RS Aggarwal 2019,2020 solution class 7 chapter 20 Mensuration Exercise 20A

Exercise 20A

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Question 1:

Find the area of the rectangle whose dimensions are:
(i) length = 24.5 m, breadth = 18 m
(ii) length = 12.5 m, breadth - 8 dm.

Answer 1:

(i) Length = 24.5 m
Breadth = 18 m
   
   ∴ Area of the rectangle = Length $\times$ Breadth
                                 = 24.5 m $\times$ 18 m
                                 = 441 m²
              
  (ii) Length = 12.5 m
Breadth = 8 dm = (8 $\times$ 10) = 80 cm = 0.8 m     [since 1 dm = 10 cm and 1 m = 100 cm]
      
      ∴ Area of the rectangle = Length $\times$ Breadth
                                    = 12.5 m $\times$ 0.8 m
                                    = 10 m²

Question 2:

Find the area of a rectangular plot, one side of which is 48 m and its diagonal is 50 m.

Answer 2:

We know that all the angles of a rectangle are 90° and the diagonal divides the rectangle into two right angled triangles.
So, 48 m will be one side of the triangle and the diagonal, which is 50 m, will be the hypotenuse.
According to the Pythagoras theorem:
(Hypotenuse)² = (Base)² + (Perpendicular)²
Perpendicular = $\sqrt{{\left(\mathrm{Hypotenuse}\right)}^2-(\mathrm{Base}{)}^2}$
Perpendicular = $\sqrt{{\left(50\right)}^2-{\left(48\right)}^2}=\sqrt{2500-2304}=\sqrt{196}=14$ m
∴ Other side of the rectangular plot = 14 m
Length = 48m
Breadth = 14m

∴ Area of the rectangular plot = 48 m $\times$ 14 m = 672 m²
Hence, the area of a rectangular plot is 672 m².

Question 3:

The sides of a rectangular park are in the ratio 4 : 3. If its area is 1728 m², find the cost of fencing it at Rs 30 per metre.

Answer 3:

Let the length of the field be 4x m.
Breadth = 3x m
∴ Area of the field = (4x $\times$ 3x) m² = 12x² m²
But it is given that the area is 1728 m².
∴ 12x² = 1728
⇒ x² = $\left(\frac{1728}{12}\right)$ = 144
⇒ x = $\sqrt{144}$ = 12
∴ Length = (4 $\times$ 12) m = 48 m
Breadth = (3 $\times$ 12) m =36 m
∴ Perimeter of the field = 2(l + b) units
                                       = 2(48 + 36) m = (2 $\times$ 84) m = 168 m
∴ Cost of fencing = Rs (168 $\times$ 30) = Rs 5040

Question 4:

The area of a rectangular field is 3584 m² and its length is 64 m. A boy runs around the field at the rate of 6 km/h. How long will he take to go 5 times around it?

Answer 4:

Area of the rectangular field = 3584 m²
Length of the rectangular field = 64 m
Breadth of the rectangular field = $\left(\frac{\mathrm{Area}}{\mathrm{Length}}\right)$ = $\left(\frac{3584}{64}\right)$ m = 56 m
Perimeter of the rectangular field = 2 (length + breadth)
                                                      = 2(64+ 56) m = (2 $\times$ 120) m = 240 m
Distance covered by the boy = 5 $\times$ Perimeter of the rectangular field
                                              = 5 $\times$ 240 = 1200 m

The boy walks at the rate of 6 km/hr.
or
Rate = $\left(\frac{6\times 1000}{60}\right)$ m/min = 100 m/min.

∴ Required time to cover a distance of 1200 m = $\left(\frac{1200}{100}\right)$ min = 12 min
Hence, the boy will take 12 minutes to go five times around the field.

Question 5:

A verandah is 40 m long and 15 m broad. It is to be paved with stones, each measuring 6 dm by 5 dm. Find the number of stones required.

Answer 5:

Given:
Length of the verandah = 40 m = 400 dm    [since 1 m = 10 dm ]
Breadth of the verandah = 15 m = 150 dm
∴ Area of the verandah= (400 $\times$ 150) dm² = 60000 dm²

Length of a stone = 6 dm
Breadth of a stone = 5 dm
∴ Area of a stone = (6 $\times$ 5) dm² = 30 dm²

∴ Total number of stones needed to pave the verandah = $\frac{\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{verandah}}{\mathrm{Area} \mathrm{of} \mathrm{each} \mathrm{stone}}$

                                                                                    = $\left(\frac{60000}{30}\right)$ = 2000

Question 6:

Find the cost of carpeting a room 13 m by 9 m with a carpet of width 75 cm at the rate of Rs 105 per metre.

Answer 6:

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

Now, width of the carpet = 75 cm    (given)
                                         = 0.75 m      [since 1 m = 100 cm]

Length of the carpet = $\left(\frac{\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{carpet}}{\mathrm{Width} \mathrm{of} \mathrm{the} \mathrm{carpet}}\right)$ = $\left(\frac{117}{0.75}\right)$ m = 156 m
Rate of carpeting = Rs 105 per m
∴ Total cost of carpeting = Rs (156 $\times$105) = Rs 16380
Hence, the total cost of carpeting the room is Rs 16380.

Question 7:

The cost of carpeting a room 15 m long with a carpet of width 75 cm at Rs 80 per metre is Rs 19200. Find the width of the room.

Answer 7:

Given:
Length of the room = 15 m
Width of the carpet = 75 cm = 0.75 m          (since 1 m = 100 cm)

Let the length of the carpet required for carpeting the room be x m.
Cost of the carpet = Rs. 80 per m
∴ Cost of x m carpet = Rs. (80 $\times$ x) = Rs. (80x)
Cost of carpeting the room = Rs. 19200
∴ 80x = 19200 ⇒ x = $\left(\frac{19200}{80}\right)$ = 240
Thus, the length of the carpet required for carpeting the room is 240 m.
Area of the carpet required for carpeting the room = Length of the carpet $\times$ Width of the carpet
                                                                           = ( 240 $\times$ 0.75) m² = 180 m²
Let the width of the room be b m.
Area to be carpeted = 15 m $\times$ b m = 15b m²
∴ 15b m² = 180 m²
⇒ b = $\left(\frac{180}{15}\right)$ m = 12 m
Hence, the width of the room is 12 m.

Question 8:

The length and breadth of a rectangular piece of land are in the ratio of 5 : 3. If the total cost of fencing it at Rs 24 per metre is Rs 9600, find its length and breadth.

Answer 8:

Total cost of fencing a rectangular piece = Rs. 9600 
Rate of fencing = Rs. 24
∴ Perimeter of the rectangular field = $\left(\frac{\mathrm{Total} \mathrm{cost} \mathrm{of} \mathrm{fencing}}{\mathrm{Rate} \mathrm{of} \mathrm{fencing}}\right)$ m = $\left(\frac{9600}{24}\right)$ m = 400 m

Let the length and breadth of the rectangular field be 5x and 3x, respectively.
Perimeter of the rectangular land =  2(5x + 3x) = 16x
But the perimeter of the given field is 400 m.
∴ 16x = 400
 x = $\left(\frac{400}{16}\right)$ = 25
Length of the field = (5 $\times$ 25) m = 125 m
Breadth of the field = (3 $\times$ 25) m = 75 m

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Question 9:

Find the length of the largest pole that can be placed in a hall 10 m long, 10 m wide and 5 m high.

Answer 9:

Length of the diagonal of the room = $\sqrt{{\mathrm{l}}^2+{\mathrm{b}}^2+{\mathrm{h}}^2}$
                                                         = $\sqrt{{\left(10\right)}^2+{\left(10\right)}^2+{\left(5\right)}^2}$ m
                                                         = $\sqrt{100+100+25}$m
                                                         = $\sqrt{225}$m = 15 m

Hence, length of the largest pole that can be placed in the given hall is 15 m.

Question 10:

Find the area of a square each of whose sides measures 8.5 m.

Answer 10:

Side of the square = 8.5 m
∴ Area of the square = (Side)²
                               = (8.5 m)²
                                = 72.25 m²

Question 11:

Find the area of the square, the length of whose diagonal is

(i) 72 cm
(ii) 2.4 m

Answer 11:

(i) Diagonal of the square = 72 cm
     ∴ Area of the square = $\left[\frac{1}{2}\times {\left(D\mathrm{iagonal}\right)}^2\right]$ sq. unit
                                    = $\left[\frac{1}{2}\times {\left(72\right)}^2\right]$ cm²
                                    = 2592 cm²

(ii)Diagonal of the square = 2.4 m
     ∴ Area of the square = $\left[\frac{1}{2}\times {\left(D\mathrm{iagonal}\right)}^2\right]$ sq. unit
                                    = $\left[\frac{1}{2}\times {\left(2.4\right)}^2\right]$ m²
                                    = 2.88 m²

Question 12:

The area of a square is 16200 m². Find the length of its diagonal.

Answer 12:

We know:
Area of a square = $\left\{\frac{1}{2}\times {\left(D\mathrm{iagonal}\right)}^2\right\}$ sq. units
Diagonal of the square = $\sqrt{2\times \mathrm{Area} \mathrm{of} \mathrm{square}}$ units
                                  = $\left(\sqrt{2\times 16200}\right)$m = 180 m
∴ Length of the diagonal of the square = 180 m

                             

Question 13:

The area of a square field is $\frac{1}{2}$ hectare. Find the length of its diagonal in metres.

Answer 13:

Area of the square = $\left\{\frac{1}{2}\times {\left(D\mathrm{iagonal}\right)}^2\right\}$ sq. units
Given:
Area of the square field = $\frac{1}{2}$ hectare
                                  = $\left(\frac{1}{2}\times 10000\right)$ m² = 5000 m²             [since 1 hectare = 10000 m² ]

Diagonal of the square = $\sqrt{2\times \mathrm{Area} \mathrm{of} the \mathrm{square}}$ 

                                  = $\left(\sqrt{2\times 5000}\right)$m = 100 m

∴ Length of the diagonal of the square field = 100 m

Question 14:

The area of a square plot is 6084 m². Find the length of the wire which can go four times along the boundary of the plot.

Answer 14:

Area of the square plot = 6084 m²
Side of the square plot = $\left(\sqrt{\mathrm{Area}}\right)$
                                         = $\left(\sqrt{6084}\right)$ m
                                         = $\left(\sqrt{78\times 78}\right)$m = 78 m

∴ Perimeter of the square plot = 4 $\times$ side = (4 $\times$ 78) m = 312 m
312 m wire is needed to go along the boundary of the square plot once.

Required length of the wire that can go four times along the boundary = 4 $\times$ Perimeter of the square plot
                                                                                                           = (4 $\times$ 312) m = 1248 m

Question 15:

A wire is in the shape of a square of side 10 cm. If the wire is rebent into a rectangle of length 12 cm, find its breadth. Which figure encloses more area and by how much?

Answer 15:

Side of the square = 10 cm
Length of the wire = Perimeter of the square = 4 $\times$ Side = 4 $\times$ 10 cm = 40 cm
Length of the rectangle (l) = 12 cm
Let b be the breadth of the rectangle.

Perimeter of the rectangle = Perimeter of the square

⇒ 2(l + b) = 40
⇒ 2(12 + b) = 40
⇒ 24 + 2b = 40
⇒ 2b = 40 - 24 = 16
⇒ b = $\left(\frac{16}{2}\right)$ cm = 8 cm
∴ Breadth of the rectangle = 8 cm
Now, Area of the square = (Side)² = (10 cm $\times$ 10 cm) = 100 cm²
Area of the rectangle = l $\times$ b = (12 cm $\times$ 8 cm) = 96 cm²

Hence, the square encloses more area.
It encloses 4 cm² more area.

Question 16:

A godown is 50 m long, 40 m broad and 10 m high. Find the cost of whitewashing its four walls and ceiling at Rs 20 per square metre.

Answer 16:

Given:
Length = 50 m
Breadth = 40 m
Height = 10 m

Area of the four walls  = {2h(l + b)} sq. unit
                                   = {2 $\times$ 10 $\times$ (50 + 40)}m²
                                   = {20 $\times$ 90} m² = 1800 m²
Area of the ceiling = l $\times$ b = (50 m $\times$ 40 m) = 2000 m²
∴ Total area to be white washed = (1800 + 2000) m² = 3800 m²
Rate of white washing = Rs 20/sq. metre
∴ Total cost of white washing = Rs (3800 $\times$ 20) = Rs 76000

Question 17:

The area of the 4 walls of a room is 168 m². The breadth and height of the room are 10 m and 4 m respectively. Find the length of the room.

Answer 17:

Let the length of the room be l m.
Given:
Breadth of the room = 10 m
Height of the room = 4 m
Area of the four walls = [2(l + b)h] sq units.
                              = 168 m²
∴ 168 = [2(l + 10) $\times$ 4]
⇒ 168 = [8l + 80]
⇒ 168 - 80 = 8l
⇒  88 = 8l
⇒ l = $\left(\frac{88}{8}\right)$ m = 11 m
∴ Length of the room = 11 m

Question 18:

The area of the 4 walls of a room is 77 m². The length and breadth of the room are 7.5 m and 3.5 m respectively. Find the height of the room.

Answer 18:

Given:
Length of the room = 7.5 m
Breadth of the room = 3.5 m
Area of the four walls = [2(l + b)h] sq. units.
                              = 77 m²
∴ 77 = [2(7.5 + 3.5)h]
⇒ 77 = [(2 $\times$ 11)h]
⇒ 77 = 22h
⇒ h =  $\left(\frac{77}{22}\right)$ m = $\left(\frac{7}{2}\right)$ m = 3.5 m
∴ Height of the room = 3.5 m

Question 19:

The area of four walls of a room is 120 m². If the length of the room is twice its breadth and the height is 4 m, find the area of the floor.

Answer 19:

Let the breadth of the room be x m.
Length of the room = 2x m
Area of the four walls = {2(l + b) $\times$ h} sq. units
           120 m²  = {2(2x + x) $\times$ 4} m²  
⇒ 120 = {8 $\times$ 3x }
⇒ 120 = 24x
⇒ x =$\left(\frac{120}{24}\right)$ = 5
∴ Length of the room = 2x = (2 $\times$ 5) m = 10 m
Breadth of the room = x = 5 m
∴ Area of the floor = l $\times$ b = (10 m $\times$ 5 m) = 50 m²

Question 20:

A room is 8.5 m long, 6.5 m broad and 3.4 m high. It has two doors, each measuring (1.5 m by 1 m) and two windows, each measuring (2 m by 1 m). Find the cost of painting its four walls at Rs 160 per m².

Answer 20:

Length = 8.5 m
Breadth = 6.5 m
Height = 3.4 m

Area of the four walls = {2(l + b) $\times$ h} sq. units
                          = {2(8.5 + 6.5) $\times$ 3.4}m² = {30 $\times$ 3.4} m² = 102 m²
Area of one door = (1.5 $\times$ 1) m² = 1.5 m²
∴ Area of two doors = (2 $\times$ 1.5) m² = 3 m²
Area of one window = (2 $\times$ 1) m² = 2 m²
 ∴ Area of two windows = (2 $\times$ 2) m² = 4 m²
Total area of  two doors and two windows = (3 + 4) m²
                                                               = 7 m²
Area to be painted = (102 - 7) m² = 95 m²
Rate of painting = Rs 160 per m²
Total cost of painting = Rs (95 $\times$ 160) = Rs 15200