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

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

**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:**

Outer length of plot (L) = 75 m

and breadth (B) = 60 m

Width of path inside = 2 m

Inner length (l) = 75 – 2 x 2 = 75 – 4 = 71 m

and width (b) = 60 – 2 x 2 = 60 – 4 = 56 m

Area of the path = L x B – l x b = (75 x 60 – 71 x 56) m² = 4500 – 3976 = 524 m²

Rate of constructing it = Rs. 125 per m²

Total cost = Rs. 524 x 125 = Rs. 65500

**Question 2.**

**Solution:**

Outer length of the plot (L) = 95 m

and breadth (B) = 72 m

Width of path = 3.5 m

Inner length (l) = 95 – 2 x 3.5 = 95 – 7 = 88 m

and breadth = 72 – 2 x 3.5 = 72 – 7 = 65 m

Outer area = L x B = 95 x 72 m² = 6840 m²

and inner area = l x b = 88 x 65 m² = 5720 m²

Area of path = outer area – inner area = 6840 – 5720 = 1120 m²

Rate of constructing it = Rs. 80 per m²

Total cost = Rs. 1120 x 80 = Rs. 89600

and rate of laying grass = Rs. 40 per m²

Total cost = Rs. 40 x 5720 = Rs. 228800

Total cost = Rs. 89600 + Rs. 228800 = Rs. 318400

**Question 3.**

**Solution:**

Length of saree (L) = 5 m

and breadth (B) = 1.3 m

Width of border = 25 cm

Inner length (l) = 5 – \(\frac { 2 x 25 }{ 100 }\) = 5 – 0.5 = 4.5 m

and inner breadth (b) = 1.3 – \(\frac { 2 x 25 }{ 100 }\) = 1.3 – 0.5 = 0.8 m

Now area of the boarder = L x B – l x b

= (5 x 1.3 – 4.5 x 0.8) m²

= 6.50 – 3.60

= 2.90 m²

= 2.90 x 100 x 100 = 29000 cm²

Cost of 10 cm² = Re. 1

Total cost = Rs. \(\frac { 29000 x 1 }{ 10 }\) = Rs. 2900

**Question 4.**

**Solution:**

Inner length of lawn (l) = 38 m

and breadth (b) = 25 m.

Width of path = 2.5 m.

Outer length (L) = 38 + 2 x 2.5 = 38 + 5 = 43 m

and outer breadth (B) = 25 + 2 x 2.5 = 25 + 5 = 30 m

Area of path = Outer area – Inner area

= (43 x 30 – 38 x 25) m²

=(1290 – 950) m² = 340 m²

Rate of gravelling the path = Rs. 120 per m²

Total cost = Rs. 120 x 340 = Rs. 40800

**Question 5.**

**Solution:**

Length’ of room (l) = 9.5m

Breadth (b) = 6m

Width of outer verandah = 1.25m

Outer length (L) = 9.5 + 2 x 1.25 = 9.5 + 2.5 = 12.0 m

and breadth (B) = 6 + 2 x 1.25 = 6 + 2.5 = 8.5 m

Area of verandah = Outer area – Inner area = L x B – l x b

= (12.0 x 8.5 – 9.5 x 6) m² – (102.0 – 57.0) m² = 45 m²

Rate of cementing = Rs. 15 per m²

Total cost = Rs. 80 x 45 = Rs. 3600

**Question 6.**

**Solution:**

Each side of square bed (a) = 2m 80 cm = 2.8 m

Width of strip = 30cm

Outer side (A) = 2.8 m + 2 x 30cm = 2.8 + 0.6 = 3.4m

Outer area = (3.4 m)² = 11.56 m²

Inner area = (2.8)² = 7.84 m²

Area of increased bed flower = 11.56 – 7.84 = 3.72 m²

**Question 7.**

**Solution:**

Ratio in length and breadth of the park = 2 : 1

Its perimeter = 240 m

Let length = 2x

then breadth = x

Perimeter = 2 (2x + x) = 2 x 3x = 6x

6x = 240

⇒ x = \(\frac { 240 }{ 6 }\) = 40

Length = 2x = 2 x 40 = 80m

and breadth = x = 40m

Area = L x B = 80 x 40 m² = 3200 m²

Width of path inside the park = 2m

Inner length (l) = 80 – 2 x 2 = 80 – 4 = 76 m

and breadth (b) = 40 – 2 x 2 = 40 – 4 = 36 m

Inner area = 76 x 36 = 2736 m²

Area of path = Outer area – Inner area = 3200 – 2736 = 464 m²

Rate of paving the path = Rs. 80 per m²

Total cost = Rs. 80 x 464 = Rs. 37120

**Question 8.**

**Solution:**

Length of hall (l) = 22m

Breadth (b) = 15.5 m

Space left along the walls = 75m = \(\frac { 3 }{ 4 }\) m

Inner length (l) = 22 – 2 x \(\frac { 3 }{ 4 }\) = 20.5 m

Inner breadth (b) = 15.5 – 2 x \(\frac { 3 }{ 4 }\) = 15.5 – 1.5 = 14m

Area of carpet = Inner area = 20.5 x 14 m² = 287 m²

Outer area = 22 x 15.5 = 341 m²

Area of strip left out = 341 – 287 = 54 m²

Width of carpet = 82 cm = \(\frac { 82 }{ 100 }\) m

Length of carpet = 287 ÷ \(\frac { 82 }{ 100 }\)

= \(\frac { 287 x 100 }{ 82 }\) = 350m

Rate of carpet = Rs. 60 per metre

Total cost = Rs. 60 x 350 = Rs. 21000

**Question 9.**

**Solution:**

Area of path = 165 m²

Width of path = 2.5m.

Let side of square lawn = x m.

Outer side = x + 2 x 2.5 = (x + 5) m

Area of path = (x + 5)² – x²

⇒ x² + 10x + 25 – x² = 165

⇒ 10x = 165 – 25 = 140

⇒ x = \(\frac { 140 }{ 10 }\) = 14m

Side of lawn = 14m

and area of lawn = (14)² m² = 196 m²

**Question 10.**

**Solution:**

Ratio in length and breadth of a park = 5 : 2

Width of path outside it = 2.5 m

Area of path = 305 m2

Let Inner length (l) = 5x

and breadth (b) = 2x

Outer length (L) = 5x + 2 x 2.5 = (5x + 5) m

Width (B) = 2x + 2 x 2.5 = (2x + 5) m

Area of path = Outer area – Inner area

⇒ (5x + 5) (2x + 5) – 5x x 2x = 305

⇒ 10x² + 10x + 25x + 25 – 10x² = 305

⇒ 35x = 305 – 25 = 280

⇒ x = 8

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

and breadth = 2x = 2 x 8 = 16m

Dimensions of park = 40m by 16m

**Question 11.**

**Solution:**

Length of lawn (l) = 70 m

Breadth (b) = 50 m

Width of crossing roads = 5m

Area of roads = 70 x 5 + 50 x 5 – (5)²

= 350 + 250 – (5)²

= 600 – 25 = 575 m²

Cost of constructing = Rs. 120 per m²

Total cost Rs. 120 x 575 = Rs. 69000

**Question 12.**

**Solution:**

Length of lawn (l) = 115m

and breadth (b) = 64m.

Width of road parallel to length = 2m

and width of road parallel to breadth= 2.5m

Area of roads = (115 x 2 + 64 x 2.5 – 2 x 2.5) m²

= (230 + 160 – 5) m² = (390 – 5) m² = 385 m²

Cost of gravelling = Rs. 60 m²

Total cost = Rs. 60 x 385 = Rs. 23100

**Question 13.**

**Solution:**

Length of field (l) = 50 m

and breadth (b) = 40 m

Width of road parallel to length = 2 m

and width of road parallel to breadth = 2.5 m

Area of roads = 50 x 2 + 40 x 2.5 – 2.5 x 2 = (100 + 100 – 5) m² = 195 m²

and area of remaining portion = 50 x 40 – 195 = 2000 – 195 = 1805 m²

**Question 14.**

**Solution:**

(i) Outer length = 43m

and breadth = 27m

Area = 43 x 27 = 1161 m²

Inner length = 43 – 2 x 1.5 = 43 – 3 = 40m

and breadth = 27 – 2 x 1 = 27 – 2 = 25m

Inner area = 40 x 25 = 1000 m²

Area of shaded portion = 1161 – 1000 = 161 m²

(ii) Side of square (a) = 40m

Area = (a)² = 40 x 40 = 1600 m²

Area of larger road = 40 x 3 = 120 m²

and area of shorter road = 40 x 2 = 80 m²

Area of roads = (120 + 80) – 3 x 2 = 200 – 6 = 194 m²

Area of shaded portion = (1600 – 194) m² = 1406 m²

**Question 15.**

**Solution:**

(i) Outer length = 24m

and breadth = 19m

Area = 24 x 19 = 456 m²

Length of unshaded portion = 24 – 4 = 20m

and breadth = 16.5m

Area of unshaded portion = 20 x 16.5 m² = 330.0 m²

Area of shaded portion = 456 – 330 = 126 m²

(ii) Dividing the figure an shown

Area of rectangle I = 15 x 3 cm² = 45 cm²

Area of rectangle II = (12 – 3) x 3 = 9 x 3 = 27 cm²

Area of rectangle III = 5 x 3 = 15 cm²

and area of rectangle IV = (12 – 3) x 3 = 9 x 3 = 27 cm²

Total area of shaded portion = 45 + 27 + 15 + 27 = 114 cm²

**Question 16.**

**Solution:**

Dividing the figure an shown

Area of rectangle I = 3.5 x 0.5 m² = 1.75 m²

Area of rectangle II = (3.5 – 2 x 0.5) x 0.5 = (3.5 – 1) x 0.5 = 2.5 x 0.5 = 1.25 m²

Area of rectangle III = (2.5 – 1) x 0.5 = 1.5 x 0.5 = 0.75 m²

Area of rectangle IV = (1.5 – 1.0) x 0.5 x 0.5 = 0.25 m²

Total area of shaded portion = (1.75 + 1.25 + 0.75 + 0.25) m² = 4 m²

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

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Suraj Kumar says

very good book

rahul kumar says

very nice solution of all lessons by rahul

SHUBHAM says

very nice sokutions this solutions are very samall and very easy to remember