RD Sharma Solutions Class 7 Mensuration I Exercise 20.2

RD Sharma Class 7 Solutions Chapter 20 Ex 20.2 PDF Free Download

RD Sharma Solutions Class 7 Chapter 20 Exercise 20.2

Exercise 20.2

Q1: A rectangular grassy lawn measuring 40 m by 25 m is to be surrounded externally by a path which is 2 m wide. Calculate the cost of leveling the path at the rate of Rs 8.25 per square metre.

 Solution: Let PQRS be the rectangular grassy lawn which has l = 40 m and b = 25 m.

RD Sharma Solutions Class 7 Maths Chapter 10 Exercise 20.2 Qs 1

Area of lawn PQRS = 40 m x 25 m = 1000 m2

Let WXYZ be the external boundaries of the path.

So, Length WX = (40 + 2 + 2 ) m = 44 m

Breadth YZ = ( 25 + 2 + 2 ) m = 29 m

Area of WXYZ = 44 m x 29 m = 1276 m2

Now, Area of the path = Area of WXYZ – Area of the lawn PQRS

= 1276 m2 – 1000 m2 = 276 m2

Rate of leveling the path = Rs. 8.25 per m2

Cost of leveling the path = Rs.( 8.25 x 276) = Rs. 2277

Q2: One metre wide path is built inside a square park of side 30 m along its sides. The remaining part of the park is covered by grass. If the total cost of covering by grass is Rs 1176, find the rate per square metre at which the park is covered by the grass.

Solution: Let the side of square park (s) = 30 m

RD Sharma Solutions Class 7 Maths Chapter 10 Exercise 20.2 Qs 2

Area of the square garden including the path = s2= (30)2 = 900 m2

From the figure, the side of the side of the square garden without including path = 28 m

Area of the square garden not including the path = (28)² = 784 m2

Total cost of covering the park with grass = Area of the park covering with green grass x Rate per square metre

1176 = 784 x Rate per square metre

Rate per square metre at which the park is covered with grass = Rs. (1176 ÷ 784 ) = Rs. 1.50

Q3: Through a rectangular field of sides 90 m x 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the field. If the width of the roads is 3 m, find the total area covered by the two roads.

Solution: Let us consider the rectangular sheet

l = 90 m, b= 60 m

Area of the rectangular field = 90 m x 60 m = 5400 m2

RD Sharma Solutions Class 7 Maths Chapter 10 Exercise 20.2 Qs 3

Width of the road = 3 m

Considering the rectangle road PQRS

l = 90 m, b = 3 m

Area of the road PQRS = 90 m x 3 m = 270 m2

Considering the rectangle road ABCD

l = 60 m, b = 3 m

Area of the road ABCD = 60 m x 3 m = 180 m2

Finding the area covered by the two roads

Clearly, area of KLMN is common to the two roads.

Thus, area of KLMN = 3 m x 3 m = 9 m2

Hence, Area of the roads = Area (PQRS) + Area (ABCD) – Area (KLMN)

= 270 + 180 – 9 = 441 m2

Q4: From a rectangular sheet of tin, of size 100 cm by 80 cm, are cut four squares of side 10 cm from each corner. Find the area of the remaining sheet.

Solution: Let us consider the rectangular sheet of tin

l = 100 cm, b = 80 cm

Area of the rectangular sheet of tin = 100 cm x 80 cm = 8000 cm2

RD Sharma Solutions Class 7 Maths Chapter 10 Exercise 20.2 Qs 4

Side of the square at the corner of the sheet = 10 cm

Area of one square at the corner of the sheet = (10 cm)2 = 100 cm2

Area of 4 squares which are cut from the sheet = 4 x 100 cm2 = 400 cm2

Hence, Area of the remaining sheet of tin = Area of the rectangular sheet – Area of the 4 squares

Area of the remaining sheet of tin = (8000 – 400) cm2 = 7600 cm2

Q 5:  A painting 8 cm long and 5 cm wide is painted on a cardboard such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.

Solution: Let us consider the cardboard

l = 8 cm and b = 5 cm

Area of the cardboard including the margin = 8 cm x 5 cm = 40 cm2

RD Sharma Solutions Class 7 Maths Chapter 10 Exercise 20.2 Qs 5

From the figure, it can be observed that,

New length of the painting when the margin is not included = 8 cm – (1.5 cm + 1.5 cm)

= 8 – 3

= 5 cm

New breadth of the painting when the margin is not included = 5 cm – (1.5 cm + 1.5 cm)

= 5 – 3

= 2 cm

Area of the painting (not including the margin) = 5 cm x 2 cm = 10 cm2

Hence, Area of the margin = Area of the cardboard including the margin – Area of the painting

= 40 – 10

= 30 cm2

Q6: Rakesh has a rectangular field of length 80 m and breadth 60 m. In it, he wants to make a garden 10 m long and 4 m broad at one of the corners and at another corner, he wants to grow flowers in two floor-beds each of size 4 m by 1.5 m. In the remaining part of the field, he wants to apply manures. Find the cost of applying the manures at the rate of Rs 300 per area.

 Solution: Let us consider the rectangular field

l = 80 m, b = 60 m

Area of the rectangular field = 80 m x 60 = 4800 m2

RD Sharma Solutions Class 7 Maths Chapter 10 Exercise 20.2 Qs 6

From the figure, we can easily find out the

Area of the garden = 10 m x 4 m = 40 m2

Area of one flower bed = 4 m x 1.5 m = 6 m2

Thus, Area of two flower beds = 2 x 6 m2 = 12 m2

Remaining area of the field for applying manure = Area of the rectangular field – (Area of the garden + Area of the two flower beds)

Remaining area of the field for applying manure = 4800 m2 – (40 + 12) m2

= (4800 – 52 ) m2 = 4748 m2 = 47.48 are [Since 100 m2 = 1 are]

So, cost of applying manure at the rate of Rs. 300 per are will be Rs. (300 x 47.48) = Rs. 14244

Q7: 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.

Solution: It is given that 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

RD Sharma Solutions Class 7 Maths Chapter 10 Exercise 20.2 Qs 7

Side of the digging strip = 30 cm

= 0.3 m [since 100 cm = 1 m ]

As we can see in the figure;

Side of the flower bed with the digging strip = 2.80 m + 0.3 m + 0.3 m

= 3.4 m

Area of the enlarged flower bed with the digging strip = (Side)2 = (3.4)2 = 11.56 m2

Thus, Increase in the area of the flower bed = 11.56 m2 – 7.84 m2 = 3.72 m2

Q8: A room 5 m long and 4 m wide is surrounded by a verandah. If the verandah occupies an area of 22 m2, find the width of the varandah.

Solution: Let us consider the rectangular room

AB = 5 m and BC = 4 m

Area of the room = 5 m x 4 m = 20 m2

RD Sharma Solutions Class 7 Maths Chapter 10 Exercise 20.2 Qs 8

Considering the verandah

Let the width of the verandah be w meter.

Length of the verandah, PQ = (5 + w + w) = (5 + 2w) m

Breadth of the verandah, QR = ( 4 + w + w) = (4 + 2w) m

Area of verandah PQRS = (5 + 2w) x (4 + 2w) = (4w2 + 18w + 20 ) m2

Area of verandah = Area of PQRS – Area of ABCD

=> 22 = 4w2 + 18w + 20 – 20

22 = 4w2 + 18w

11 = 2w2 + 9w

2w2 + 9w – 11 = 0

2w2 + 11w – 2w – 11 =0

w(2w+11)-1(2w+11)=0

(w – 1)(2w+11)= 0

When w – 1 = 0, w = 1

When 2w + 11 = 0, w = -11/2

The width cannot be a negative value. So, width of the verandah = w = 1 m.

Q9: A square lawn has a 2 m wide path surrounding it. If the area of the path is 136 m2 , find the area of the lawn.

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

Let side of the lawn AB be s meter.

RD Sharma Solutions Class 7 Maths Chapter 10 Exercise 20.2 Qs 9

Length PQ = (s + 2 + 2) = (s + 4) m

Area of PQRS = (s + 4)2 = (s2 + 8s + 16) m2

Now, Area of the path = Area of PQRS – Area of the square lawn

136 = s2 + 8s + 16 – s2

136 = 8s + 16

136 – 16 = 8s

120 = 8s

x = 120/8 = 15

Side of the lawn = 15 m

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

Q 10: A poster of size 10 cm by 8 cm is pasted on a sheet of cardboard such that there is a margin of width 1.75 cm along each side of the poster. Find (i) the total area of the margin (ii) the cost of the cardboard used at the rate of Re 0.60 per c m2 .

Solution: Let us consider the poster

Length of poster = 10 cm and breadth of poster = 8 cm

Area of the poster = Length x Breadth = 10 cm x 8 cm = 80 cm2

RD Sharma Solutions Class 7 Maths Chapter 10 Exercise 20.2 Qs 10

Considering the cardboard

From the figure, we can see that

Length of the cardboard when the margin is included = 10 cm + 1.75 cm + 1.75 cm = 13.5 cm

Breadth of the cardboard when the margin is included = 8 cm + 1.75 cm + 1.75 cm = 11.5 cm

Area of the cardboard = Length x Breadth = 13.5 cm x 11.5 cm = 155.25 c m2

Hence,

(i) Area of the margin = Area of cardboard including the margin – Area of the poster

= 155.25 cm2 – 80 cm2

= 75.25 cm2

(ii) Cost of the cardboard = Area of cardboard x Rate of the cardboard Rs 0.60 per cm2

= Rs. (155.25 x 0.60)

= Rs. 93.15

Q11: A rectangular field is 50 m by 40 m. It has two roads through its centre, running parallel to its sides. The widths of the longer and shorter roads are 1.8 m and 2.5 m respectively. Find the area of the roads and the area of the remaining portion of the field.

Solution: 

Let ABCD be the rectangular field. The PQRS and KLMN are two rectangular roads with width 1.8 m and 2.5 m, respectively.

RD Sharma Solutions Class 7 Maths Chapter 10 Exercise 20.2 Qs 11

Let us consider the rectangle ABCD

Length = 50 cm and breadth = 40 m

Area of the rectangular field ABCD = 50 m x 40 m = 2000 m2

Considering the road KLMN

l = 40 m, b = 2.5 m

Area of the road KLMN = 40 m x 2.5 m = 100 m2

Considering the road PQRS

l = 50 m, b = 1.8 m

Area of the road PQRS = 50 m x 1.8 m = 90 m2

Finding the area of the road

Clearly area of EFGH is common to the two roads.

Thus, Area of EFGH = 2.5 m x 1.8 m = 4.5 m2

Hence, Area of the roads = Area (KLMN) + Area (PQRS) – Area (EFGH)

= (100 m2 + 90 m2) – 4.5 m2

= 185.5 m2

Finding area of the remaining portion of the field

Area of the remaining portion of the field = Area of the rectangular field ABCD – Area of the roads

= (2000 – 185.5) m2

= 1814.5 m2

Q12: There is a rectangular field of size 94 m x 32 m. Three roads each of 2 m width pass through the field such that two roads are parallel to the breadth of the field and the third is parallel to the length. Calculate: (i) area of the field covered by the three roads (ii) area of the field not covered by the roads.

Solution: Let ABCD be the rectangular field.

RD Sharma Solutions Class 7 Maths Chapter 10 Exercise 20.2 Qs 12

Here, Two roads which are parallel to the breadth of the field are KLMN and EFGH with width 2 m each. One road which is parallel to the length of the field PQRS with width 2 m.

Considering the rectangle ABCD

Length = 94 m and breadth = 32 m

Area of the rectangular field = Length x Breadth = 94 m x 32 m = 3008 m2

Considering the rectangle KLMN

l = 32 m and b = 2 m

Area of the road KLMN = 32 m x 2 m = 64 m2

Considering the rectangle EFGH

l = 32 m and b = 2 m

Area of the road EFGH = 32 m x 2 m = 64 m2

Considering the rectangle PQRS

l = 94 m and b = 2 m

Area of the road PQRS = 94 m x 2 m = 188 m2

Clearly area of TUVI and WXYZ is common to these three roads.

Thus, Area of TUVI = 2 m x 2 m = 4 m2

Area of WXYZ = 2 m x 2 m = 4 m2

Hence,

(i) Area of the field covered by the three roads = Area (KLMN) + Area (EFGH) + Area (PQRS) – {Area (TUVI) + Area (WXYZ)}

= [ 64+ 64 + 188 – (4 + 4 )] m2

= 316 m2 – 8 m2

= 308 m2

(ii) Area of the field not covered by the roads = Area of the rectangular field ABCD – Area of the field covered by the three roads

= 3008 m2 – 308 m2

= 2700 m2

Q13: 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 the cost at the rate of Rs 18 per metre.

Solution: Let PQRS be the school hall and ABCE is the carpet laid inside the hall.

Considering the school hall, PQRS

l = 22 m and b = 15.5 m

Area of the school hall = 22 m x 15.5 m = 341 m2

RD Sharma Solutions Class 7 Maths Chapter 10 Exercise 20.2 Qs 13

Considering the carpet, ABCD

AB = 22 m – ( 0.75 m + 0.75 m) = 20.5 m [ Since 100 cm = 1 m]

BC = 15.5 m – ( 0.75 m + 0.75 m) = 14 m

Area of the carpet ABCD = 20.5 m x 14 m = 287 m2

Finding Area of the strip

Area of the strip = Area of the school hall PQRS – Area of the carpet ABCD

= 341 m2 – 287 m2 = 54 m2

Finding the cost of the carpet

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

Thus, Length of the carpet whose area is 287 m2 = 287 m2+ 0.82 m2 = 350 m

Cost of the 350 m long carpet = Rs. 18 x 350 = Rs. 6300

Q14: Two cross roads, each of width 5 m, run at right angles through the centre of a rectangular park of length 70 m and breadth 45 m parallel to its sides. Find the area of the roads. Also, find the cost of constructing the roads at the rate of Rs 105 per m2 .

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

RD Sharma Solutions Class 7 Maths Chapter 10 Exercise 20.2 Qs 14

Considering the rectangle ABCD

Length of the rectangular park = 70 cm

Breadth of the rectangular park = 45 m

Area of the rectangular park = Length x Breadth = 70 m x 45 m = 3150 m2

Area of the road EFGH

l = 70 m, b = 5 m

Area of the road EFGH = 70 m x 5 m = 350 m2

Area of the road JKIL

l = 45 m, b = 5 m

Area of the road JKIL = 45 m x 5 m = 225 m2

Finding the area of the road

Clearly area of MNOP is common to the two roads.

Thus, Area of MNOP = 5 m x 5 m = 25 m2

Hence,

Area of the roads = Area (EFGH) + Area (JKIL) – Area (MNOP)

= (350 + 225) m2– 25 m2 = 550 m2

Finding the cost of constructing the roads

Again, it is given that the cost of constructing the roads = Rs. 105 per m2

Therefore, Cost of constructing 550 m2 area of the roads = Rs. (105 x 550)

= Rs. 57750

Q15: 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 the park has an area 305 m2 . Find the dimensions of the park.  

Solution: It is given that, Area of path = 305 m2

Let the length of the park = 5a meter

Breadth of the park = 2a meter

RD Sharma Solutions Class 7 Maths Chapter 10 Exercise 20.2 Qs 15

Area of the rectangular park = (5a) x (2a) = 10a2 m2

Width of the path = 2.5 m

Outer length PQ = 5a m + 2.5 m + 2.5 m = (5a + 5) m

Outer breadth QR = 2a + 2.5 m + 2.5 m = (2a + 5) m

Area of PQRS = (5a + 5) m x (2a + 5) m = (10a2 + 25a + 10a + 25) m2= (10a2 + 35a + 25) m2

Area of the path = [(10a2 + 35a + 25) – 10a2] m2

=>           305 = 35a + 25

=>           305 – 25 = 35a

=>           280 = 35a

=>           a = 280 + 35 = 8

Therefore,

Length of the park = 5a = 5 x 8 = 40 m

Breadth of the park = 2a = 2 x 8 = 16 m

Q16: 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.

SolutionLet the side of the lawn be s meter.

RD Sharma Solutions Class 7 Maths Chapter 10 Exercise 20.2 Qs 16

The width of the path is given as 2.5 m

Side of the lawn including the path = (s + 2.5 + 2.5) m = (s + 5 ) m

So, area of lawn = (Area of the lawn including the path) – (Area of the path)

(s2) = (s + 5)2 – 165

=>           s2 = (s2 + 10s + 25) – 165

=>           165 = 10s + 25

=>        165 – 25 =10s

=>           140 = 10s

Therefore s = 140 / 10 = 14

Thus the side of the lawn = 14 m

Hence, the area of the lawn = (14 m) 2 = 196 m2

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