Rs Aggarwal 2017 Solutions for Class 7 Math Chapter 20 Mensuration are provided here with simple step-by-step explanations. These solutions for Mensuration are extremely popular among Class 7 students for Math Mensuration Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 7 Math Chapter 20 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 7 Math are prepared by experts and are 100% accurate.

Page No 229:

Answer:

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

Page No 229:

Answer:

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 = Hypotenuse2-(Base)2
Perpendicular = 502-482=2500-2304=196=14 m
∴ Other side of the rectangular plot = 14 m
Length = 48m
Breadth = 14m

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

Page No 229:

Answer:

Let the length of the field be 4x m.
Breadth = 3x m
∴ Area of the field = (4x × 3x) m2 = 12x2 m2
But it is given that the area is 1728 m2.
∴ 12x2 = 1728
x2 = 172812 = 144
x = 144 = 12
∴ Length = (4 × 12) m = 48 m
Breadth = (3 × 12) m =36 m
∴ Perimeter of the field = 2(l + b) units
                                       = 2(48 + 36) m = (2 × 84) m = 168 m
∴ Cost of fencing = Rs (168 × 30) = Rs 5040

Page No 229:

Answer:

Area of the rectangular field = 3584 m2
Length of the rectangular field = 64 m
Breadth of the rectangular field = AreaLength = 358464 m = 56 m
Perimeter of the rectangular field = 2 (length + breadth)
                                                      = 2(64+ 56) m = (2 × 120) m = 240 m
Distance covered by the boy = 5 × Perimeter of the rectangular field
                                              = 5 × 240 = 1200 m

The boy walks at the rate of 6 km/hr.
or
Rate = 6×100060 m/min = 100 m/min.

∴ Required time to cover a distance of 1200 m = 1200100 min = 12 min
Hence, the boy will take 12 minutes to go five times around the field.

Page No 229:

Answer:

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 × 150) dm2 = 60000 dm2

Length of a stone = 6 dm
Breadth of a stone = 5 dm
∴ Area of a stone = (6 × 5) dm2 = 30 dm2

∴ Total number of stones needed to pave the verandah = Area of the verandahArea of each stone

                                                                                    = 6000030 = 2000

Page No 229:

Answer:

Area of the carpet = Area of the room
                              = (13 m × 9 m) = 117 m2

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

Length of the carpet = Area of the carpetWidth of the carpet1170.75 m = 156 m
Rate of carpeting = Rs 105 per m
∴ Total cost of carpeting = Rs (156 ×105) = Rs 16380
Hence, the total cost of carpeting the room is Rs 16380.

Page No 229:

Answer:

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 × x) = Rs. (80x)
Cost of carpeting the room = Rs. 19200
∴ 80x = 19200 ⇒ x = 1920080 = 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 × Width of the carpet
                                                                           = ( 240 × 0.75) m2 = 180 m2
Let the width of the room be b m.
Area to be carpeted = 15 m × b m = 15b m2
∴ 15b m2 = 180 m2
b = 18015 m = 12 m
Hence, the width of the room is 12 m.

Page No 229:

Answer:

Total cost of fencing a rectangular piece = Rs. 9600 
Rate of fencing = Rs. 24
∴ Perimeter of the rectangular field = Total cost of fencingRate of fencing m = 960024 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
 x40016 = 25
Length of the field = (5 × 25) m = 125 m
Breadth of the field = (3 × 25) m = 75 m



Page No 230:

Answer:

Length of the diagonal of the room = l2+b2+h2
                                                         = 102+102+52 m
                                                         = 100+100+25m
                                                         = 225m = 15 m

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

Page No 230:

Answer:

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

Page No 230:

Answer:

(i) Diagonal of the square = 72 cm
     ∴ Area of the square = 12×Diagonal2 sq. unit
                                    = 12×722 cm2
                                    = 2592 cm2

(ii)Diagonal of the square = 2.4 m
     ∴ Area of the square = 12×Diagonal2 sq. unit
                                    = 12×2.42 m2
                                    = 2.88 m2

Page No 230:

Answer:

We know:
Area of a square = 12×Diagonal2 sq. units
Diagonal of the square = 2×Area of square units
                                  = 2×16200m = 180 m
∴ Length of the diagonal of the square = 180 m

                             

Page No 230:

Answer:

Area of the square = 12×Diagonal2 sq. units
Given:
Area of the square field = 12 hectare
                                  = 12×10000 m2 = 5000 m2             [since 1 hectare = 10000 m2 ]

Diagonal of the square = 2×Area of the square 

                                  = 2×5000m = 100 m

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

Page No 230:

Answer:

Area of the square plot = 6084 m2
Side of the square plot = Area
                                         = 6084 m
                                         = 78×78m = 78 m

∴ Perimeter of the square plot = 4 × side = (4 × 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 × Perimeter of the square plot
                                                                                                           = (4 × 312) m = 1248 m

Page No 230:

Answer:

Side of the square = 10 cm
Length of the wire = Perimeter of the square = 4 × Side = 4 × 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 = 162 cm = 8 cm
∴ Breadth of the rectangle = 8 cm
Now, Area of the square = (Side)2 = (10 cm × 10 cm) = 100 cm2
Area of the rectangle = × b = (12 cm × 8 cm) = 96 cm2

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

Page No 230:

Answer:

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

Area of the four walls  = {2h(l + b)} sq. unit
                                   = {2 × 10 × (50 + 40)}m2
                                   = {20 × 90} m2 = 1800 m2
Area of the ceiling = × b = (50 m × 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 × 20) = Rs 76000

Page No 230:

Answer:

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) × 4]
⇒ 168 = [8l + 80]
⇒ 168 - 80 = 8l
⇒  88 = 8l
l = 888 m = 11 m
∴ Length of the room = 11 m

Page No 230:

Answer:

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 × 11)h]
⇒ 77 = 22h
h7722 m = 72 m = 3.5 m
∴ Height of the room = 3.5 m

Page No 230:

Answer:

Let the breadth of the room be x m.
Length of the room = 2x m
Area of the four walls = {2(l + b× h} sq. units
           120 m2  = {2(2x + x) × 4} m2  
⇒ 120 = {8 × 3x }
⇒ 120 = 24x
x =12024 = 5
∴ Length of the room = 2x = (2 × 5) m = 10 m
Breadth of the room = x = 5 m
∴ Area of the floor = l × b = (10 m × 5 m) = 50 m2

Page No 230:

Answer:

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

Area of the four walls = {2(l + b× h} sq. units
                          = {2(8.5 + 6.5) × 3.4}m2 = {30 × 3.4} m2 = 102 m2
Area of one door = (1.5 × 1) m2 = 1.5 m2
∴ Area of two doors = (2 × 1.5) m2 = 3 m2
Area of one window = (2 × 1) m2 = 2 m2
 ∴ Area of two windows = (2 × 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 × 160) = Rs 15200



Page No 232:

Answer:

Let PQRS be the given grassy plot and ABCD be the inside boundary of the path.

Length = 75 m
Breadth = 60 m
Area of the plot = (75 × 60) m2 = 4500 m2
Width of the path = 2 m
∴ AB = (75 - 2 × 2) m = (75 - 4) m =71 m
 AD = (60 - 2 × 2) m = (60 - 4) m = 56 m
Area of rectangle ABCD = (71 x 56) m2 = 3976 m2
Area of the path = (Area of PQRS - Area of ABCD)
                           = (4500 - 3976) m2 = 524 m2
Rate of constructing the path = Rs 125 per m2
∴ Total cost of constructing the path = Rs (524 × 125) = Rs 65,500

Page No 232:

Answer:

Let PQRS be the given rectangular plot and ABCD be the inside boundary of the path.

Length = 95 m
Breadth = 72 m
Area of the plot = (95 × 72) m2 = 6,840 m2
Width of the path = 3.5 m
∴ AB = (95 - 2 × 3.5) m = (95 - 7) m = 88 m
AD = (72 - 2 × 3.5) m = (72 - 7) m = 65 m
Area of the grassy rectangle plot ABCD = (88 × 65) m2 = 5,720 m2
Area of the path = (Area PQRS - Area ABCD)
                           = (6840 - 5720) m2 = 1,120 m2
Rate of constructing the path = Rs. 80 per m2
∴ Total cost of  constructing the path = Rs. (1,120 × 80) = Rs. 89,600
Rate of laying the grass on the plot ABCD = Rs. 40 per m2
∴ Total cost of laying the grass on the plot = Rs. (5,720 × 40) = Rs. 2,28,800
∴ Total expenses involved = Rs. ( 89,600 + 2,28,800) = Rs. 3,18,400

Page No 232:

Answer:

Let ABCD be the saree and EFGH be the part of saree without border.

Length, AB= 5 m
Breadth, BC = 1.3 m
Width of the border of the saree = 25 cm = 0.25 m

∴ Area of ABCD = 5 m × 1.3 m = 6.5 m2

Length, GH = {5 -( 0.25 + 0.25} m = 4.5 m
Breadth, FG = {1.3 - 0.25 + 0.25} m = 0.8 m
∴ Area of EFGH = 4.5 m × .8 m = 3.6 m2

Area of the border = Area of ABCD − Area of EFGH
                              =  6.5 m2  − 3.6 m2
                              = 2.9 m2 = 29000 cm2     [since 1 m2 = 10000 cm2]
Rate of printing the border = Rs 1 per 10 cm2
∴ Total cost of printing the border = Rs 1×2900010
                                                       = Rs 2900



Page No 233:

Answer:

Length, EF = 38 m
Breadth, FG = 25 m

∴ Area of EFGH = 38 m ×  25 m = 950 m2

Length, AB = (38  + 2.5  + 2.5 ) m = 43 m
Breadth, BC = ( 25 + 2.5 + 2.5 ) m = 30 m
∴ Area of ABCD = 43 m × 30 m = 1290 m2

Area of the path = Area of ABCD − Area of PQRS
                          = 1290 m2 − 950 m2
                         = 340 m2
Rate of gravelling the path = Rs 120 per m2

∴ Total cost of gravelling the path = Rs (120 × 340)
                                                    = Rs 40800

Page No 233:

Answer:

Let EFGH denote the floor of the room.
The white region represents the floor of the 1.25 m verandah.

Length, EF = 9.5 m
Breadth, FG = 6 m

∴ Area of EFGH = 9.5 m  ×  6 m = 57 m2

Length, AB = (9.5  + 1.25  + 1.25 ) m = 12 m
Breadth, BC = ( 6 + 1.25 + 1.25 ) m = 8.5 m
∴ Area of ABCD = 12 m × 8.5 m = 102 m2

Area of the verandah = Area of ABCD − Area of EFGH
                                  = 102 m2 − 57 m2
                                  = 45 m2
Rate of cementing the verandah = Rs 80 per m2

∴ Total cost of cementing the verandah = Rs ( 80 × 45)
                                                            = Rs 3600

Page No 233:

Answer:

Side of the flower bed = 2 m 80 cm = 2.80 m      [since 100 cm = 1 m]

∴ Area of the square flower bed = (Side)2 = (2.80 m )2 = 7.84 m2
Side of the flower bed with the digging strip = 2.80 m + 30 cm + 30 cm
                                                                        = (2.80 + 0.3 + 0.3) m = 3.4 m
Area of the enlarged flower bed with the digging strip = (Side )2 = (3.4 )2 = 11.56 m2

∴ Increase in the area of the flower bed = 11.56 m2 − 7.84 m2
                                                            = 3.72 m2

Page No 233:

Answer:

Let the length and the breadth of the park be 2x m and x m, respectively.
Perimeter of the park = 2(2x + x) = 240 m
⇒ 2(2x + x) = 240
⇒ 6x = 240
x = 2406 m =40 m
∴ Length of the park = 2x = (2 × 40) = 80 m
Breadth = x = 40 m
Let PQRS be the given park and ABCD be the inside boundary of the path.

Length = 80 m
Breadth = 40 m
Area of the park = (80 × 40) m2 = 3200 m2
Width of the path = 2 m
∴ AB = (80 - 2 × 2) m = (80 - 4) m =76 m
 AD = (40 - 2 × 2) m = (40 - 4) m = 36 m
Area of the rectangle ABCD = (76 × 36) m2 = 2736 m2
Area of the path = (Area of PQRS - Area of ABCD)
                           = (3200 - 2736) m2 = 464 m2
Rate of paving the path = Rs. 80 per m2
∴ Total cost of  paving the path = Rs. (464 × 80) = Rs. 37,120

Page No 233:

Answer:

Length of the hall, PQ = 22 m
Breadth of the hall, QR = 15.5 m

∴ Area of the school hall PQRS = 22 m × 15.5 m = 341 m2
Length of the carpet, AB = 22 m − ( 0.75 m + 0.75 m) = 20.5 m         [since 100 cm = 1 m]
Breadth of the carpet, BC = 15.5 m − ( 0.75 m + 0.75 m) = 14 m
∴ Area of the carpet ABCD = 20.5 m × 14 m = 287 m2
Area of the strip = Area of the school hall (PQRS) − Area of the carpet (ABCD)
                           = 341 m2 − 287 m2
                           = 54 m2

Area of 1 m length of the carpet = 1 m × 0.82 m = 0.82 m2

∴ Length of the carpet whose area is 287 m2 = 287 m2 ÷ 0.82 m2 = 350 m
Cost of the 350 m long carpet = Rs 60 × 350 = Rs 21000

Page No 233:

Answer:

Let ABCD be the square lawn and PQRS be the outer boundary of the square path.

Let a side of the lawn (AB) be x m.
Area of the square lawn = x2
Length, PQ = (x m + 2.5 m + 2.5 m) = (x + 5) m
∴ Area of PQRS = (x + 5)2 = (x2 + 10x + 25) m2

Area of the path = Area of PQRS − Area of the square lawn (ABCD)
⇒ 165 = x2 + 10x + 25 x2
⇒ 165 = 10x + 25
⇒ 165 − 25 = 10x
⇒ 140 = 10x
x = 140 ÷ 10 = 14
∴ Side of the lawn = 14 m

∴ Area of the lawn = (Side)2 = (14 m)2 = 196 m2

Page No 233:

Answer:

Area of the path = 305 m2

Let the length of the park be 5x m and the breadth of the park be 2x m.

∴ Area of the rectangular park = 5x × 2x = 10x2 m2
Width of the path = 2.5 m
Outer length, PQ = 5x m + 2.5 m + 2.5 m = (5x + 5) m
Outer breadth, QR = 2x + 2.5 m + 2.5 m = (2x + 5) m
Area of PQRS = (5x + 5) × (2x + 5)  = (10x2 + 25x + 10x + 25) = (10x2 + 35x + 25) m2
∴ Area of the path = [(10x2 + 35x + 25) − 10x2 ] m2
⇒  305 = 35x + 25
⇒ 305 − 25 = 35x  
⇒ 280 = 35x
x = 280 ÷ 35 = 8

∴ Length of the park = 5x = × 8 = 40 m
Breadth of the park = 2x = 2 × 8 = 16 m

Page No 233:

Answer:

Let ABCD be the rectangular park.
Let EFGH and IJKL be the two rectangular roads with width 5 m.

Length of the rectangular park, AD = 70 m 
  Breadth of the rectangular park, CD = 50 m
∴ Area of the rectangular park = Length × Breadth = 70 m × 50 m = 3500 m2
Area of road EFGH = 70 m × 5 m = 350 m2
Area of road IJKL = 50 m × 5 m = 250 m2

Clearly, area of MNOP is common to both the two roads.

∴ Area of MNOP = 5 m × 5 m = 25 m2

Area of the roads = Area (EFGH) + Area (IJKL) − Area (MNOP)
                            = (350  + 250 ) m2− 25 m2 = 575 m2
It is given that the cost of constructing the roads is Rs. 120/m2.

Cost of constructing 575 m2 area of the roads = Rs. (120 × 575)
                                                                    = Rs. 69000

Page No 233:

Answer:

Let ABCD be the rectangular field and PQRS and KLMN be the two rectangular roads with width 2 m and 2.5 m, respectively.

Length of the rectangular field, CD = 115 cm
Breadth of the rectangular field, BC = 64 m
∴ Area of the rectangular lawn ABCD = 115 m × 64 m = 7360 m2
Area of the road PQRS = 115 m × 2 m = 230 m2
Area of the road KLMN = 64 m × 2.5 m = 160 m2

Clearly, the area of EFGH is common to both the two roads.

∴ Area of EFGH = 2 m × 2.5 m = 5 m2

∴ Area of the roads = Area (KLMN) + Area (PQRS) − Area (EFGH)
                            = (230 m2 + 160 m2) − 5 m2 = 385 m2

Rate of gravelling the roads = Rs 60 per m2
∴ Total cost of gravelling the roads = Rs (385 × 60)
                                                     = Rs 23,100

Page No 233:

Answer:

Let ABCD be the rectangular field and KLMN and PQRS be the two rectangular roads with width 2.5 m and 2 m, respectively.

Length of the rectangular field CD = 50 cm
Breadth of the rectangular field BC = 40 m
∴ Area of the rectangular field ABCD = 50 m × 40 m = 2000 m2
Area of road KLMN = 40 m × 2.5 m = 100 m2
Area of road PQRS = 50 m × 2 m = 100 m2

Clearly, area of EFGH is common to both the two roads.

∴ Area of EFGH = 2.5 m × 2 m = 5 m2

∴ Area of the roads = Area (KLMN) + Area (PQRS) − Area (EFGH)
                                = (100 m2 + 100 m2) − 5 m2 = 195 m2

Area of the remaining portion of the field = Area of the rectangular field (ABCD) − Area of the roads
                                                                   = (2000 − 195) m2
                                                                                  = 1805 m2

Page No 233:

Answer:

(i) Complete the rectangle as shown below:
    
    Area of the shaded region = [Area of rectangle ABCD - Area of rectangle EFGH] sq. units
                                              = [(43 m × 27 m) - {(43 - 2 × 1.5) m x (27 - 1 × 2) m}]
                                              = [(43 m × 27 m) - {40 m × 25 m}]
                                              = 1161 m2 - 1000 m2
                                              = 161 m2

(ii) Complete the rectangle as shown below:
     
     Area of the shaded region = [Area of square ABCD - {(Area of EFGH) + (Area of IJKL) - (Area of MNOP)}] sq. units
                                               = [(40 × 40) - {(40 × 2) + (40 × 3) - (2 × 3)}] m2
                                               = [1600 - {(80 + 120 - 6)] m2
                                               = [1600 - 194] m2
                                               = 1406 m2



Page No 234:

Answer:

(i) Complete the rectangle as shown below:
    
Area of the shaded region = [Area of rectangle ABCD - Area of rectangle EFGD] sq. units
                                          = [(AB × BC) - (DG × GF)] m2
                                          = [(24 m × 19 m) - {(24 - 4) m × 16.5 m} ]
                                          = [(24 m × 19 m) - (20 m × 16.5) m]
                                          = (456 - 330) m2 = 126 m2

(ii) Complete the rectangle by drawing lines as shown below:
     
Area of the shaded region ={(12 × 3) + (12 × 3) + (5× 3) + {(15 - 3 - 3) ×3)} cm2
                                    = { 36 + 36 + 15 + 27} cm2
                                    = 114 cm2

Page No 234:

Answer:

Divide the given figure in four parts shown below:

Given:
Width of each part = 0.5 m

Now, we have to find the length of each part.

Length of part I = 3.5 m
Length of part II = (3.5 - 0.5 - 0.5) m = 2.5 m
Length of part III = (2.5 - 0.5 - 0.5) = 1.5 m
Length of part IV = (1.5 - 0.5 - 0.5) = 0.5 m
∴ Area of the shaded region = [Area of part (I) + Area of part (II) + Area of part (III) + Area of part (IV)] sq. units
                                        = [(3.5 × 0.5) + (2.5 × 0.5) + ( 1.5 × 0.5) + (0.5 × 0.5)] m2
                                        = [1.75 + 1.25 + 0.75 + 0.25] m2
                                        = 4 m2



Page No 237:

Answer:

Base = 32 cm
Height = 16.5 cm

∴ Area of the parallelogram = Base × Height
                                          =  32 cm × 16.5 cm
                                          = 528 cm2

Page No 237:

Answer:

Base = 1 m 60 cm = 1.6 m             [since 100 cm = 1 m]
Height = 75 cm = 0.75 m

∴ Area of the parallelogram = Base × Height
                                              = 1.6 m × 0.75 m
                                              = 1.2 m2

Page No 237:

Answer:

(i) Base = 14 dm = (14 × 10) cm = 140 cm                  [since 1 dm  = 10 cm]
     Height = 6.5 dm = (6.5 × 10) cm = 65 cm
    
     Area of the parallelogram  = Base × Height
                                                = 140 cm × 65 cm
                                                = 9100 cm2

(ii) Base = 14 dm = (14 × 10) cm                            [since 1 dm = 10 cm and 100 cm  = 1 m]
              = 140 cm = 1.4 m            
      Height = 6.5 dm = (6.5 × 10) cm
                  = 65 cm = 0.65 m
    
 ∴ Area of the parallelogram = Base × Height
                                                = 1.4 m × 0.65 m
                                                = 0.91 m2

Page No 237:

Answer:

Area of the given parallelogram = 54 cm2
Base of the given parallelogram = 15 cm
∴ Height of the given parallelogram = AreaBase5415 cm = 3.6 cm

Page No 237:

Answer:

Base of the parallelogram = 18 cm
Area of the parallelogram = 153 cm2
∴ Area of the parallelogram = Base × Height
⇒ Height = Area of the parallelogramBase = 15318 cm = 8.5 cm
Hence, the distance of the given side from its opposite side is 8.5 cm.

Page No 237:

Answer:

Base, AB = 18 cm
Height, AL = 6.4 cm
∴ Area of the parallelogram ABCD = Base × Height
                                                    = (18 cm × 6.4 cm) = 115.2 cm2                    ... (i)

Now, taking BC as the base:
Area of the parallelogram ABCD = Base × Height
                                                      = (12 cm × AM)                          ... (ii)
From equation (i) and (ii):
12 cm × AM = 115.2 cm2
⇒ AM = 115.212cm
= 9.6 cm

Page No 237:

Answer:

ABCD is a parallelogram with side AB of length 15 cm and the corresponding altitude AE of length 4 cm.
The adjacent side AD is of length 8 cm and the corresponding altitude is CF.

Area of a parallelogram = Base × Height

We have two altitudes and two corresponding bases.

AD × CF = AB × AE
⇒ 8 cm × CF = 15 cm ×4 cm

CF =15×48 cm = 152 cm = 7.5 cm
Hence, the distance between the shorter sides is 7.5 cm.

Page No 237:

Answer:

Let the base of the parallelogram be x cm.
Then, the height of the parallelogram will be 13x cm.
It is given that the area of the parallelogram is 108 cm2.

Area of a parallelogram =  Base × Height
                     ∴ 108 cm2 = x × 13x
                        108 cm2 = 13x2
x2 = (108 × 3) cm2 = 324 cm2
x2 = (18 cm)2
x = 18 cm

∴ Base = x = 18 cm
Height = 13x 13×18 cm
            = 6 cm

Page No 237:

Answer:

Let the height of the parallelogram be x cm.
Then, the base of the parallelogram will be 2x cm.
It is given that the area of the parallelogram is 512 cm2.

Area of a parallelogram =  Base × Height
                     ∴ 512 cm2 = 2x × x
                        512 cm2 = 2x2
x2 =5122 cm2 = 256 cm2
x2 = (16 cm)2
x = 16 cm

∴ Base = 2x = × 16
             = 32 cm
Height = x = 16 cm

Page No 237:

Answer:

A rhombus is a special type of a parallelogram.

The area of a parallelogram is given by the product of its base and height.
∴ Area of the given rhombus = Base × Height
(i) Area of the rhombus = 12 cm × 7.5 cm = 90 cm2

(ii) Base = 2 dm = (2 × 10) = 20 cm    [since 1 dm = 10 cm]
     Height = 12.6 cm
      ∴ Area of the rhombus = 20 cm × 12.6 cm = 252 cm2

Page No 237:

Answer:

(i)
     Length of one diagonal = 16 cm
     Length of the other diagonal = 28 cm
     ∴ Area of the rhombus = 12 × (Product of the diagonals)
                                           = 12×16×28 cm2 = 224 cm2
(ii)
      Length of one diagonal = 8 dm 5 cm = (8 × 10 + 5) cm = 85 cm              [since 1 dm = 10 cm]
      Length of the other diagonal = 5 dm 6 cm = (5 × 10 + 6) cm = 56 cm
      ∴ Area of the rhombus = 12 × (Product of the diagonals)
                                            = 12×85×56 cm2
                                            = 2380 cm2
    

Page No 237:

Answer:

Let ABCD be the rhombus, whose diagonals intersect at O.

 AB = 20 cm and AC = 24 cm
The diagonals of a rhombus bisect each other at right angles.

Therefore, ΔAOB is a right angled triangle, right angled at O. 

Here, OA =12AC = 12 cm
AB = 20 cm

By Pythagoras theorem:
(AB)2 = (OA)2 + (OB)2
⇒ (20)2 = (12)2 + (OB)2
⇒ (OB)2 = (20)2 − (12)2
⇒ (OB)2 = 400 − 144 = 256
⇒ (OB)2 = (16)2
⇒ OB = 16 cm
∴ BD = 2 × OB = 2 × 16 cm = 32 cm

∴  Area of the rhombus ABCD = 12×AC×BD cm2
                                                  = 12×24×32 cm2
                                                  = 384 cm2

Page No 237:

Answer:

Area of a rhombus =  12 × (Product of the diagonals)
Given:
Length of one diagonal = 19.2 cm
Area of the rhombus = 148.8 cm2

∴ Length of the other diagonal = 148.8×219.2 cm = 15.5 cm

Page No 237:

Answer:

Perimeter of the rhombus = 56 cm
Area of the rhombus = 119 cm2
Side of the rhombus = Perimeter4 = 564 cm = 14 cm
Area of a rhombus = Base × Height

∴  Height of the rhombus = AreaBase= 11914 cm
                                          = 8.5 cm

Page No 237:

Answer:

Given:
Height of the rhombus = 17.5 cm
Area of the rhombus = 441 cm2

We know:
Area of a rhombus = Base × Height

∴  Base of the rhombus =AreaHeight44117.5 cm = 25.2 cm
Hence, each side of a rhombus is 25.2 cm.

Page No 237:

Answer:

Area of a triangle = 12 × Base × Height
                          = 12×24.8×16.5 cm2 = 204.6 cm2
Given:
Area of the rhombus = Area of the triangle
Area of the rhombus = 204.6 cm2

Area of the rhombus = 12 × (Product of the diagonals)
Given:
Length of one diagonal = 22 cm 
∴ Length of the other diagonal = 204.6×222 cm
                                                  = 18.6 cm

                          



Page No 242:

Answer:

We know:
Area of a triangle = 12×Base×Height
(i) Base = 42 cm
Height = 25 cm
     ∴ Area of the triangle = 12×42×25 cm2 = 525 cm2
(ii) Base = 16.8 m    
     Height = 75 cm = 0.75 m      [since 100 cm = 1 m]
     ∴ Area of the triangle = 12×16.8×0.75 m2 = 6.3 m2
(iii) Base = 8 dm = (8 × 10) cm = 80 cm     [since 1 dm = 10 cm]
       Height = 35 cm
     ∴ Area of the triangle = 12×80×35 cm2 = 1400 cm2

Page No 242:

Answer:

Height of a triangle = 2×AreaBase
Here, base = 16 cm and area = 72 cm2

∴ Height = 2×7216 cm = 9 cm

Page No 242:

Answer:

Height of a triangle = 2×AreaBase
Here, base = 28 m and area = 224 m2

∴ Height = 2×22428 m = 16 m

Page No 242:

Answer:

Base of a triangle = 2×AreaHeight
Here, height = 12 cm and area = 90 cm2

∴ Base = 2×9012 cm = 15 cm

Page No 242:

Answer:

Total cost of cultivating the field = Rs. 14580

Rate of cultivating the field = Rs. 1080 per hectare

Area of the field = Total costRate per hectare hectare
                           = 145801080 hectare
                           =  13.5 hectare
                           = (13.5 × 10000) m2 = 135000 m2       [since 1 hectare = 10000 m2 ]
Let the height of the field be x m.
Then, its base will be 3x m.
Area of the field = 12×3x×x m2 = 3x22 m2
3x22 = 135000
x2 =135000×23=90000
x = 90000 = 300
∴ Base = (3 × 300) = 900 m
Height = 300 m

Page No 242:

Answer:

Let the length of the other leg be h cm.
Then, area of the triangle = 12×14.8×h cm2 = (7.4 h) cm2
But it is given that the area of the triangle is 129.5 cm2.
∴ 7.4h = 129.5
h = 129.57.4 = 17.5 cm
∴ Length of the other leg = 17.5 cm

Page No 242:

Answer:

Here, base = 1.2 m and hypotenuse = 3.7 m

In the right angled triangle:

Perpendicular = (Hypotenuse)2-(Base)2

                        =3.72-1.22=13.69-1.44=12.25=3.5
 Area = 12×base×perpendicular sq. units
          = 12×1.2×3.5 m2
∴ Area of the right angled triangle = 2.1 m2

Page No 242:

Answer:

In a right angled triangle, if one leg is the base, then the other leg is the height.
Let the given legs be 3x and 4x, respectively.
Area of the triangle = 12×3x×4x cm2
⇒ 1014 = (6x2)
⇒ 1014 = 6x2
x2 = 10146 = 169
x = 169 = 13
∴ Base = (3 × 13) = 39 cm
Height = (4 × 13) = 52 cm



Page No 243:

Answer:

Consider a right-angled triangular scarf (ABC).
Here, ∠B= 90°
BC = 80 cm
AC = 1 m = 100 cm

Now, AB2 + BC2 = AC2
⇒ AB2 = AC2 - BC2 = (100)2 - (80)2
            = (10000 - 6400) = 3600
 ⇒ AB = 3600  = 60 cm
Area of the scarf ABC = 12×BC×AB sq. units
                               = 12×80×60 cm2
                               = 2400 cm2 = 0.24 m2     [since 1 m2 = 10000 cm2]
Rate of the cloth = Rs 250 per m2
∴ Total cost of the scarf = Rs (250 × 0.24) = Rs 60
Hence, cost of the right angled scarf is Rs 60.
                            

Page No 243:

Answer:

(i) Side of the equilateral triangle = 18 cm
     Area of the equilateral triangle = 34Side2  sq. units
                                                      =  34182 cm2 = 3×81 cm2
                                                      = (1.73 × 81) cm2 = 140.13 cm2

(ii) Side of the equilateral triangle = 20 cm
     Area of the equilateral triangle = 34Side2  sq. units
                                                      =  34202 cm2 = 3×100 cm2
                                                      = (1.73 × 100) cm2 = 173 cm2

Page No 243:

Answer:

It is given that the area of an equilateral triangle is 163 cm2.

We know:
Area of an equilateral triangle = 34side2 sq. units

∴ Side of the equilateral triangle = 4×Area3 cm
                                                      =   4×1633cm = 4×16cm = 64cm = 8 cm

Hence, the length of the equilateral triangle is 8 cm.

Page No 243:

Answer:

Let the height of the triangle be h cm.
Area of the triangle = 12× Base × Height sq. units
                          = 12×24×h cm2

Let the side of the equilateral triangle be a cm.
Area of the equilateral triangle = 34a2 sq. units
                                            = 34×24×24 cm2 = 3×144 cm2
12×24×h = 3×144
⇒ 12 h = 3×144
h = 3×14412=3×12=1.73×12=20.76 cm
∴ Height of the equilateral triangle = 20.76 cm

Page No 243:

Answer:

(i) Let a = 13 m, b = 14 m and c = 15 m
     s = a+b+c2 = 13+14+152=422m = 21 m
∴  Area of the triangle = ss-as-bs-c sq. units
                               = 2121-1321-1421-15m2
                               =   21×8×7×6 m2
                               = 3×7×2×2×2×7×2×3 m2
                               = (2 ××× 7) m2
                               = 84 m2

(ii) Let a = 52 cm, b = 56 cm and c = 60 cm
      s = a+b+c2 = 52+56+602=1682 cm = 84 cm
∴  Area of the triangle = ss-as-bs-c sq. units
                               = 8484-5284-5684-60cm2
                               =   84×32×28×24 cm2
                               = 12×7×4×8×4×7×3×8 cm2
                               = 2×2×3×7×2×2×2×2×2×2×2×7×3×2×2×2 cm2
                               = (2 ×××××××× 7) m2
                               = 1344 cm2

(iii) Let a = 91 m, b = 98 m and c = 105 m
      s = a+b+c2 = 91+98+1052=2942 m = 147 m
∴  Area of the triangle = ss-as-bs-c sq. units
                               = 147147-91147-98147-105m2
                               =  147×56×49×42 m2
                               = 3×49×8×7×49×6×7 m2
                               = 3×7×7×2×2×2×7×7×7×2×3×7 m2
                               = ( 2 ××××× 7) m2
                               = 4116 m2

Page No 243:

Answer:

Let a = 33 cm, b = 44 cm and c = 55 cm
Then, s = a+b+c2 = 33+44+552 cm = 1322 cm = 66 cm
∴  Area of the triangle = ss-as-bs-c sq. units
                                      = 6666-3366-4466-55 cm2
                                      =  66×33×22×11 cm2
                                      = 6×11×3×11×2×11×11 cm2
                                      = (6 × 11 × 11) cm2 = 726 cm2

Let the height on the side measuring 44 cm be h cm.
Then, Area  = 12×b×h
⇒ 726 cm2 = 12×44×h
h = 2×72644 cm = 33 cm.
∴  Area of the triangle = 726 cm2
Height corresponding to the side measuring 44 cm = 33 cm

Page No 243:

Answer:

Let a = 13x cm, b = 14x cm and c = 15x cm
Perimeter of the triangle = 13x + 14x + 15x = 84 (given)
⇒ 42x = 84
x = 8442=2
a = 26 cm , b = 28 cm and c = 30 cm
               
s = a+b+c2 = 26+28+302cm = 842cm = 42 cm
∴  Area of the triangle = ss-as-bs-c sq. units
                               = 4242-2642-2842-30 cm2
                               =  42×16×14×12 cm2
                               = 6×7×4×4×2×7×6×2 cm2
                               = (2 ××× 7) cm2 = 336 cm2
Hence, area of the given triangle is 336 cm2.

Page No 243:

Answer:

Let a = 42 cm, b = 34 cm and c = 20 cm
Then, s = a+b+c2 = 42+34+202 cm = 962 cm = 48 cm
∴  Area of the triangle = ss-as-bs-c sq. units
                               = 4848-4248-3448-20 cm2
                               =  48×6×14×28 cm2
                               = 6×2×2×2×6×14×2×14 cm2
                               = (2 ××× 14) cm2 = 336 cm2

Let the height on the side measuring 42 cm be h cm.
Then, Area  = 12×b×h
⇒ 336 cm2 = 12×42×h
h = 2×33642 cm = 16 cm
∴ Area of the triangle = 336 cm2
Height corresponding to the side measuring 42 cm = 16 cm

Page No 243:

Answer:

Let each of the equal sides be a cm.
b = 48 cm
a = 30 cm
Area of the triangle = 12×b×a2-b24 sq. units
                                = 12×48×302-4824 cm2 = 24×900-23044 cm2
                                = 24×900-576 cm2 = 24×324 cm2 = (24 × 18) cm2 = 432 cm2
∴ Area of the triangle = 432 cm2

Page No 243:

Answer:

Let each of the equal sides be a cm.
a + a + 12 = 32 ⇒ 2a = 20  ⇒ a = 10
b = 12 cm and a = 10 cm
Area of the triangle = 12×b×a2-b24 sq. units
                                = 12×12×100-1444 cm2 = 6-100-36 cm2
                                = 6×64 cm2 = (6 × 8) cm2
                                = 48 cm2

Page No 243:

Answer:

We have:
AC = 26 cm, DL = 12.8 cm and BM = 11.2 cm

Area of ΔADC 12 × AC × DL
                          = 12 × 26 cm × 12.8 cm = 166.4 cm2
Area of ΔABC = 12 × AC × BM
                         = 12 × 26 cm × 11.2 cm = 145.6 cm2

∴ Area of the quadrilateral ABCD = Area of ΔADC  + Area of ΔABC
                                              = (166.4 + 145.6) cm2
                                                        = 312 cm2

Page No 243:

Answer:

First, we have to find the area of ΔABC and ΔACD.
For ΔACD:
Let a = 30 cm, b = 40 cm and c = 50 cm
     s = a+b+c2=30+40+502=1202=60 cm
∴  Area of triangle ACD = ss-as-bs-c  sq. units
                                        = 6060-3060-4060-50 cm2
                                        =  60×30×20×10 cm2
                                        = 360000 cm2
                                        = 600 cm2
For ΔABC:
Let a = 26 cm, b = 28 cm and c = 30 cm
     s = a+b+c2=26+28+302=842=42 cm
∴  Area of triangle ABC = ss-as-bs-c  sq. units
                                        = 4242-2642-2842-30 cm2
                                        =  42×16×14×12 cm2
                                        = 2×3×7×2×2×2×2×2×7×3×2×2 cm2
                                        = (2 ××××× 7) cm2
                                        = 336 cm2
∴ Area of the given quadrilateral ABCD =  Area of ΔACD + Area of ΔABC
                                                                           = (600 + 336) cm2 = 936 cm2

Page No 243:

Answer:

Area of the rectangle = AB × BC
                                   = 36 m × 24 m
                                   = 864 m2
Area of the triangle  = 12 × AD × FE
                                 = 12 × BC × FE       [since AD = BC]
                                 = 12 × 24 m × 15 m
                                 = 12 m ×15 m = 180 m2
∴ Area of the shaded region = Area of the rectangle − Area of the triangle
                                              = (864 − 180) m2
                                                         =  684 m2



Page No 244:

Answer:

Join points PR and SQ.
These two lines bisect each other at point O.

Here, AB = DC = SQ = 40 cm
AD = BC =RP = 25 cm

Also, OP = OR RP2=252 = 12.5 cm
From the figure we observe:
Area of ΔSPQ = Area of ΔSRQ
∴ Area of the shaded region   = 2 × (Area of ΔSPQ)
                                                       = 2 × (12× SQ ×OP)
                                                       = 2 × (12 × 40 cm × 12.5 cm)
                                                       = 500 cm2

Page No 244:

Answer:

(i) Area of rectangle ABCD = (10 cm x 18 cm) = 180 cm2
    
    Area of triangle I = 12×6×10 cm2 = 30 cm2
    Area of triangle II = 12×8×10 cm2 = 40 cm2
    ∴ Area of the shaded region = {180 - ( 30 + 40)} cm2 = { 180 - 70}cm2 = 110 cm2

(ii) Area of square ABCD = (Side)2 = (20 cm)2 = 400 cm2
     
     Area of triangle I = 12×10×20 cm2 = 100 cm2
    Area of triangle II = 12×10×10 cm2 = 50 cm2
    Area of triangle III = 12×10×20 cm2 = 100 cm2
   ∴ Area of the shaded region = {400 - ( 100 + 50 + 100)} cm2 = { 400 - 250}cm2 = 150 cm2

Page No 244:

Answer:

Let ABCD be the given quadrilateral and let BD be the diagonal such that BD is of the length 24 cm.
Let AL ⊥ BD and CM ⊥ BD
Then, AL = 5 cm and CM = 8 cm
Area of the quadrilateral ABCD = (Area of ΔABD + Area of ΔCBD)
                                              =  12×BD×AL+12×BD×CM sq. units
                                              =  12×24×5+12×24×8 cm2
                                              = ( 60 + 96) cm2 = 156 cm2

∴ Area of the given quadrilateral = 156 cm



Page No 248:

Answer:

Here, r = 15 cm
∴ Circumference = 2πr
                             = ( 2 × 3.14 × 15) cm
                             = 94.2 cm
Hence, the circumference of the given circle is 94.2 cm

Page No 248:

Answer:

(i) Here, r = 28 cm
     ∴ Circumference = 2π r
                                  = 2×227×28cm
                                  =  176 cm
    Hence, the circumference of the given circle is 176 cm.

(ii) Here, r = 1.4 m
      ∴ Circumference = 2π r
                                   = 2×227×1.4 m
                                   =  2×22×0.2 m = 8.8 m
    Hence, the circumference of the given circle is 8.8 m.

Page No 248:

Answer:

(i) Here, d = 35 cm
     Circumference = 2π r
                              = πd    [since 2r = d]
                              =  227×35 cm = (22 × 5) = 110 cm
     Hence, the circumference of the given circle is 110 cm.

(ii) Here, d = 4.9 m
      Circumference =2π r
                              = πd    [since 2r = d]
                              =  227×4.9 m = (22 × 0.7) = 15.4 m
      Hence, the circumference of the given circle is 15.4 m.

Page No 248:

Answer:

Circumference of the given circle = 57.2 cm
∴ C = 57.2 cm
Let the radius of the given circle be r cm.
C = 2πr
r = C2π cm
r = 57.22×722 cm = 9.1 cm
Thus, radius of the given circle is 9.1 cm.

Page No 248:

Answer:

Circumference of the given circle = 63.8 m
∴ C = 63.8 m
Let the radius of the given circle be r cm.
C = 2πr
r = C2π
r = 63.82×722m =10.15 m
∴ Diameter of the given circle = 2r = (2 × 10.15) m = 20.3 m

Page No 248:

Answer:

Let the radius of the given circle be r cm.
Then, its circumference = 2πr

Given:
(Circumference) - (Diameter) = 30 cm     
∴ (2πr - 2r ) =  30
2rπ-1=30
2r227-1=30
2r×157=30
r=30×730=7
∴ Radius of the given circle = 7 cm

Page No 248:

Answer:

Let the radii of the given circles be 5x and 3x, respectively.
Let their circumferences be C1 and C2, respectively.

C1 = 2×π×5x=10πx

C2 = 2×π×3x=6πx
C1C2=10πx6πx=53
⇒ C1:C2 = 5:3
Hence, the ratio of the circumference of the given circle is 5:3.

Page No 248:

Answer:

Radius of the circular field, r = 21 m.
Distance covered by the cyclist = Circumference of the circular field
                                             = 2πr
                                             = 2×227×21 m = 132 m
Speed of the cyclist = 8 km per hour = 8000 m(60×60) s=80003600m/s = 209m/s
                                                                                                
Time taken by the cyclist to cover the field = Distance covered by the cyclistSpeed of the cyclist
                                                                     = 132209s
                                                                     = 132×920s
                                                                     = 59.4 s

Page No 248:

Answer:

Let the inner and outer radii of the track be r metres and R metres, respectively.

Then, 2πr = 528
2πR = 616
2×227×r=528 
2×227×R=616
r = 528×744=84 
R = 616×744=98
⇒ (R - r) = (98 - 84) m = 14 m
Hence, the width of the track is 14 m.

Page No 248:

Answer:

Let the inner and outer radii of the track be r metres and (r + 10.5) metres, respectively.

Inner circumference = 330 m
2πr=3302×227×r=330
                      ⇒ r = 330×744=52.5 m
Inner radius of the track = 52.5 m
∴ Outer radii of the track = (52.5 + 10.5) m = 63 m


∴ Circumference of the outer circle = 2×227×63 m=396 m
Rate of fencing = Rs. 20 per metre
∴ Total cost of fencing the outer circle = Rs. (396 × 20) = Rs. 7920

Page No 248:

Answer:

We know that the concentric circles are circles that form within each other, around a common centre point.
Radius of the inner circle, r = 98 cm
∴  Circumference of the inner circle = 2πr
                                                         = 2×227×98 cm = 616 cm

Radius of the outer circle, R = 1 m 26 cm = 126 cm              [since 1 m = 100 cm]
∴  Circumference of the outer circle = 2πR
                                                           = 2×227×126 cm = 792 cm
∴ Difference in the lengths of the circumference of the circles = (792 - 616) cm = 176 cm
Hence, the circumference of the second circle is 176 cm larger than that of the first circle.

Page No 248:

Answer:


Length of the wire = Perimeter of the equilateral triangle
                              = 3 × Side of the equilateral triangle = (3 × 8.8) cm = 26.4 cm
Let the wire be bent into the form of a circle of radius r cm.
Circumference of the circle = 26.4 cm
2πr=26.4
2×227×r=26.4
r = 26.4×72×22 cm = 4.2 cm

∴ Diameter = 2r = (2 × 4.2) cm = 8.4 cm
Hence, the diameter of the ring is 8.4 cm.

Page No 248:

Answer:

Circumference of the circle = Perimeter of the rhombus
                                           = 4  × Side of the rhombus = (4  ×  33) cm = 132 cm

∴ Circumference of the circle = 132 cm
⇒ 2πr=132
2×227×r=132
r = 132×72×22cm = 21 cm
Hence, the radius of the circle is 21 cm.

Page No 248:

Answer:


Length of the wire = Perimeter of the rectangle
                               = 2(l + b) = 2  × (18.7 + 14.3) cm = 66 cm

Let the wire be bent into the form of a circle of radius r cm.

Circumference of the circle = 66 cm

2πr=66
2×227×r=66
r = 66×72×22 cm = 10.5 cm

Hence, the radius of the circle formed is 10.5 cm.

Page No 248:

Answer:

It is given that the radius of the circle is 35 cm.
Length of the wire = Circumference of the circle
⇒ Circumference of the circle  = 2πr = 2×227×35 cm = 220 cm
Let the wire be bent into the form of a square of side a cm.
Perimeter of the square = 220 cm
⇒ 4a = 220
a = 2204cm = 55 cm
Hence, each side of the square will be 55 cm.

Page No 248:

Answer:

Length of the hour hand (r)= 4.2 cm.
Distance covered by the hour hand in 12 hours = 2πr = 2×227×4.2 cm = 26.4 cm

∴ Distance covered by the hour hand in 24 hours = (2  × 26.4) = 52.8 cm
Length of the minute hand (R)= 7 cm
Distance covered by the minute hand in 1 hour = 2πR = 2×227×7 cm = 44 cm

∴ Distance covered by the minute hand in 24 hours = (44  × 24) cm = 1056 cm

∴ Sum of the distances covered by the tips of both the hands in 1 day = (52.8 + 1056) cm
                                                                                                                = 1108.8 cm

Page No 248:

Answer:

Given:
Diameter of the well (d) = 140 cm.
Radius of the well (r) = 1402cm = 70 cm

Let the radius of the outer circle (including the stone parapet) be R cm.
Length of the outer edge of the parapet = 616 cm
2πR=616
2×227×R=616
R = 616×72×22 cm = 98 cm

Now, width of the parapet = {Radius of the outer circle (including the stone parapet) - Radius of the well}
                                           = {98 -70} cm = 28 cm
Hence, the width of the parapet is 28 cm.



Page No 249:

Answer:

It may be noted that in one rotation, the bus covers a distance equal to the circumference of the wheel.
Now, diameter of the wheel = 98 cm
∴ Circumference of the wheel = πd = 227×98 cm = 308 cm
Thus, the bus travels 308 cm in one rotation.

∴ Distance covered by the bus in 2000 rotations = (308  × 2000) cm
                                                                          = 616000 cm 
                                                                          = 6160 m          [since 1 m = 100 cm]

Page No 249:

Answer:

It may be noted that in one revolution, the cycle covers a distance equal to the circumference of the wheel.
Diameter of the wheel = 70 cm
∴ Circumference of the wheel = πd = 227×70 cm = 220 cm
Thus, the cycle covers 220 cm in one revolution.

∴ Distance covered by the cycle in 250 revolutions = (220  × 250) cm
                                                                               = 55000 cm 
                                                                               = 550 m              [since 1 m = 100 cm]

Hence, the cycle will cover 550 m in 250 revolutions.

Page No 249:

Answer:

Diameter of the wheel = 77 cm
⇒ Radius of the wheel = 772 cm
Circumference of the wheel = 2πr
                                              = 2×227×772cm = (22  × 11) cm = 242 cm
                                                                                                        = 242100m = 12150m
Distance covered by the wheel in 1 revolution = 12150 m
Now, 12150 m is covered by the car in 1 revolution.
(121  × 1000) m will be covered by the car in 1×50121×121×1000 revolutions, i.e. 50000 revolutions.
∴ Required number of revolutions = 50000

Page No 249:

Answer:

It may be noted that in one revolution, the bicycle covers a distance equal to the circumference of the wheel.
Total distance covered by the bicycle in 5000 revolutions = 11 km

⇒ 5000 × Circumference of the wheel = 11000 m                [since 1 km = 1000 m]

Circumference of the wheel = 110005000 m =2.2 m = 220 cm               [since 1 m = 100 cm]

Circumference of the wheel = π×Diameter of the wheel
⇒ 220 cm = 227×Diameter of the wheel
⇒ Diameter of the wheel = 220×722 cm = 70 cm
Hence, the circumference of the wheel is 220 cm and its diameter is 70 cm.



Page No 252:

Answer:

(i) Given:
r = 21 cm
     
∴ Area of the circle = πr2 sq. units
                                   = 227×21×21 cm2 = 22×3×21 cm2 = 1386 cm2

(ii) Given:
r = 3.5 m
     
Area of the circle = πr2 sq. units
                                   = 227×3.5×3.5 m2 = 22×0.5×3.5 m2 = 38.5 m2

Page No 252:

Answer:

(i) Given:
d = 28 cm ⇒ r = d2 = 282 cm = 14 cm
Area of the circle = πr2 sq. units
                                   = 227×14×14 cm2 = 22×2×14 cm2 = 616 cm2

(ii) Given:
r = 1.4 m ⇒ r = d2 = 1.42m = 0.7 m
      Area of the circle = πr2 sq. units
                                   = 227×0.7×0.7 m2 = 22×0.1×0.7 m2 = 1.54 m2

Page No 252:

Answer:

Let the radius of the circle be r cm.
Circumference = 2πrcm
2πr = 264
2×227×r=264
r = 264×72×22 = 42
∴ Area of the circle = πr2
                                 = 227×42×42 cm2
                                = 5544 cm2

Page No 252:

Answer:

Let the radius of the circle be r m.
Then, its circumference will be 2πrm.
2πr = 35.2
2×227×r=35.2
r = 35.2×72×22 = 5.6
∴ Area of the circle = πr2
                                 = 227×5.6×5.6 m2 = 98.56 m2

Page No 252:

Answer:

Let the radius of the circle be r cm.
Then, its area will be πr2 cm2.
∴ πr2 = 616
227×r×r = 616
r2 = 616×722 = 196
r = 196 = 14
⇒  Circumference of the circle = 2πr cm
                                                  = 2×227×14 cm = 88 cm

Page No 252:

Answer:

Let the radius of the circle be r m.
Then, area = πr2 m2
∴ πr2 = 1386
227×r×r = 1386
r2 = 1386×722 = 441
r = 441 = 21
⇒  Circumference of the circle = 2πr m
                                                  = 2×227×21 m = 132 m

Page No 252:

Answer:

Let r1 and r2 be the radii of the two given circles and A1 and A2 be their respective areas.

r1r2=45
A1A2=πr12πr22=r12r22=r1r22=452=1625
Hence, the ratio of the areas of the given circles is 16:25.

Page No 252:

Answer:

If the horse is tied to a pole, then the pole will be the central point and the area over which the horse will graze will be a circle. The string by which the horse is tied will be the radius of the circle.
Thus,
Radius of the circle (r) = Length of the string = 21 m

Now, area of the circle = πr2 = 227×21×21 m2 = 1386 m2
∴ Required area = 1386 m2

Page No 252:

Answer:

Let a be one side of the square.
Area of the square = 121 cm2                (given)
                 ⇒ a2 = 121
                 ⇒ a = 11 cm   (since 11  × 11 = 121)
Perimeter of the square = 4  × side = 4a = (4  × 11) cm = 44 cm
Length of the wire = Perimeter of the square
                               = 44 cm
The wire is bent in the form of a circle.
Circumference of a circle = Length of the wire
∴ Circumference of a circle = 44 cm
2πr=44
2×227×r=44
r = 44×72×22= 7 cm
∴ Area of the circle = πr2
                                 = 227×7×7 cm2
                                 = 154 cm2

Page No 252:

Answer:

It is given that the radius of the circle is 28 cm.

Length of the wire = Circumference of the circle
⇒ Circumference of the circle  = 2πr=2×227×28 cm = 176 cm
Let the wire be bent into the form of a square of side a cm.

Perimeter of the square = 176 cm

⇒ 4a = 176
a = 1764cm = 44 cm
Thus, each side of the square is 44 cm.

Area of the square = (Side)2 = (a)2 = (44 cm)2
                                                        = 1936 cm2
∴ Required area of the square formed = 1936 cm2

Page No 252:

Answer:

Area of the acrylic sheet = 34 cm  × 24 cm = 816 cm2
Given that the diameter of a circular button is 3.5 cm.
∴ Radius of the circular button (r)= 3.52 cm = 1.75 cm
∴ Area of 1 circular button = πr2
                                                 = 227×1.75×1.75 cm2
                                                 = 9.625 cm2
∴ Area of 64 such buttons = (64  × 9.625) cm2 = 616 cm2
Area of the remaining acrylic sheet = (Area of the acrylic sheet - Area of 64 circular buttons)
                                                                    = (816 - 616) cm2 = 200 cm2



Page No 253:

Answer:

Area of the rectangular ground = 90 m  × 32 m = (90  × 32) m2 = 2880 m2
Given:
Radius of the circular tank (r) = 14 m
∴ Area covered by the circular tank = πr2  = 227×14×14 m2
                                                               = 616 m2

∴ Remaining portion of the rectangular ground for turfing = (Area of the rectangular ground - Area covered by the circular tank)
                                                                                          = (2880 - 616) m2 = 2264 m2
Rate of turfing = Rs 50 per sq. metre
∴ Total cost of turfing the remaining ground = Rs (50  × 2264) = Rs 1,13,200

Page No 253:

Answer:

Area of each of the four quadrants is equal to each other with radius 7 cm.

Area of the square ABCD = (Side)2 = (14 cm)2 = 196 cm2
Sum of the areas of the four quadrants = 4×14×227×7×7 cm2
                                                        = 154 cm2
∴ Area of the shaded portion  = Area of square ABCD - Areas of the four quadrants
                                          = (196 - 154) cm2
                                          = 42 cm2

Page No 253:

Answer:

Let ABCD be the rectangular field.

Here, AB = 60 m
BC = 40 m

Let the horse be tethered to corner A by a 14 m long rope.

Then, it can graze through a quadrant of a circle of radius 14 m.
∴ Required area of the field = 14×227×14×14 m2 = 154 m2
Hence, horse can graze 154 m2 area of the rectangular field.

Page No 253:

Answer:

Diameter of the big circle = 21 cm
Radius = 212 cm = 10.5 cm
∴ Area of the bigger circle = πr2 = 227×10.5×10.5 cm2
                                          = 346.5 cm2


Diameter of circle I  = 23 of the diameter of the bigger circle
                                 = 23 of 21 cm = 23×21 cm = 14 cm
Radius of circle I (r1) = 142 cm = 7 cm
∴ Area of circle I = πr12 = 227×7×7 cm2
                                         = 154 cm2

Diameter of circle II  = 13 of the diameter of the bigger circle
                                   = 13 of 21 cm = 13×21 cm = 7 cm
Radius of circle II (r2) = 72 cm = 3.5 cm
∴ Area of circle II = πr22 = 227×3.5×3.5 cm2
                                          = 38.5 cm2

∴ Area of the shaded portion = {Area of the bigger circle - (Sum of the areas of circle I and II)}
                                      = {346.5 - (154 + 38.5)} cm2
                                      = {346.5 - 192.5} cm2
                                            
  = 154 cm2
Hence, the area of the shaded portion is 154 cm2

Page No 253:

Answer:



Let ABCD be the rectangular plot of land that measures 8 m by 6 m.
∴ Area of the plot = (8 m  × 6 m) = 48 m2
Area of the four flower beds = 4×14×227×2×2 m2 = 887 m2
Area of the circular flower bed in the middle of the plot = πr2
                                                                                       = 227×2×2 m2 = 887 m2

Area of the remaining part = 48-887+887 m2
                                           = 48-1767 m2
                                           = 336-1767 m2 = 1607 m2 = 22.86 m2
∴ Required area of the remaining plot = 22.86 m2

Page No 253:

Answer:

(c) 192 cm2

Let ABCD be the rectangular plot.
Then, AB = 16 cm
          AC = 20 cm

Let BC = x cm
From right triangle ABC:
AC2 = AB2 + BC2
⇒ (20)2 = (16)2 + x2
x2 = (20)2 - (16)2 ⇒ {400 - 256} = 144
x = 144 = 12
∴ BC = 12 cm
∴ Area of the plot = (16 × 12) cm2 = 192 cm2

Page No 253:

Answer:

(b) 72 cm2

Given:
Diagonal of the square = 12 cm
∴ Area of the square = 12×Diagonal2 sq. units.
                                  = 12×122 cm2
                                  = 72 cm2



Page No 254:

Answer:

(b) 20 cm

Area of the square = 12×Diagonal2 sq. units.
Area of the square field = 200 cm2

Diagonal of a square = 2×Area of the square 
                                  = 2×200 cm = 400 cm = 20 cm
∴ Length of the diagonal of the square = 20 cm

Page No 254:

Answer:

(a) 100 m

Area of the square = 12×Diagonal2sq. units.
Given:
Area of square field = 0.5 hectare
                                 = 0.5×10000m2                     [since 1 hectare = 10000 m2]
                                 = 5000 m2                  

Diagonal of a square = 2×Area of the square 
                                  = 2×5000m = 100 m
Hence, the length of the diagonal of a square field is 100 m.

Page No 254:

Answer:

(c) 90 m

Let the breadth of the rectangular field be x m.
Length = 3x m
Perimeter of the rectangular field = 2(l + b)
⇒ 240 = 2( x + 3x)
⇒ 240 = 2(4x)
⇒ 240 = 8x     ⇒ x = 2408=30
∴ Length of the field = 3x = (3 × 30) m = 90 m

Page No 254:

Answer:

(d) 56.25%

Let the side of the square be a cm.
Area of the square = (a)2 cm2
Increased side = (a + 25% of a) cm
                        = a+25100a cm = a+14acm=54a cm
Area of the square = 54a2cm2=2516a2 cm2
Increase in the area = 2516a2-a2 cm2= 25a2-16a216 cm2 = 9a216 cm2
% increase in the area = Increased areaOld area×100
                              = 916a2a2×100 = 9×10016=56.25

Page No 254:

Answer:

(b) 1:2

Let the side of the square be a.
Length of its diagonal = 2a
∴ Required ratio = a22a2=a22a2=12=1:2

Page No 254:

Answer:

(c) A > B

We know that a square encloses more area even though its perimeter is the same as that of the rectangle.

∴ Area of a square  > Area of a rectangle

Page No 254:

Answer:

(b) 13500 m2

Let the length of the rectangular field be 5x.
Breadth = 3x
Perimeter of the field = 2(l + b) = 480 m          (given)
⇒ 480 = 2(5x + 3x)  ⇒ 480 = 16x
x48016 = 30
∴ Length = 5x = (5 × 30) = 150 m
Breadth = 3x = (3 × 30) = 90 m
∴ Area of the rectangular park = 150 m × 90 m = 13500 m2

Page No 254:

Answer:

(a) 6 m

Total cost of carpeting = Rs 6000
Rate of carpeting = Rs 50 per m
∴ Length of the carpet = 600050 m = 120 m
∴ Area of the carpet = 120×75100 m2 =  90 m2     [since 75 cm = 75100 m]
Area of the floor = Area of the carpet = 90 m2
∴ Width of the room = AreaLength=9015 m=6 m

Page No 254:

Answer:

(a) 84 cm2

Let a = 13 cm, b = 14 cm and c = 15 cm
Then, s = a+b+c2 = 13+14+152 cm = 21 cm
∴  Area of the triangle = ss-as-bs-c sq. units
                               = 2121-1321-1421-15 cm2
                               =  21×8×7×6 cm2
                               = 3×7×2×2×2×7×2×3 cm2
                               = (2 × 2 × 3 × 7) cm2
                               = 84 cm2

Page No 254:

Answer:

(b) 48 m2

Base = 12 m
Height = 8 m
Area of the triangle = 12× Base× Height sq. units
                          = 12×12×8 m2
                          = 48 m2

Page No 254:

Answer:

(b) 4 cm

Area of the equilateral triangle = 43 cm2

We know:
Area of an equilateral triangle = 34side2 sq. units
∴ Side of the equilateral triangle = 4×Area3 cm
                                                    =   4×433cm = 4×4 cm = 16cm = 4 cm

Page No 254:

Answer:

(c) 163 cm2

It is given that one side of an equilateral triangle is 8 cm.
∴ Area of the equilateral triangle = 34Side2 sq. units
                                                      =  3482 cm2
                                                      = 34×64 cm2 = 163 cm2

Page No 254:

Answer:

(b) 23 cm2

Let ΔABC be an equilateral triangle with one side of the length a cm.
Diagonal of an equilateral triangle = 32a cm
32a=6
a = 6×23=3×2×23=22 cm
Area of the equilateral triangle = 34a2
                                                 = 34222 cm2 = 34×8 cm2 = 23 cm2

Page No 254:

Answer:

(b) 72 cm2

Base of the parallelogram = 16 cm
Height of the parallelogram = 4.5 cm
∴ Area of the parallelogram = Base × Height
                                              = (16 × 4.5) cm2  = 72 cm2



Page No 255:

Answer:

(b) 216 cm2

Length of one diagonal = 24 cm
 Length of the other diagonal = 18 cm
     ∴ Area of the rhombus = 12 × (Product of the diagonals)
                                           = 12×24×18 cm2 = 216 cm2

Page No 255:

Answer:

(c) 154 cm2

Let the radius of the circle be r cm.
Circumference = 2πr

(Circumference) - (Radius) = 37
2πr -r=37
r2π-1=37
r = 372π-1 = 372×227-1=37447-1=3744-77=37×737=7
∴ Radius of the given circle is 7 cm.
∴ Area = πr2 = 227×7×7 cm2 = 154 cm2

Page No 255:

Answer:

(c) 54 m2

Given:
Perimeter of the floor = 2(l + b) = 18 m      
Height of the room = 3 m        
                    
∴ Area of the four walls = {2(l + b) × h}
                                        = Perimeter × Height
                                        = 18 m × 3 m = 54 m2

Page No 255:

Answer:

(a) 200 m

Area of the floor of a room = 14 m × 9 m = 126 m2

Width of the carpet = 63 cm = 0.63 m                (since 100 cm = 1 m)

∴ Required length of the carpet = Area of the floor of a roomWidth of the carpet
                                                = 1260.63 m=200 m

Page No 255:

Answer:

(c) 120 cm2

Let the length of the rectangle be x cm and the breadth be y cm.
Area of the rectangle = xy cm2
Perimeter of the rectangle = 2( x + y) = 46 cm          (given)
⇒ 2( x + y) = 46
⇒ ( x + y) = 462 cm = 23 cm

Diagonal of the rectangle = x2+y2 = 17 cm
⇒  x2+y2 = 17

Squaring both the sides, we get:
x2 + y2 = (17)2
x2 + y2 = 289

 Now, (x2 + y2) = ( x + y)2 - 2xy
⇒ 2xy = ( x + y)2 - (x2 + y2)
           = (23)2 - 289
           = 529 - 289 = 240
xy = 2402 cm2 = 120 cm2

Page No 255:

Answer:

(b) 3:1

Let a side of the first square be a cm and that of the second square be b cm.
Then, their areas will be a2 and b2, respectively.
Their perimeters will be 4a and 4b, respectively.

According to the question:
a2b2=91ab2=91=312ab=31

∴ Required ratio of the perimeters = 4a4b=4×34×1=31= 3:1

Page No 255:

Answer:

(d) 4:1

Let the diagonals be 2d and d.
Area of the square = 12×Diagonal2 sq. units
 Required ratio = A1A2=122d212d2=4d2d2=41=4:1

Page No 255:

Answer:

(c) 49 m

Let the width of the rectangle be x m.

Given:
Area of the rectangle = Area of the square
⇒ Length × Width = Side × Side
⇒ (144 × x) = 84 × 84
∴ Width (x) = 84×84144 m = 49 m
Hence, width of the rectangle is 49 m.

Page No 255:

Answer:

(d) 4:3

Let one side of the square and that of an equilateral triangle be the same, i.e. a units.
Then, Area of the square = (Side)2 = (a)2
Area of the equilateral triangle = 34Side2 = 34a2
∴ Required ratio = a234a2=43=4:3

Page No 255:

Answer:

(a) π:1

Let the side of the square be x cm and the radius of the circle be r cm.
Area of the square = Area of the circle
⇒ (x)2 = πr2
∴ Side of the square (x) = πr
Required ratio = Side of the squareRadius of the circle
                        = xr=πrr=π1=π:1

Page No 255:

Answer:

(b) 4934 cm2

Let the radius of the circle be r cm.
Then, its area = πr2 cm2
∴  πr2 = 154
227×r×r=154
r2154×722 = 49
r49cm = 7 cm

Side of the equilateral triangle =  Radius of the circle
                                                 = 7 cm
∴ Area of the equilateral triangle = 34side2 sq. units

                                                     = 3472 cm2

                                                     = 4934 cm2

Page No 255:

Answer:

(c) 12 cm

Area of the rhombus = 12 × (Product of the diagonals)
Given:
Length of one diagonal = 6 cm
Area of the rhombus = 36 cm2

∴ Length of the other diagonal = 36×26 cm = 12 cm



Page No 256:

Answer:

(d) 24 cm

Let the length of the shorter diagonal of the rhombus be x cm.
∴ Longer diagonal = 2x

Area of the rhombus = 12× (Product of its diagonals)
⇒ 144 = 12×x×2x

⇒ 144 = 2x22 = x2

x = 144cm = 12 cm

∴ Length of the longer diagonal = 2x
                                                     = (2 × 12) cm
                                                     = 24 cm

Page No 256:

Answer:

(c) 17.60 m

Let the radius of the circle be r m.
Area = πr2 m2
∴ πr2 = 24.64
227×r×r = 24.64
r2 = 24.64×722 = 7.84
r = 7.84 = 2.8 m
⇒  Circumference of the circle = 2πr m
                                                  = 2×227×2.8 m = 17.60 m

Page No 256:

Answer:

(c) 3 cm

Suppose the radius of the original circle is r cm.
Area of the original circle = πr2

Radius of the circle = (r +1) cm
According to the question:
πr+12 =πr2+22
πr2+1+2r=πr2+22
πr2+π+2πr=πr2+22
π+2πr=22      [cancel πr2 from both the sides of the equation]
π1+2r=22
1+2r=22π=22×722=7
⇒ 2r = 7 -1 = 6
r = 62 cm = 3 cm
∴ Original radius of the circle = 3 cm

Page No 256:

Answer:

(c) 1000

Radius of the wheel = 1.75 m
Circumference of the wheel = 2πr
                                              = 2×227×1.75cm = (2 × 22 × 0.25) m = 11 m
                                                                                                     
Distance covered by the wheel in 1 revolution is 11 m.
Now, 11 m is covered by the car in 1 revolution.
(11 × 1000) m will be covered by the car in 1×111×11×1000 revolutions, i.e. 1000 revolutions.
∴ Required number of revolutions = 1000



Page No 257:

Answer:

We know that all the angles of a rectangle are 90° and the diagonal divides the rectangle into two right angled triangles.
So, one side of the triangle will be 48 m and the diagonal, which is 50 m, will be the hypotenuse.
According to Pythagoras theorem:
(Hypotenuse)2 = (Base)2 + (Perpendicular)2
Perpendicular = Hypotenuse2-(Base)2
Perpendicular = 502-482=2500-2304=196=14 m
∴ Other side of the rectangular plot = 14 m

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

Page No 257:

Answer:

Length = 9 m
Breadth = 8 m
Height = 6.5 m

Area of the four walls = {2(l + b) × h} sq. units
                          = {2(9 + 8) × 6.5} m2 = {34 × 6.5} m2 = 221 m2
Area of one door = (2 × 1.5) m2 = 3 m2
Area of one window = (1.5 × 1) m2 = 1.5 m2
 ∴ Area of four windows = (4 × 1.5) m2 = 6 m2
Total area of one door and four windows = (3 + 6) m2
                                                           = 9 m2
Area to be painted = (221 - 9) m2 = 212 m2
Rate of painting = Rs 50 per m2
Total cost of painting = Rs ( 212 × 50) = Rs 10600

Page No 257:

Answer:

Given that the diagonal of a square is 64 cm.

Area of the square = 12×diagonal2 sq. units.
                               = 12×642 cm2 =12×4096 cm2 = 2048 cm2
∴ Area of the square = 2048 cm2

Page No 257:

Answer:


Let ABCD be the square lawn and PQRS be the outer boundary of the square path.

Let one side of the lawn (AB) be x m.
Area of the square lawn = x2
Length PQ = (x m + 2 m + 2 m) = (x + 4) m
∴ Area of PQRS = (x + 4)2 = (x2 + 8x + 16) m2

Now, Area of the path = Area of PQRS − Area of the square lawn
⇒ 136 = x2 + 8x + 16 x2
⇒ 136 = 8x + 16
⇒ 136 16 = 8x
⇒ 120 = 8x
x = 120 ÷ 8 = 15
∴ Side of the lawn = 15 m

∴ Area of the lawn = (Side)2 = (15 m)2 = 225 m2

Page No 257:

Answer:

Let ABCD be the rectangular park.
EFGH and IJKL are the two rectangular roads with width 2 m.

Length of the rectangular park AD = 30 cm 
 Breadth of the rectangular park CD = 20 cm
Area of the road EFGH = 30 m × 2 m = 60 m2
Area of the road IJKL = 20 m × 2 m = 40 m2
Clearly, area of MNOP is common to the two roads.
∴ Area of MNOP = 2 m × 2 m = 4 m2

∴ Area of the roads = Area (EFGH) + Area (IJKL) − Area (MNOP)
                            = (60  + 40 ) m2 − 4 m2 = 96 m2

Page No 257:

Answer:

Let ABCD be the rhombus whose diagonals intersect at O.

Then, AB = 13 cm
        AC = 24 cm
The diagonals of a rhombus bisect each other at right angles.
Therefore, ΔAOB is a right-angled triangle, right angled at O, such that:
OA = 12AC = 12 cm
AB = 13 cm

By Pythagoras theorem:
(AB)2 = (OA)2 + (OB)2
⇒ (13)2 = (12)2 + (OB)2
⇒ (OB)2 = (13)2 − (12)2
⇒ (OB)2 = 169 − 144 = 25
⇒ (OB)2 = (25)2
⇒ OB = 5 cm
∴ BD = 2 × OB = 2 × 5 cm = 10 cm

∴ Area of the rhombus ABCD = 12 × AC × BD cm2
                                              = 12×24×10 cm2 = 120 cm2

Page No 257:

Answer:

Let the base of the parallelogram be x m.
Then, the altitude of the parallelogram will be 2x m.
It is given that the area of the parallelogram is 338 m2.

Area of a parallelogram =  Base × Altitude
                        ∴ 338 m2x × 2x
                            338 m2 = 2x2
x2 = 3382 m2 = 169 m2
x2 = 169 m2
x = 13 m
∴ Base = x m = 13 m
Altitude = 2x m = (2 × 13)m = 26 m

Page No 257:

Answer:

Consider ΔABC.
Here, ∠B = 90°
          AB = 24 cm
          AC = 25 cm

Now, AB2 + BC2 = AC2
⇒ BC2 = AC2 - AB2
               = (252 - 242)
               = (625 - 576)
               = 49
⇒ BC = 49cm = 7 cm
Area of ΔABC = 12×BC×AB sq. units
                         = 12×7×24 cm2 = 84 cm2
Hence, area of the right angled triangle is 84 cm2.

Page No 257:

Answer:

Radius of the wheel = 35 cm
Circumference of the wheel = 2πr
                                              = 2×227×35cm = (44 × 5) cm = 220 cm
                                                                                                       = 220100 m = 115 m
Distance covered by the wheel in 1 revolution = 115 m
Now, 115 m is covered by the car in 1 revolution.
Thus, (33 × 1000) m will be covered by the car in 1×511×33×1000 revolutions, i.e. 15000 revolutions.
∴ Required number of revolutions = 15000

Page No 257:

Answer:

Let the radius of the circle be r cm.
∴ Area = πr2 cm2
∴ πr2 = 616
227×r×r = 616
r2 = 616×722 = 196
r = 196 = 14 cm
Hence, the radius of the given circle is 14 cm.

Page No 257:

Answer:

(a) 14 cm

Let the radius of the circle be r cm.
Then, its area will be πr2 cm2.
∴ πr2 = 154
227×r×r = 154
r2 = 154×722 = 49
r = 49 = 7 cm
∴ Diameter of the circle = 2r = (2 × 7) cm = 14 cm

Page No 257:

Answer:

(b) 154 cm2

Let the radius of the circle be r cm.
Circumference = 2πrcm
2πr = 44
2×227×r=44
r = 44×72×22 = 7 cm
∴ Area of the circle = πr2
                                 = 227×7×7 cm2 = 154 cm2

Page No 257:

Answer:

(c) 98 cm2

Given that the diagonal of a square is 14 cm.

Area of a square = 12×Diagonal2sq. units.
                               = 12×142 cm2 =12×196 cm2 = 98 cm2
Hence, area of the square is 98 cm2.

Page No 257:

Answer:

(b) 10 cm

Given that the area of the square is 50 cm2.
We know:
Area of a square = 12×Diagonal2sq. units
∴ Diagonal of the square = 2×Area of the square = 2×50cm = 100cm = 10 cm
Hence, the diagonal of the square is 10 cm.

Page No 257:

Answer:

(a) 192 m2

Let the length of the rectangular park be 4x.
∴ Breadth = 3x
Perimeter of the park = 2(l + b) = 56 m (given)
⇒ 56 = 2(4x + 3x)  
⇒ 56 = 14x
x = 5614= 4
Length = 4x = (4 × 4) = 16 m
Breadth = 3x = (3 × 4) = 12 m
∴ Area of the rectangular park = 16 m × 12 m = 192 m2

Page No 257:

Answer:

(a) 84 cm2

Let a = 13 cm, b = 14 cm and c = 15 cm
 s = a+b+c2 = 13+14+152cm = 21 cm
∴  Area of the triangle = ss-as-bs-c sq. units
                               = 2121-1321-1421-15 cm2
                               =  21×8×7×6 cm2
                               = 3×7×2×2×2×7×2×3 cm2
                               = (2 × 2 × 3 × 7) cm2 = 84 cm2

Page No 257:

Answer:

(a) 163 cm2

Given that each side of an equilateral triangle is 8 cm.
∴ Area of the equilateral triangle = 34side2 sq. units

                                                      =  3482 cm2

                                                      = 34×64 cm2 = 163 cm2

Page No 257:

Answer:

(b) 91 cm2

Base  = 14 cm
Height = 6.5 cm
∴ Area of the parallelogram = Base × Height
                                               = (14 × 6.5) cm2
                                               =  91 cm2

Page No 257:

Answer:

(b) 135 cm2

Area of the rhombus = 12 ×  (Product of the diagonals)
                                  = 12×18×15 cm2 = 135 cm2
Hence, the area of the rhombus is 135 cm2.



Page No 258:

Answer:

(i) If d1 and d2 be the diagonals of a rhombus, then its area is 12d1d2 sq. units.
      Area of a rhombus = 12× (Product of its diagonals)

(ii) If l, b and h are the length, breadth and height respectively of a room, then area of its 4 walls = 2h(l + b) sq. units.
     
(iii) 1 hectare = (10000) m2
      (since 1 hectometre = 100 m)
        ∴1 hectare = (100 × 100) m2
(iv) 1 acre = 100 m2
(v) If each side of a triangle is a cm, then its area = 34a2 cm2.
    Area of equilateral triangle with side a = 34a2sq. units.

Page No 258:

Answer:

(i) F
   Area of a triangle = 12×Base×Height

(ii) T
      Area of a parallelogram = Base × Height

(iii) F
        Area of a circle = πr2

(iv) T
       Circumference of a circle = 2πr2πr2



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