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

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TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 17
Chapter NamePerimeter and Areas of Plane Figures
Exercise17 B
CategoryRS Aggarwal Solutions

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
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 1.1
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
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 3.1
By Pythagoras theorem, we have
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 3.2
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
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 4.1
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
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 5.1
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
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 6.1
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
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 7.1
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 = RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 8.1
= 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 = RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 9.1
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 – l= 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
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 11.1
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
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 12.1
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.
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 14.1
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 = RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17 Q14.1
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 14.2
(98x – 49) = 1911
98x = 1911 + 49
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 14.3
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
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 16.1
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 + 4x)] m2
=[40 – 40 + 26x – 4x] m2
= (26x – 4x) m2
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 16.2

Question 17:
Area of road ABCD
= ( 80 × 5 ) m2
= 400 m2
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 17.1
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) m= 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) m= 6 m2
Area of four walls to be painted = [Area of 4 walls – Area of two doors – Area of two windows]
= [312 – 6 – 6] m= 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
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 19.1
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
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 19.2

Question 20:
(i) Area of the square =  12(diagonal)2 Sq.unit
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 20.1
(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
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 21.1
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)
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 22.1
Time taken to cross the field along the diagonal
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 22.2
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
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 26.1
For area of ∆DBC
a = 29 cm, b = 21 cm, c = 20 cm
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 26.2

Question 27:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 27.1
Area of quad. ABCD = Area of ∆ABC + Area of ∆ACD
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 27.2
Now, we find area of a ∆ACD
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 27.3
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
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 28.1
For area of equilateral ∆DBC, we have
a = 26 cm
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17b 28.2
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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