In this chapter, we provide RD Sharma Class 7 ex 20.1 Solutions Chapter 20 Mensuration I (Perimeter and Area of Rectilinear Figures) for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 7 ex 20.1 Solutions Chapter 20 Mensuration I (Perimeter and Area of Rectilinear Figures) Maths pdf, Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 7 |

Subject | Maths |

Chapter | Chapter 20 |

Chapter Name | Mensuration I (Perimeter and Area of Rectilinear Figures) |

Exercise | Ex 20.1 |

Table of Contents

**RD Sharma Solutions for Class 7 Chapter 20 Mensuration I (Perimeter and Area of Rectilinear Figures)** **Ex 20.1 Download PDF**

**Mensuration I (Perimeter and Area of Rectilinear Figures) Exercise 20.1**

**Question: 1**

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

**Solution:**

We have,

(i) Length = 5.5 m, Breadth = 2.4 m Therefore, Area of rectangle = Length x Breadth = 5.5 m x 2.4 m = 13.2 m^{2}

(ii) Length = 180 cm = 1.8 m, Breadth = 150 cm = 1.5 m [ Since 100 cm = 1 m] Therefore, Area of rectangle = Length x Breadth = 1.8 m x 1.5 m = 2.7 m^{2}

**Question: 2**

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

(i) 2.6 cm

(ii) 1.2 dm

**Solution:**

We have,

(i) Side of the square = 2.6 cm

Therefore, area of the square = (Side)^{2} = (2.6 cm)2= 6.76 cm^{2}

(ii) Side of the square = 1.2 dm = 1.2 x 10 cm = 12 cm

Therefore, area of the square = (Side)^{2} = (12 cm)^{2}= 144 cm^{2} [ Since 1 dm = 10 cm]

**Question: 3**

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

**Solution:**

We have,

Side of the square = 16.5

dam = 16.5 x 10 m = 165 m

Area of the square = (Side)^{2} = (165 m)^{2} = 27225 m^{2}

[Since 1 dam/dm (decameter) = 10 m]

**Question: 4**

Find the area of a rectangular field in acres whose sides are:

(1) 200 m and 125 m

(ii) 75 m 5 dm and 120 m

**Solution:**

We have,

(i) Length of the rectangular field = 200 m

Breadth of the rectangular field = 125 m

Therefore, Area of the rectangular field = Length x Breadth = 200 m x 125 m

= 25000 m^{2} = 250 acres [Since 100 m^{2} = 1 are]

(ii) Length of the rectangular field =75 m 5 dm = (75 + 0.5) m

= 75.5 m [Since 1 dm = 10 cm = OA m]

Breadth of the rectangular field = 120 m

Therefore, Area of the rectangular field = Length x Breadth

= 75.5 m x 120 m = 9060 m^{2} = 90.6 acres [Since 100 m^{2} = 1 are]

**Question: 5**

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

**Solution:**

We have,

(i) Length of the rectangular field = 125 m

Breadth of the rectangular field = 400 m

Therefore, Area of the rectangular field = Length x Breadth

= 125 m x 400 m = 50000 m^{2} = 5 hectares [Since 10000 m^{2} = 1 hectare]

(ii) Length of the rectangular field =75 m 5 dm = (75 + 0.5) m

= 75.5 m [Since 1 dm = 10 cm = 0.1 m]

Breadth of the rectangular field = 120 m

Therefore, Area of the rectangular field = Length x Breadth

= 75.5 m x 120 m = 9060 m^{2} = 0.906 hectares [Since 10000 m^{2} = 1 hectare]

**Question: 6**

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:**

We have,

Length of the door = 3 m

Breadth of the door = 2 m

Side of the wall = 10 m

Area of the wall = Side x Side = 10 m x 10 m

= 100 m^{2}

Area of the door = Length x Breadth = 3 m x 2 m = 6 m

Thus, required area of the wall for painting = Area of the wall – Area of the door

= (100 – 6) m^{2}= 94 m^{2}

Rate of painting per square metre = Rs. 2.50

Hence, the cost of painting the wall = Rs. (94 x 2.50) = Rs. 235

**Question: 7**

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:**

We have,

Perimeter of the rectangle = 2(Length + Breadth)

= 2(40 cm + 22 cm) = 124 cm

It is given that the wire which was in the shape of a rectangle is now bent into a square.

Therefore, the perimeter of the square = Perimeter of the rectangle

→ Perimeter of the square = 124 cm

4 x side = 124 cm

Side = 124/4 = 31 cm

Now, Area of the rectangle = 40 cm x 22 cm = 880 cm^{2}

Area of the square = (Side)^{2} = (31 cm)^{2} = 961 cm^{2}.

Therefore, the square-shaped wire encloses more area.

**Question: 8**

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:**

We have,

Length of the glass pane = 25 cm

Breadth of the glass pane = 16 cm

Area of one glass pane = 25 cm x 16 cm

= 400 cm^{2} = 0.04 m^{2}

[Since 1 m^{2} = 10000 cm^{2} ]

Thus, Area of 12 such panes = 12 x 0.04 = 0.48 m^{2}

**Question: 9**

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:**

We have,

Area of the wall = 3 m x 4 m = 12 m^{2}

Area of one marble tile = 10 cm x 12 cm

= 120 c m^{2} = 0.012 m^{2} [Since 1 m^{2} = 10000 c m^{2} ]

Thus, Number of tiles = Area of wall

Area of one tile=12 m^{2} = 0.012 m^{2}=1000

Cost of one tile = Rs. 2

Total cost = Number of tiles x Cost of one tile

= Rs. (1000 x 2)

= Rs. 2000

**Question: 10**

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:**

We have,

Length of the table top = 9 dm 5 cm = (9 x 10 + 5) cm = 95 cm [ Since 1 dm = 10 cm]

Breadth of the table top = 6 dm 5 cm = (6 x 10 + 5) cm = 65 cm

Area of the table top = Length x Breadth = (95 cm x 65 cm) = 6175 c m^{2}

Rate of polishing per square centimetre = 20 paise = Rs. 0.20

Total cost = Rs. (6175 x 0.20) = Rs. 1235

**Question: 11**

A room is 9.68 m long and 6.2 m wide. Its floor is to be 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:**

We have,

Length of the floor of the room = 9.68 m

Breadth of the floor of the room = 6.2 m

Area of the floor = 9.68 m x 6.2 m = 60.016 m^{2}

Length of the tile = 22 cm

Breadth of the tile = 10 cm

Area of one tile = 22 cm x 10 cm = 220 c m^{2} = 0.022 m^{2} [Since 1 m^{2} = 10000 c m^{2}]

Thus, Number of tiles = 60.016 m^{2}/0.022 m^{2}=2728

Cost of one tile = Rs. 2.50

Total cost = Number of tiles x Cost of one tile = Rs. (2728 x 2.50) = Rs. 6820

**Question: 12**

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:**

We have,

Side of the square field = 179 m

Area of the field = (Side)^{ 2} = (179 m)^{ 2} = 32041 m^{2}

Rate of raising a lawn on the field per square metre = Rs. 1.50 Thus,

Total cost of raising a lawn on the field = Rs. (32041 x 1.50) = Rs. 48061.50

**Question: 13**

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:**

We have,

Length of the rectangular field = 290 m

Breadth of the rectangular field = 210 m

Perimeter of the rectangular field = 2(Length + Breadth) = 2(290 + 210) = 1000 m

Distance covered by the girl = 2 x Perimeter of the rectangular field = 2 x 1000 = 2000 m

The girl walks at the rate of 1.5 m/sec. Or, Rate = 1.5 x 60 m/min = 90 m/min

Thus, required time to cover a distance of 2000 m = 2000 m/90 m/min

= 2229min

Hence, the girl will take 2229 min to go two times around the field.

**Question: 14**

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:**

We have,

Length of the corridor = 8 m

Breadth of the corridor = 6 m

Area of the corridor of a school = Length x Breadth = (8 m x 6 m) = 48 m^{2}

Length of the canvas sheet = 2 m

Breadth of the canvas sheet = 1 m

Area of one canvas sheet = Length x Breadth = (2 m x 1 m) = 2 m^{2}

Thus, Number of canvas sheets = 48 m^{2} /2m^{2}=24

Cost of one canvas sheet = Rs. 8

Total cost of the canvas sheets = Rs. (24 x 8) = Rs. 192

**Question: 15**

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:**

We have,

Length of a playground = 62 m 60 cm = 62.6 m [ Since 10 cm = 0.1 m]

Breadth of a playground = 25 m 40 cm = 25.4 m

Area of a playground = Length x Breadth= 62.6 m x 25.4 m = 1590.04 m^{2}

Rate of turfing = Rs. 2.50/ m^{2} Total cost of turfing = Rs. (1590.04 x 2.50) = Rs. 3975.10

Again, Perimeter of a rectangular field = 2(Length + Breadth) = 2(62.6 + 25.4) = 176 m

Distance covered by the man in 3 rounds of a field = 3 x Perimeter of a rectangular field

= 3 x 176 m = 528 m

The man walks at the rate of 2 m/sec. Or, Rate = 2 x 60 m/min = 120 m/min

Thus, required time to cover a distance of 528 m = 528 m120 m/min=4.4 min

= 4 minutes 24 seconds [Since 0.1 minutes = 6 seconds]

**Question: 16**

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:**

We have,

Length of the lane = 180 m

Breadth of the lane = 5 m

Area of a lane = Length x Breadth = 180 m x 5 m = 900 m^{2}

Length of the brick = 20 cm

Breadth of the brick = 15 cm

Area of a brick = Length x Breadth = 20 cm x 15 cm

= 300 cm^{2} = 0.03 m^{2} [Since 1 m^{2} = 10000 cm^{2}]

Required number of bricks = 900 m^{2}/0.03 m^{2 }= 30000

Cost of 1000 bricks = Rs. 750

Total cost of 30,000 bricks = Rs. 750×30,000/1000 = Rs. 22,500

**Question: 17**

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:**

We have,

Length of the sheet of paper = 125 cm

Breadth of the sheet of paper = 85 cm

Area of a sheet of paper = Length x Breadth = 125 cm x 85 cm = 10,625 cm^{2}

Length of sheet required for an envelope = 17 cm

Breadth of sheet required for an envelope = 5 cm

Area of the sheet required for one envelope = Length x Breadth

= 17 cm x 5 cm = 85 c m^{2}

Thus, required number of envelopes = 10,625 cm^{2}/85 c m^{2 }= 125

**Question: 18**

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:**

We have,

Length of the diaper = 50 cm

Breadth of the diaper = 17 cm

Area of cloth to make 1 diaper = Length x Breadth = 50 cm x 17 cm = 850 cm^{2}

Thus, Area of 25 such diapers = (25 x 850) c m^{2} = 21,250 cm^{2}

Area of total cloth = Area of 25 diapers = 21,250 cm^{2}

It is given that width of a cloth = 170 cm

Length of the cloth = Area of cloth

Width of a cloth = 21,250 cm 2170 cm = 125 cm

Hence, length of the cloth will be 125 cm.

**Question: 19**

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:**

We have,

Length of a room = 6.6 m

Breadth of a room = 5.6 m

Area of a room = Length x Breadth = 6.6 m x 5.6 m = 36.96 m^{2}

Width of a carpet = 70 cm = 0.7 m [Since 1 m = 100 cm]

Length of a carpet = Area of a room

Width of a carpet = 36.96 m 20.7 m = 52.8 m

Cost of 52.8 m long roll of carpet = Rs. 3960

Therefore, Cost of 1 m long roll of carpet = Rs. 396052.8 = Rs. 75

**Question: 20**

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 T4L washing the walls at Rs 3.80 per square metre.

**Solution:**

We have,

Length of a room = 9 m

Breadth of a room = 8 m

Height of a room = 6.5 m

Area of 4 walls = 2(1 + b)h = 2(9 m + 8 m) x 6.5 m = 2 x 17 m x 6.5 m = 221 m^{2}

Length of a door = 2 m

Breadth of a door = 1.5 m

Area of a door = Length x Breadth = 2 m x 1.5 m = 3 m^{2}

Length of a window = 1.5 m

Breadth of a window = 1 m

Since, area of one window = Length x Breadth = 1.5 m x 1 m = 1.5 m^{2}

Thus, Area of 3 such windows = 3 x 1.5 m^{2} = 4.5 m^{2}

Area to be white-washed = Area of 4 walls – (Area of one door + Area of 3 windows)

Area to be white-washed = [221 – (3 + 4.5)] m^{2} = (221 – 7.5) m^{2} = 213.5 m^{2}

Cost of white-washing for 1 m^{2} area = Rs. 3.80

Cost of white-washing for 213.5 m^{2} area = Rs. (213.5 x 3.80) = Rs. 811.30

**Question: 21**

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:**

We have,

Length of the hall = 36 m

Breadth of the hall = 24 m

Let h be the height of the hall.

Now, in papering the wall, we need to paper the four walls excluding the floor and roof of the hall. So, the area of the wall which is to be papered = Area of 4 walls

= 2h(I + b)

= 2h (36 + 24)

= 120h m^{2}

Now, area left for the door and the windows = 80 m^{2}

So, the area which is actually papered = (120h – 80) m^{2}

Again, The cost of papering the walls at Rs 8.40 per m^{2} = Rs. 9408

→ (120h – 80) m^{2} x Rs. 8.40 per m^{2}= Rs. 9408

→ (120h – 80) m^{2} = Rs. 9408/Rs. 8.40

→ (120h – 80) m^{2} = 1120 m^{2}

→ 120h m^{2} = (1120 + 80) m^{2}

→ 120h m^{2}= 1200 m^{2}

h = 1200 m^{2} 120 m = 10 m

Hence, the height of the wall would be 10 m.

**All Chapter RD Sharma Solutions For Class 7 Maths**

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