RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17 B

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RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17 B

Question 1:
Let the length of plot be x meters
Its perimeter = 2 [length + breadth]
=2(x + 16) = (2x + 32) meters

Length of the rectangle is 21. 5 meter
Area of the rectangular plot = length × breadth = ( 16 × 21.5 ) m² = 344 m²
The length = 21.5 m and the area = 344 m²

Question 2:
Let the breadth of a rectangular park be x meter
Then, its length = 2x meter
∴ perimeter = 2(length + breadth)
=2(2x + x) = 6x meters
∴ 6x = 840 m [ ∵ 1 km = 1000 m]
⇒ x = 140 m
Then, breadth = 140 m and length = 280 m
Area of rectangular park = (length × breadth) = ( 140 × 280 ) m² = 39200 m²
Hence, area of the park = 39200 m²

Question 3:
Let ABCD be the rectangle in which AB = 12 cm and AC = 37 m

By Pythagoras theorem, we have

Thus, length = 35 cm and breadth = 12 cm
Area of rectangle = (12 × 35) cm² = 420 cm²
Hence, the other side = 35 cm and the area = 420 cm²

Question 4:
Let the breadth of the plot be x meter
Area = Length × Breadth = (28 × x) meter
= 28x m²

Breadth of plot is = 16. 5 m
Perimeter of the plot is = 2(length + breadth)
= 2 (28 + 16.5 ) m = 2 ( 44.5) m = 89 m

Question 5:
Let the breadth of rectangular hall be x m
Then, Length = (x + 5) m

Breadth = 25 m and length = (25 + 5) m = 30 m
Perimeter of rectangular hall = 2(length + breadth)
= 2(30 + 25)m = (2 × 55) m = 110 m

Question 6:
Let the length of lawn be 5x m and breadth of the lawn be 3x m
Area of rectangular lawn = (5x × 3x) m² = (15x²) m²
Area of lawn = 3375 m²

Length = 5 × 15 = 75
Breadth = (3 × 15)m = 45 m
Perimeter of lawn = 2(length + breadth)
=2 (75 + 45)m = 240 m
Cost of fencing the lawn per meter = Rs. 8.50 per meter
Cost of fencing the lawn = Rs 8. 50 × 240 = Rs. 2040

Question 7:
Length of the floor = 16 m
Breadth of the floor = 13.5 m
Area of floor = (16 x 13.5) m²

Cost of carpet = Rs. 15 per meter
Cost of 288 meters of carpet = Rs. (15 × 288) = Rs. 4320

Question 8:
Area of floor = Length × Breadth
= (24 x 18) m²
Area of carpet = Length × Breadth
= (2.5 x 0.8) m²
Number of carpets = 
= 216
Hence the number of carpet pieces required = 216

Question 9:
Area of verandah = (36 × 15) m² = 540 m²
Area of stone = (0.6 × 0.5) m²   [10 dm = 1 m]
Number of stones required = 
Hence, 1800 stones are required to pave the verandah.

Question 10:
Perimeter of rectangle = 2(l + b)
2(l + b) = 56 ⇒ l + b = 28 cm
b = (28 – l) cm
Area of rectangle = 192 m²
l × (28 – l) = 192
28l – l² = 192
l² – 28l + 192 = 0
l² – 16l – 12l + 192 = 0
l(l – 16) – 12(l – 16) = 0
(l – 16) (l – 12) = 0
l = 16 or l = 12
Therefore, length = 16 cm and breadth = 12 cm

Question 11:
Length of the park = 35 m
Breadth of the park = 18 m

Area of the park = (35 × 18) m² = 630 m²
Length of the park with grass =(35 – 5) = 30 m
Breadth of the park with grass = (18- 5) m = 13 m
Area of park with grass = (30 × 13) m² = 390 m²
Area of path without grass = Area of the whole park – area of park with grass
= 630 – 390 = 240 m²
Hence, area of the park to be laid with grass = 240 m²

Question 12:
Length of the plot = 125 m
Breadth of the plot = 78 m

Area of plot ABCD = (125 × 78) m² = 9750 m²
Length of the plot including the path = (125 + 3 + 3) m = 131 m
Breadth of the plot including the path = (78 + 3 + 3) m = 84 m
Area of plot PQRS including the path
= (131 × 84) m² = 11004 m²
Area of path = Area of plot PQRS – Area of plot ABCD
= (11004 – 9750) m²
= 1254 m²
Cost of gravelling = Rs 75 per m²
Cost of gravelling the whole path = Rs. (1254 × 75) = Rs. 94050
Hence, cost of gravelling the path = Rs 94050

Question 13:
Area of rectangular field including the foot path = (54 × 35) m²
Let the width of the path be x m
Then, area of rectangle plot excluding the path = (54 × 2x) × (35 × 2x)
Area of path = (54 × 35) + (54 × 2x) (35 × 2x)
(54 × 35) + (54 × 2x) (35 × 2x) = 420
1890 – 1890 + 108x + 70x – 4x² = 420
178x – 4x² = 420
4x² – 178x + 420 = 0
2x² – 89x + 210 = 0
2x² – 84x – 5x + 210 = 0
2x(x – 42) – 5(x – 42) = 0
(x – 42) (2x – 5) = 0

Question 14:
Let the length and breadth of a rectangular garden be 9x and 5x.

Then, area of garden = (9x × 5x) m² = 45 x²m²
Length of park excluding the path = (9x – 7) m
Breadth of the park excluding the path = (5x – 7) m
Area of the park excluding the path = (9x – 7)(5x – 7)
Area of the path = 

(98x – 49) = 1911
98x = 1911 + 49

Length = 9x = 9 × 20 = 180 m
Breadth = 5x = 5 × 20 = 100 m
Hence, length = 180 m and breadth = 100 m

Question 15:
Area of carpet = (4.9 – 0.5) (3.5 – 0.5) m²
= 4.4 × 3.0 = 13.2 m²
Length of the carpet = (13.20.80)m  = 16.5m
Cost of carpet = Rs. 40 per meter
Cost of 16.5 m carpet = Rs. (40 × 16.5) = Rs. 660

Question 16:
Let the width of the carpet = x meter

Area of floor ABCD = (8 × 5) m²
Area of floor PQRS without border
= (8 – 2x)(5 – 2x)
= 40 – 16x – 10x + 4x²
= 40 – 26x + 4x²
Area of border = Area of floor ABCD – Area of floor PQRS
= [40 – (40 – 26x + 4x² )] m²
=[40 – 40 + 26x – 4x² ] m²
= (26x – 4x² ) m²

Question 17:
Area of road ABCD
= ( 80 × 5 ) m²
= 400 m²

Area of road EFGH
= ( 64 × 5 ) m²
= 320 m²
Area of common road PQRS
= ( 5 × 5 ) m²
= 25 m²
Area of the road to be gravelled
=(400 + 320 – 25) m² = 695 m²
Cost of gravelling the roads
=Rs. (695 × 24) m² = Rs. 16680

Question 18:
Area of four walls of room = 2(l + b) × h
= 2(14 + 10) × 6.5 = 2 × 24 × 6.5
= 312 m²
Area of two doors = 2 × (2.5 × 1.2) m² = 6 m²
Area of four windows = 4 (1.5 × 1) m² = 6 m²
Area of four walls to be painted = [Area of 4 walls – Area of two doors – Area of two windows]
= [312 – 6 – 6] m² = 300 m²
Cost of painting the walls = Rs 38 per m²
Cost of painting 300 m² of walls = Rs 38 x 300
= Rs. 11400

Question 19:
Cost of papering the wall at the cost of Rs. 30 m² per in Rs. 7560

Let h meter be the height and b m be the breadth of the room
Length of the room = 12 m
Area of four walls = 2 ×(12 + b) × h
2(12 + b) × h = 252
Or (12 + b) h = 126 —–(1)
The cost of covering the floor with mat at the cost of Rs. 15 per m² is Rs. 1620

Question 20:
(i) Area of the square =  12(diagonal)2 Sq.unit

(ii) Side of the square = 288−−−√m = 16.97 m
Perimeter of the square = (4 × side) units
= (4 × 16.97)m
= 67.88 m

Question 21:
Area of the square =  12(diagonal)2 Sq.unit
Let diagonal of square be x

Length of diagonal = 16 m
Side of square =  128−−−√m = 11.31 m
Perimeter of square = [4 × side] sq. units
=[ 4 × 11.31] cm = 45.24 cm

Question 22:
Let d meter be the length of diagonal
Area of square field =  12(diagonal)2 Sq.unit = 80000 m² (given)

Time taken to cross the field along the diagonal

Hence, man will take 6 min to cross the field diagonally.

Question 23:
Rs. 180 is the cost of harvesting an area = 1 hectare = 10000 m²
Re 1 is the cost of harvesting an area = 10000180 m²
Rs. 1620 is the cost of harvesting an area = (10000180×1620) m²
Area = 90000 m²
Area of square = (side)² = 90000 m²
side = 90000−−−−−√m = 300 m
Perimeter of square = 4 × side = 4 × 300 = 1200 m
Cost of fencing = Rs 6.75 per meter.
Cost of fencing 1200 m long border = 1200 × Rs 6.75 = Rs. 8100

Question 24:
Rs. 14 is the cost of fencing a length = 1m
Rs. 28000 is the cost of fencing the length =  2800014 m = 2000 m
Perimeter = 4 × side = 2000
side = 500 m
Area of a square = (side)²  = (500)²  m
= 250000 m²
Cost of mowing the lawn = Rs. (250000×54100) = Rs. 135000

Question 25:
Largest possible size of square tile = HCF of 525 cm and 378 cm
= 21 cm
Number of tiles =  AreaofrectangleAreaofsquaretiles
=  (525×378)(21×21) cm²
Number of tiles = 450

Question 26:
Area of quad. ABCD = Area of ∆ABD + Area of ∆DBC
For area of ∆ABD
Let a = 42 cm, b = 34 cm, and c = 20 cm

For area of ∆DBC
a = 29 cm, b = 21 cm, c = 20 cm

Question 27:

Area of quad. ABCD = Area of ∆ABC + Area of ∆ACD

Now, we find area of a ∆ACD

Area of quad. ABCD = Area of ∆ABC + Area of ∆ACD
= ( 60+54 ) cm² = 114 cm²
Perimeter of quad. ABCD = AB + BC + CD + AD
=(17 + 8 + 12 + 9) cm
= 46 cm
Perimeter of quad. ABCD = 46 cm

Question 28:
ABCD be the given quadrilateral in which AD = 24 cm, BD = 26 cm, DC = 26 cm and BC = 26 cm
By Pythagoras theorem

For area of equilateral ∆DBC, we have
a = 26 cm

Area of quad. ABCD = Area of ∆ABD + Area of ∆DBC
= (120 + 292.37) cm² = 412.37 cm²
Perimeter ABCD = AD + AB + BC + CD
= 24 cm + 10 cm + 26 cm + 26 cm
= 86 cm

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