RD Sharma Solutions Class 7 Mensuration I Exercise 20.1

RD Sharma Class 7 Solutions Chapter 20 Ex 20.1 PDF Free Download

Class 7 Mensuration – Exercise 20.1

Q1) Find the area, in square meters, of a rectangle whose

i) Length = 5.5 m, breadth = 2.4 m

ii) Length = 180 cm, breadth = 150 cm

Solutions:

i) Length = 5.5 m, breadth = 2.4 m

Area of the rectangle = Length × Breadth

= 5.5 × 2.4

= 13.20 m2

ii) Length = 180 cm, breadth = 150 cm

Converting length and breadth in meter.

100 cm = 1m

180 cm = 180/100

= 1.8 m

150 cm = 150/100

= 1.5 m

Area of rectangle = Length × Breadth

= 1.8 × 1.5

= 2.70 m2

Q2) Find the area, in square centimeters, of a square whose side is

i) 2.6 cm

ii) 1.2 cm

Solutions:

i) 2.6 cm

Area of a square = (Side)2

= (2.6)2

= 6.76 cm2

ii) 1.2 cm

Area of a square = (Side)2

= (1.2)2

= 1.44 cm2

Q3) Find in square metres, the area of a square of side 16.5 dam.

Solution: 1 dam = 10 m

16.5 dam = 16.5 × 10 = 165 m

Area of a square = (Side)2

= (165)2

= 27225 m2

Q4) Find the area of a rectangular field in ares whose sides are:

i) 200 m and 125 m

ii) 75 m 5 dm and 120 m

Solutions:

i) 200 m and 125 m

Area of the rectangle = Length × Breadth

= 200 × 125

= 25000 m2

Converting the meter square into are.

100 m2 = 1 are

25000 m2 = 25000/100 = 250 are

ii) 75 m 5 dm and 120 m

L = 75 m 5 dm, B = 120 m

Converting 5 dm into m

10 dm = 1 m

5 dm = 5/10 = 0.5 m

L = 75 m + 0.5 m = 75.5 m

Area of the rectangle = Length × Breadth

= 75.5 × 120

= 9060 m2

Converting the meter square into are.

100 m2 = 1 are

9060 m2 = 9060/100 = 90.60 are

Q5) Find the area of a rectangular field in hectares whose sides are:

i) 125 m and 400 m

ii) 75 m 5 dm and 120 m

Solutions:

i) 125 m and 400 m

Area of the rectangle = Length × Breadth

= 125 × 400

= 50000 m2

Converting the meter square into hectare.

104 m2 = 1 hectare

50000 = 50000/104 = 5 hectare

ii) 75 m 5 dm and 120 m

L = 75 m 5 dm, B = 120 m

Converting 5 dm into m

10 dm = 1 m

5 dm = 5/10 = 0.5 m

L = 75 m + 0.5 m = 75.5 m

Area of the rectangle = Length × Breadth

= 75.5 × 120

= 9060 m2

Converting the meter square into hectare.

104 m2 = 1 hectare

9060 = 9060/104 = 0.906 hectare

Q6) A door of dimensions 3 m x 2m is on the wall of dimension 10 m x 10 m. Find the cost of painting the wall if rate of painting is Rs 2.50 per sq. m.

Solution: Let’s consider the Door.

L = 3, B = 2

Area of door = L × B

= 3 × 2

= 6 m2

Now, let’s take wall

Area of wall = L × B

= 10 × 10

= 100 m2

Area of the wall required for painting = Area of wall – Area of door

= 100 – 6

= 94 m2

Cost of painting the wall = 94 × 2.50

= Rs. 235

Q7) A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is bent in the shape of a square, what will be the measure of each side? Also, find which side encloses more area?

Solution: let us consider the wire which is in the shape of rectangle originally.

l = 40 cm, b = 22 cm

Perimeter of rectangle = 2 (40 + 22)

= 124 cm

When the wire is bent in the shape of a square, the perimeter will remain the same.

So, perimeter of square = perimeter of rectangle

4 × side = 124

side = 31 cm

Area of rectangle = 40 × 22

= 880 cm2

Area of square = (31)2

= 961 cm2

Therefore the area of square is more than the area of rectangle.

Q8) How many square metres of glass will be required for a window, which has 12 panes, each pane measuring 25 cm by 16 cm?

Solution: Area of a single pane = 25 × 16

= 400 cm2

Area of 12 panes = 12 × 400

= 4800 cm2

Converting centimeter square into meter square

10000 cm2 = 1 m2

4800 cm2 = 4800/10000 = 0.48 m2

Q9) A marble tile measures 10 cm x 12 cm. How many tiles will be required to cover a wall of size 3 m x 4 m? Also, find the total cost of the tiles at the rate of Rs 2 per tile.

Solution: Area of one marble tile = 10 × 12

= 120 cm2

Converting centimeter square into meter square

10000 cm2 = 1 m2

120 cm2 = 120/10000 = 0.012 m2

Area of the wall = 3 × 4

= 12 m2

No. of tiles required to cover the wall = Area of wall/ Area of one marble tile

= 12/0.012

= 1000 tiles

Total cost of a tile = cost of 1 tile × total number of tiles required

= 2 × 1000

= Rs. 2000

Q10) A table top is 9 dm 5 cm long 6 dm 5 cm broad. What will be the cost to polish it at the rate of 20 paise per square centimetre?

Solution: l = 9 dm 5 cm

= (9 × 10) + 5

= 95 cm

b = 6 dm 5 cm

= (6 × 10) + 5

= 65 cm

Area of the table = 95 × 65

= 6175 cm2

Cost of polishing per square centimetre of table = 20 paise

= 20/100

= Rs. 0.2

Cost of polishing the table = Area of table × cost of polishing per square centimetre

= 33.04 × 0.2

= Rs. 1235

Q11) A room is 9.68 m long and 6.2 m wide. Its floor is covered with rectangular tiles of size 22 cm by 10 cm. Find the total cost of the tiles at the rate of Rs 2.50 per tile.

Solution: Area of room = 9.68 × 6.2

= 60.016 m2

Area of 1 rectangular tile = 22 × 10

= 220 cm2

= 0.022 m2 [Since 1 m2 = 10000 cm2]

Total number of tiles required to cover the floor = Area of room/Area of 1 tile

= 60.016/0.022

= 2728

Total cost of tiles required to cover the floor = Total number of tiles required × cost of 1 tile

= 2728 × 2.5

= Rs. 6820

Q12) One side of a square field is 179 m. Find the cost of raising a lawn on the field at the rate of Rs 1.50 per square metre.

Solution: Area of the square field is = (179)2

= 32041 m2

Cost of raising a lawn on the field = Area of square field × cost of raising per square metre

= 32041 × 1.5

= Rs. 48061.5

Q13) A rectangular field is measured 290 m by 210 m. How long will it take for a girl to go two times round the field, if she walks at the rate of 1.5 m/sec?

Solution: Perimeter of rectangular field = 2 (l + b)

= 2 (290 + 210)

= 1000 m

Distance covered by the girl 2 times = 2 × 1000

= 2000 m

Speed of the girl = 1.5 m/sec

= 90 m/min [Since 1 min = 60 sec]

Speed = Distance/ Time

So, time taken by the girl = Distance/Speed

= 2000/90

=22 29 min

Q14) A corridor of a school is 8 m long and 6 m wide. It is to be covered with canvas sheets. If the available canvas sheets have the size 2 m x 1 m, find the cost of canvas sheets required to cover the corridor at the rate of Rs 8 per sheet.

Solution: Area of the corridor = 8 × 6

= 48 m2

Area of 1 canvas sheet = 2 × 1

= 2 m2

Number of canvas sheets required = 48/2

= 24 canvas sheet

Cost of canvas sheet to cover the corridor = Cost of 1 canvas sheet × No. of canvas sheet

= 8 × 24

= Rs. 192

Q15) The length and breadth of a playground are 62 m 60 cm and 25 m 40 cm respectively. Find the cost of turfing it at Rs 2.50 per square metre. How long will a man take to go three times round the field, if he walks at the rate of 2 metres per second?

Solution: Length of playground = 62 m 60 cm

= 62 + 0.6 m [Since 1 m = 100 cm]

= 62.6 m

Breadth of playground = 25 m 40 cm

= 25 + 0.4 [Since 1 m = 100 cm]

= 25.4 m

Area of playground = l × b

= 62.6 × 25.4

= 1590.04 m2

Cost of turfing the playground = Area of playground × cost of turfing per square metre

= 1590.04 × 2.50

= Rs. 3975.1

Perimeter of playground = 2 (l + b)

= 2 (62.6 + 25.4)

= 2 × 88

= 176 m

Distance covered by the man in 3 times round the field = 3 × 176

= 528 m

Speed of the man = 2 m/sec

= 120 m/min [Since 1 min = 60 sec]

Speed = Distance/ Time

So, time taken by the man = Distance/Speed

= 528/120

= 4.4 min

Q16) A lane 180 m long and 5 m wide is to be paved with bricks of length 20 cm and breadth 15 cm. Find the cost of bricks that are required, at the rate of Rs 750 per thousand.

Solution: Area of the lane = 180 × 5

= 900 m2

Area of 1 brick = 20 × 15

= 300 cm2

= 0.03 m2 [As 1 m2 = 10000 cm2]

Number of bricks required = 900/0.03

= 30000 bricks

Cost of 1000 bricks = Rs. 750

Cost of 1 brick = 750/1000

= Rs. 0.75

Cost of bricks required = Cost of 1 brick × No. of bricks

= 0.75 × 30000

= Rs. 22500

Q17) How many envelopes can be made out of a sheet of paper 125 cm by 85 cm; supposing one envelope requires a piece of paper of size 17 cm by 5 cm?

Solution: Area of the sheet = 125 × 85

= 10625 cm2

Area of 1 envelope = 17 × 5

= 85 cm2

Number of envelopes can be made = Area of sheet/Area of 1 envelope

= 10625/85

= 125 envelopes

Q18) The width of a cloth is 170 cm. Calculate the length of the cloth required to make 25 diapers, if each diaper requires a piece of cloth of size 50 cm by 17 cm.

Solution: Area of 1 diaper = 50 × 17

= 850 cm2

Area of 25 diapers = 25 × 850

= 21250 cm2

From the questions, we know that

Area of 25 diapers = Area of cloth

21250 = L × 170

L = 21250/170

L = 125 cm

Q19) The carpet for a room 6.6 m by 5.6 m costs Rs 3960 and it was made from a roll 70 cm wide. Find the cost of the carpet per metre.

Solution: Area of the room = l × b

= 6.6 × 5.6

= 36.96 m2

Breadth of the roll = 70 cm

= 70/100

= 0.7 m

Area of the room = Area of the roll

36.96 = L × 0.7

L = 36.96/0.7

L = 52.8 m

Cost of the carpet per square = Total cost of the carpet/Length of the room

= 3960/52.8

= Rs. 75

Q20) A room is 9 m long, 8 m broad and 6.5 m high. It has one door of dimensions 2 m x 1.5 m and three windows each of dimensions 1.5 m x 1 m. Find the cost of white washing the walls at Rs 3.80 per square metre.

Solution: Area of the 4 walls of room = 2 (l + b) × h

= 2 (9 + 8) × 6.5

= 34 × 6.5

= 221 m2

Area of 1 door = 2 × 1.5

= 3 m2

Area of 1 window = 1.5 × 1

= 1.5 m2

Area of 3 windows = 3 × 1.5

= 4.5 m2

Area required to white wash the wall = Area of 4 walls – Area of 1door – Area of 1 window

= 221 – 3 – 4.5

= 213.5 m2

Cost of white washing the walls = 213.5 × 3.8

= Rs. 811.3

Q21) A hall 36 m long and 24 m broad allowing 80 m2 for doors and windows, the cost of papering the walls at Rs 8.40 per m2 is Rs 9408. Find the height of the hall.

Solution: Area of hall = 2 (l + b) × h

= 2 (36 + 24) × h

= 120 h

Area required for papering the wall = Area of hall – Area of door and windows

= (120 h – 80) m2

Cost of papering the wall = Area required for papering × Cost per meter square of papering wall

9408 = (120 h – 80) × 8.40

1120 = 120 h – 80

1200 = 120 h

h = 10 m

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