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

Exercise 20B

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Question 1:

A rectangular grassy plot is 75 m long and 60 m broad. If has path of width 2 m all around it on the inside. Find the area of the path and cost and of constructing it at Rs 125 per m².

Answer 1:

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 $\times$ 60) m² = 4500 m²
Width of the path = 2 m
∴ AB = (75 - 2 $\times$ 2) m = (75 - 4) m =71 m
 AD = (60 - 2 $\times$ 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 $\times$ 125) = Rs 65,500

Question 2:

A rectangular plot of land measures 95 m by 72 m. Inside the plot, a path of uniform width of 3.5 m is to be constructed all around. The rest of the plot is to be laid with grass. Find the total expenses involved in constructing the path at Rs 80 per m² and laying the grass at Rs 40 per m².

Answer 2:

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 $\times$ 72) m² = 6,840 m²
Width of the path = 3.5 m
∴ AB = (95 - 2 $\times$ 3.5) m = (95 - 7) m = 88 m
AD = (72 - 2 $\times$ 3.5) m = (72 - 7) m = 65 m
Area of the grassy rectangle plot ABCD = (88 $\times$ 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 $\times$ 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 $\times$ 40) = Rs. 2,28,800
∴ Total expenses involved = Rs. ( 89,600 + 2,28,800) = Rs. 3,18,400

Question 3:

A saree is 5 m long and 1.3 m wide. A border of width 25 cm is printed along its sides. Find the cost of printing the border at Rs 1 per 10 cm².

Answer 3:

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 $\times$ 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 $\times$ .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 $\left(\frac{1\times 29000}{10}\right)$
                                                       = Rs 2900

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Question 4:

A rectangular grassy lawn measuring 38 m by 25 m has been surrounded externally by a 2.5-m-wide path. Calculate the cost of gravelling the path at the rate of Rs 120 per m².

Answer 4:

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

∴ Area of EFGH = 38 m $\times$  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 $\times$ 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 $\times$ 340)
                                                    = Rs 40800

Question 5:

A room 9.5 m long and 6 m wide is surrounded by a 1.25-m-long verandah. Calculate cost of cementing the floor of this verandah at Rs 80 per m².

Answer 5:

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  $\times$  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 $\times$ 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 $\times$ 45)
                                                            = Rs 3600

Question 6:

Each side of a square flower bed is 2 m 80 cm long. It is extended by digging a strip 30 cm wide all around it. Find the area of the enlarged flower bed and also the increase in the area of the flower bed.

Answer 6:

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

Question 7:

The length and breadth of a park in the ratio 2 : 1 and its perimeter is 240 m. A path 2 m wide runs inside it, along its boundary. Find the cost of paving the path at Rs 80 per m².

Answer 7:

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 = $\left(\frac{240}{6}\right)$ m =40 m
∴ Length of the park = 2x = (2 $\times$ 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 $\times$ 40) m² = 3200 m²
Width of the path = 2 m
∴ AB = (80 - 2 $\times$ 2) m = (80 - 4) m =76 m
 AD = (40 - 2 $\times$ 2) m = (40 - 4) m = 36 m
Area of the rectangle ABCD = (76 $\times$ 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 $\times$ 80) = Rs. 37,120

Question 8:

A school has a hall which is 22 m long and 15.5 m broad. A carpet is laid inside the hall leaving all around a margin of 75 cm from the walls. Find the area of the carpet and the area of the strip left uncovered. If the width of the carpet is 82 cm, find its cost at the rate of Rs 60 per m.

Answer 8:

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

∴ Area of the school hall PQRS = 22 m $\times$ 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 $\times$ 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 $\times$ 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 $\times$ 350 = Rs 21000

Question 9:

A square lawn is surrounded by a path 2.5 m wide. If the area of the path is 165 m², find the area of the lawn.

Answer 9:

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

Question 10:

The length and breadth of a rectangular park are in the ratio 5 : 2. A 2.5-m-wide path running all around the outside of the aprk has an area of  305 m². Find the dimensions of the park.

Answer 10:

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 $\times$ 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) $\times$ (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 = 5 $\times$ 8 = 40 m
Breadth of the park = 2x = 2 $\times$ 8 = 16 m

Question 11:

A rectangular lawn 70 m by 50 m has two roads, each 5 m wide, running through its middle, one parallel to its length and the other parallel to its breadth. Find the cost of constructing the roads at Rs 120 per m².

Answer 11:

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 $\times$ Breadth = 70 m $\times$ 50 m = 3500 m²
Area of road EFGH = 70 m $\times$ 5 m = 350 m²
Area of road IJKL = 50 m $\times$ 5 m = 250 m²

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

∴ Area of MNOP = 5 m $\times$ 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

Question 12:

A 115-m-long and 64-m-broad lawn has two roads at right angles, one 2 m wide, running parallel to its length, and the other 2.5 m wide, running parallel to its breadth. Find the cost of gravelling the roads at Rs 60 per m².

Answer 12:

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 $\times$ 64 m = 7360 m²
Area of the road PQRS = 115 m $\times$ 2 m = 230 m²
Area of the road KLMN = 64 m $\times$ 2.5 m = 160 m²

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

∴ Area of EFGH = 2 m $\times$ 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 $\times$ 60)
                                                     = Rs 23,100

Question 13:

A rectangular field is 50 m by 40 m. It has two roads through its centre, running parallel to its sides. The width of the longer and the shorter roads are 2 m and 2.5-m-respectively. Find the area of the roads and the area of the remaining portion of the field.

Answer 13:

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 $\times$ 40 m = 2000 m²
Area of road KLMN = 40 m $\times$ 2.5 m = 100 m²
Area of road PQRS = 50 m $\times$ 2 m = 100 m²

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

∴ Area of EFGH = 2.5 m $\times$ 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

Question 14:

Calculate the area of the shaded region in each of the figures given below:

Answer 14:

(i) Complete the rectangle as shown below:
    
    Area of the shaded region = [Area of rectangle ABCD - Area of rectangle EFGH] sq. units
                                              = [(43 m $\times$ 27 m) - {(43 - 2 $\times$ 1.5) m x (27 - 1 $\times$ 2) m}]
                                              = [(43 m $\times$ 27 m) - {40 m $\times$ 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 $\times$ 40) - {(40 $\times$ 2) + (40 $\times$ 3) - (2 $\times$ 3)}] m²
                                               = [1600 - {(80 + 120 - 6)] m²
                                               = [1600 - 194] m²
                                               = 1406 m²

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Question 15:

Calculate the area of the shaded region in each of the figures given below.
Fig. (ii) has uniform width of 3 cm and it is given that AB = CD.

Answer 15:

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

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

Question 16:

In the given figure, all steps are 0.5 m high. Find the area of the shaded region.

Answer 16:

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 $\times$ 0.5) + (2.5 $\times$ 0.5) + ( 1.5 $\times$ 0.5) + (0.5 $\times$ 0.5)] m²
                                        = [1.75 + 1.25 + 0.75 + 0.25] m²
                                        = 4 m²