$

Answer:

Let the length of the room be x m.
Breadth of the room = 12 m
Area of the room = (Length × Breadth) = (x × 12) m²
It is given that the area of the room is 216 m².
⇒ x × 12 = 216
⇒ x = $\frac{216}{12}=18$
∴ Length of the rectangle = 18 m

Answer:

Radius(r) of the given circle = 7 cm
Circumference of the circle, C = 2 πr
                                                 $$\begin{gathered} = \left(2\times \frac{22}{7}\times 7\right) \mathrm{cm} \\ = 44 \mathrm{cm} \end{gathered}$$
Hence, the circumference of the given circle is 44 cm.

Answer:

Radius of the wheel =$\frac{\mathrm{Diameter} \mathrm{of} \mathrm{the} \mathrm{wheel}}{2}$
⇒ r = $\frac{77}{2}$cm
Circumference of the wheel = 2 πr
= $\left(2\times \frac{22}{7}\times \frac{77}{2}\right)$ cm
= 242 cm

In 1 revolution, the wheel covers a distance equal to its circumference.
∴ Distance covered by the wheel in 1 revolution = 242 cm
∴ Distance covered by the wheel in 500 revolutions = ( 500 × 242 ) cm
                                                                               = 121000 cm       (100 cm =1 m)
                                                                               = 1210 m

Answer:

Let the radius be r cm.
Diameter = 2 × Radius(r) = 2r cm
Circumference of the wheel = 2πr
∴ 2πr = 176
$$\begin{gathered} \Rightarrow 2r=\frac{176}{\pi } \\ \Rightarrow 2r=176\times \frac{7}{22}=56 \end{gathered}$$
⇒ 2r = 56
Thus, the diameter of the given wheel is 56 cm.

Answer:

Length of the rectangle = 36 cm 
Breadth of the rectangle = 15 cm
Area of the rectangle = (Length × Breadth) sq units
                             = (36 × 15) cm² = 540 cm²

Answer:

(b) 64 cm
Side of the square = 16 cm
 Perimeter of the square = (4 × side)
                               = (4 × 16) cm 
                               = 64 cm

Answer:

(a) 15 m

Let the breadth of the rectangle be x m.
Length of the rectangle = 16 m
Area of rectangle = (Length × Breadth) = (16 × x) m²
It is given that the area of the rectangle is 240 m².
⇒ 16 × x = 240
⇒ x = $\frac{240}{16}=15$
So, the breadth of the rectangle is 15 m.

Answer:

(b) 225 m²

$$\begin{gathered} \mathrm{Side} \mathrm{of} the \mathrm{square} \mathrm{lawn} = 15 \mathrm{m} \\ \mathrm{Area} \mathrm{of} the \mathrm{square} \mathrm{lawn} = (S\mathrm{ide}{)}^2 \mathrm{sq} \mathrm{unit}s=(15{)}^2 {\mathrm{m}}^2=225 {\mathrm{m}}^2 \end{gathered}$$

Answer:

(a) 16 cm
Let one side of the square be x cm.
Area of the square = (Side )² cm² = x² cm²
It is given that the area of the square is 256 cm².
⇒ x² = 256
⇒ x = $\sqrt{256}=\pm 16$
We know that the side of a square cannot be negative.
So, we will neglect −16.
Therefore, the side of the square is 16 cm.

Perimeter of the square = $\left(4\times \mathrm{side}\right)=\left(4\times 16\right)\mathrm{cm}=64 \mathrm{cm}$

Answer:

(b) 10.5 m
Let the breadth of the rectangle be x m.
Length of the rectangle = 12 m
Area of the rectangle = 126 m²
Area of the rectangle = $\left(\mathrm{length}\times \mathrm{breadth}\right)\mathrm{sq} \mathrm{units}=(12\times x){\mathrm{m}}^2=12x {\mathrm{m}}^2$
It is given that the area of the rectangle is 126 m².
$$\begin{gathered} \Rightarrow 12x=126 \\ \Rightarrow x=\frac{126}{12}=10.5 \\ \mathrm{So}, the \mathrm{breadth} \mathrm{of} \mathrm{the} \mathrm{rectangle} \mathrm{is} 10.5 \mathrm{m}. \end{gathered}$$

Answer:

(i) A polygon having all sides equal and all angles equal is called a regular polygon
(ii) Perimeter of a square = 4 × side
(iii) Area of a rectangle = (length) × (breadth)
(iv) Area of a square = (side)²
(v) If the length of a rectangle is 5 m and its breadth is 4 m, then its area is 20 m²
Area of a rectangle = (length) × (breadth) = (5×4) m² = 20 m²

Answer: