## RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20A.

**Other Exercises**

- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20A
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20B
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20C
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20D
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20E
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20F
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20G
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration CCE Test Paper

**Question 1.**

**Solution:**

Length of rectangle (l) = 24.5 m

Breadth (b) = 18 m

Area = l x b = 24.5 x 18 m² = 441 m²

(ii) Length of rectangle (l) = 12.5 m

Breadth (b) = 8 cm = 0.80 m

Area = l x b = 12.5 x 0.80 m² = 10 m²

**Question 2.**

**Solution:**

Length of rectangular plot (l) = 48m

and its diagonal = 50m

**Question 3.**

**Solution:**

Ratio in the sides of a rectangle = 4 : 3

Area = 1728 cm²

Let length = 4x,

then breadth = 3x

Area = l x b

1728 = 4x x 3x

⇒ 12x² = 1728

⇒ x² = 144 = (12)²

⇒ x = 12

Length = 4x = 4 x 12 = 48 m

and breadth = 3m = 3 x 12 = 36m

Now perimeter = 2 (l + b) = 2 (48 + 36) m = 2 x 84 = 168 m

Rate of fencing = Rs. 30 per metre

Total cost= 168 x 30 = Rs. 5040

**Question 4.**

**Solution:**

Area of rectangular field = 3584 m²

Length = 64 m

Area = 3584

Breadth = \(\frac { Area }{ Length }\) = \(\frac { 3584 }{ 64 }\) = 56 m

Now perimeter = 2 (l + b) = 2 (64 + 56) m = 2 x 120 = 240 m

Distance covered in 5 rounds = 240 x 5 = 1200 m

Speed = 6 km/h

Time take = \(\frac { 1200 }{ 1000 }\) x \(\frac { 60 }{ 6 }\) = 12 minutes (1 hour = 60 minutes)

**Question 5.**

**Solution:**

Length of verandah (l) = 40m

Breadth (b) = 15m

Area = l x b = 40 x 15 = 600m²

Length of one stone = 6dm = \(\frac { 6 }{ 10 }\) m

and breadth = 5 dm = \(\frac { 5 }{ 10 }\) m

Area of one stone = \(\frac { 6 }{ 10 }\) x \(\frac { 5 }{ 10 }\)

**Question 6.**

**Solution:**

Length of a room = 13 m

Breadth = 9 m

Area of floor = l x b = 13 x 9 m² = 117 m²

or area of carpet = 117 m²

Width = 75 cm = \(\frac { 75 }{ 100 }\) = \(\frac { 3 }{ 4 }\) m

Length of carpet = Area ÷ Width

= 117 ÷ \(\frac { 3 }{ 4 }\)

= 117 x \(\frac { 4 }{ 3 }\) m

= 39 x 4 = 156 m

Rate = Rs. 105 per m

Total cost = Rs. 156 x 105 = Rs. 16380

**Question 7.**

**Solution:**

Cost of carpeting a room = Rs. 19200

Rate = Rs. 80 per m

Length of carpet = \(\frac { 19200 }{ 80 }\) m = 240 m

Width of carpet = 75 cm = \(\frac { 75 }{ 100 }\) = \(\frac { 3 }{ 4 }\) m

Area of carpet = 240 x \(\frac { 3 }{ 4 }\) = 180 m²

Length of a room = 15 m

Width = \(\frac { Area }{ Length }\) = \(\frac { 180 }{ 15 }\) = 12m

**Question 8.**

**Solution:**

Ratio in length and breadth of a rectangular piece of land = 5 : 3

Cost of fencing = Rs. 9600

and rate = Rs. 24 per m

Perimeter = \(\frac { 9600 }{ 24 }\) = 400 m

Let length = 5x

Then breadth = 3x

Perimeter = 2 (l + b)

⇒ 400 = 2 (5x + 3x)

⇒ 400 = 2 x 8x = 16x

⇒ 16x = 400

⇒ x = 25

Length of the land = 5x = 5 x 25 = 125 m

and width = 3x = 3 x 25 = 75 m

**Question 9.**

**Solution:**

Length of hall (l) = 10m.

Breadth (b) = 10m

and height (h) = 5m

Longest pole which can be placed in it

**Question 10.**

**Solution:**

Side of square (a) = 8.5m

Area = a² = (8.5)² = 8.5 x 8.5 m² = 72.25 m²

**Question 11.**

**Solution:**

(i) Length of diagonal of square = 72 cm.

Let length of side = a

Then diagonal = √2 a.

√2 a = 72

= 1.2 x √2 x 1.2 x √2 m²

= 1.44 x 2 = 2.88 m²

**Question 12.**

**Solution:**

Area of a square = 16200 m²

Side = √16200 m = √(8100 x 2) m = 90√2 m

Length of diagonal = √2 (side) = √2 x 90√2 = 180 m

**Question 13.**

**Solution:**

Area of square = \(\frac { 1 }{ 2 }\) hectare

= \(\frac { 1 }{ 2 }\) x 10000 m² = 5000 m²

side (a) = √Area = √5000 m = √(2500 x 2) = 50√2 m

Length of diagonal = √2 (side) = √2 x 50√2 = 100 m

**Question 14.**

**Solution:**

Area of sphere plot = 6084 m²

Side (a) = √Area = √6084 m = 78m

Perimeter = 4a = 4 x 78 = 312 m

Length of boundary four times = 312 x 4 = 1248 m

**Question 15.**

**Solution:**

Side of a square wire = 10 cm

Perimeter = 4a = 4 x 10 cm = 40 cm

or perimeter of rectangle = 40 cm

Length of rectangle = 12 cm

Breadth = \(\frac { 40 }{ 2 }\) – 12 = 20 – 12 = 8 cm

Now area of square = a² = (10)² = 100 cm²

and area of rectangle = l x b = 12 x 8 = 96 cm²

Difference in areas = 100 – 96 = 4 cm²

Square has 4 cm² more area

**Question 16.**

**Solution:**

Length of go down (l) = 50 m

Breadth (b) = 40 m

and height (h) = 10 m

Area of 4 walls = 2 (l + b) x h

= 2 (50 + 40) x 10 m

= 2 x 90 x 10 = 1800 m²

and area of ceiling = l x b = 50 x 40 = 2000 m²

Total area of walls and ceiling = 1800 + 2000 = 3800 m²

Rate of whitewashing = Rs. 20 per m²

Total cost = Rs. 20 x 3800 = Rs. 76000

**Question 17.**

**Solution:**

Area of 4 walls of a room= 168m²

Breadth of the room (b) = 10m

and height (h) = 4m.

Let l be the length of room

2 (l + b) h = 168

⇒ 2 (l + 10) x 4 = 168

⇒ l + 10 = \(\frac { 168 }{ 2 x 4 }\) = 21

⇒ l = 21 – 10 = 11m

Length of the room = 11 m

**Question 18.**

**Solution:**

Area of 4 walls of a room = 77 m²

Length of room (l) = 7.5 m

and breadth (b) = 3.5 m

Let h be the height,

then area of four walls = 2 (l + b) h

= 2 (7.5 + 3.5) h = 77

⇒ 2 x 11 x h = 77

⇒ h = \(\frac { 77 }{ 2 x 11 }\)

Height of room = 3.5 m

**Question 19.**

**Solution:**

Area of 4 walls = 120 m²

Height (h) = 4m.

Let breadth (b) = x

and length (l) = 2x

Area of 4 walls = 2 (l + h) x h = 2(2x + x) x 4 = 8 x 3x = 24x

24x = 120

x = \(\frac { 120 }{ 24 }\) = 5

Length of room = 2x = 2 x 5 = 10m

and breadth = x = 5m

Area of floor = l x b = 10 x 5 = 50 m²

**Question 20.**

**Solution:**

Length of a room (l) = 8.5 m

Breadth (b) = 6.5 m

and height (h) = 3.4 m

Area of four walls = 2 (l + b) x h

= 2 (8.5 +6.5) x 3.4 m²

= 2 x 15 x 3.4 m²

= 30 x 3.4= 102.0 m²

Area of two doors of size 1.5 m x 1 m = 2 x 1.5 x 1 m = 3 m²

and area of two windows of size 2 m x 1 m = 2 x 2 x 1 = 4 m²

Area of remaining portion = 102 – (3 + 4) = 102 – 7 m2 = 95 m²

Rate of painting = Rs. 160 per m²

Total cost = Rs. 160 x 95 = Rs. 15200

Hope given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20A are helpful to complete your math homework.

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