Class 10 Class 10 Maths RS Aggarwal Solutions Class 10

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

RS Aggarwal Solutions Class 10 Chapter 17
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RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures is a part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures.

Exercise 17A

Question 1:
rs-aggarwal-class-10-solutions-perimeter-and-areas-of-plane-figures-17a-q1

Question 2:
If the cost of sowing the field is Rs. 58, then area = 10000 m2
If the cost of sowing is Re. 1, area =  \frac { 10000 }{ 58 }   m2
If the cost of sowing is Rs. 783, area =  (\frac { 10000 }{ 58 } \times 783)    m2
Area of the field = 135000 m2
Let the attitude of the field be x meters
Then, Base = 3x meter
Area of the field = rs-aggarwal-class-10-solutions-perimeter-and-areas-of-plane-figures-17a-q2-1
rs-aggarwal-class-10-solutions-perimeter-and-areas-of-plane-figures-17a-q2-2
Hence, the altitude = 300m and the base = 900 m

Question 3:
Let a = 42 cm, b = 34 cm and c = 20 cm
rs-aggarwal-class-10-solutions-perimeter-and-areas-of-plane-figures-17a-q3-1
(i) Area of triangle = RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures ex 17a
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures ex 17a
(ii) Let base = 42 cm and corresponding height = h cm
Then area of triangle = RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures ex 17a
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures ex 17a
Hence, the height corresponding to the longest side = 16 cm

Question 4:
Let a = 18 cm, b = 24 cm, c = 30 cm
Then, 2s = (18 + 24 + 30) cm = 72 cm
s = 36 cm
(s – a) = 18cm, (s – b) = 12 cm and (s – c) = 6 cm
(i) Area of triangle = RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures ex 17a
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures ex 17a
(ii) Let base = 18 cm and altitude = x cm
Then, area of triangle = RS Aggarwal Class 10 Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17a
RS Aggarwal Class 10 Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17a
Hence, altitude corresponding to the smallest side = 24 cm

Question 5:
On dividing 150 m in the ratio 5 : 12 : 13, we get
Length of one side =  (150\times \frac { 5 }{ 30 } )m=25m  
Length of the second side =  (150\times \frac { 12 }{ 30 } )m=60m  
Length of third side =  (150\times \frac { 13 }{ 30 } )m=65m  
Let a = 25 m, b = 60 m, c = 65 m
RS Aggarwal Class 10 Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17a
(s – a) = 50 cm, (s – b) = 15 cm, and (s – c) = 10 cm
RS Aggarwal Class 10 Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17a
Hence, area of the triangle = 750 m2

Question 6:
On dividing 540 m in ratio 25 : 17 : 12, we get
Length of one side =  (540\times \frac { 25 }{ 54 } )m=250m  
Length of second side =  (540\times \frac { 17 }{ 54 } )m=170m  
Length of third side =  (540\times \frac { 12 }{ 54 } )m=120m  
Let a = 250m, b = 170 m and c = 120 m
RS Aggarwal Class 10 Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17a
Then, (s – a) = 29 m, (s – b) = 100 m, and (s – c) = 150m
RS Aggarwal Class 10 Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17a
The cost of ploughing 100 area is = Rs. 18. 80
The cost of ploughing 1 is =  Rs. \frac { 18. 80 }{ 100 }  
The cost of ploughing 9000 area = Rs. (\frac { 18. 80 }{ 100 } \times 9000)  
= Rs. 1692
Hence, cost of ploughing = Rs 1692.

Question 7:
Let the length of one side be x cm
Then the length of other side = {40 × (17 + x)} cm = (23 – x) cm
Hypotenuse = 17 cm
Applying Pythagoras theorem, we get
Solution of RS Aggarwal Class 10 Chapter 17 Perimeter and Areas of Plane Figures ex 17a
Hence, area of the triangle = 60 cm2

Question 8:
Let the sides containing the right angle be x cm and (x × 7) cm
Solution of RS Aggarwal Class 10 Chapter 17 Perimeter and Areas of Plane Figures ex 17a
One side = 15 cm and other = (15 × 7) cm = 8 cm
Solution of RS Aggarwal Class 10 Chapter 17 Perimeter and Areas of Plane Figures ex 17a
perimeter of triangle (15 + 8 + 17) cm = 40 cm

Question 9:
Let the sides containing the right angle be x and (x × 2) cm
Solution of RS Aggarwal Class 10 Chapter 17 Perimeter and Areas of Plane Figures ex 17a
One side = 8 cm, and other (8 × 2) cm = 6 cm
= 10 cm
RS Aggarwal Solutions Class 10 2017 Chapter 17 Perimeter and Areas of Plane Figures ex 17a
Therefore, perimeter of the triangle = 8 + 6 + 10 = 24 cm

Question 10:
(i) Here a = 8 cm
Area of the triangle =  (\frac { \sqrt { 3 }  }{ 4 } \times { a }^{ 2 })   Sq.unit
RS Aggarwal Solutions Class 10 2017 Chapter 17 Perimeter and Areas of Plane Figures ex 17a
(ii) Height of the triangle=  (\frac { \sqrt { 3 }  }{ 4 } \times { a })    Sq.unit
RS Aggarwal Solutions Class 10 2017 Chapter 17 Perimeter and Areas of Plane Figures ex 17a
Hence, area = 27.71 cm2 and height = 6.93 cm

Question 11:
Let each side of the equilateral triangle be a cm
RS Aggarwal Solutions Class 10 2017 Chapter 17 Perimeter and Areas of Plane Figures ex 17a

Question 12:
Let each side of the equilateral triangle be a cm
RS Aggarwal Solutions Class 10 2017 Chapter 17 Perimeter and Areas of Plane Figures ex 17a
Perimeter of equilateral triangle = 3a = (3 × 12) cm = 36 cm

Question 13:
Let each side of the equilateral triangle be a cm
area of equilateral triangle =  \frac { \sqrt { 3 }  }{ 4 } { a }^{ 2 }  
RS Aggarwal Solutions Class 10 2017 Chapter 17 Perimeter and Areas of Plane Figures ex 17a
Height of equilateral triangle
RS Aggarwal Solutions Class 10 2017 Chapter 17 Perimeter and Areas of Plane Figures ex 17a

Question 14:
Base of right angled triangle = 48 cm
Height of the right angled triangle = Class 10 RS Aggarwal Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17a
Class 10 RS Aggarwal Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17a

Question 15:
Let the hypotenuse of right angle triangle = 6.5 m
Base = 6 cm
Class 10 RS Aggarwal Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17a
Hence, perpendicular = 2.5 cm and area of the triangle =7.5 cm2

Question 16:
The circumcentre of a right triangle is the midpoint of the hypotenuse
Class 10 RS Aggarwal Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17a
Hypotenuse = 2 ×(radius of circumcircle)
= (2 × 8) cm = 16 cm
Base = 16 cm, height = 6 cm
Area of right angled triangle
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 17 Perimeter and Areas of Plane Figures ex 17a
Hence, area of the triangle= 48 cm2

Question 17:
Let each side a = 13 cm and the base b = 20 cm
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 17 Perimeter and Areas of Plane Figures ex 17a
Hence, area of the triangle = 83.1 cm2.

Question 18:
Let each equal side be a cm in length.
Then,
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 17 Perimeter and Areas of Plane Figures ex 17a
Hence, hypotenuse = 28.28 cm and perimeter = 68.28 cm

Question 19:
Let each equal side be a cm and base = 80 cm
RS Aggarwal Class 10 Book pdf download Chapter 17 Perimeter and Areas of Plane Figures ex 17a
perimeter of triangle = (2a + b) cm
= (2 ×41 + 80) cm
= (82 + 80) cm = 162 cm
Hence, perimeter of the triangle = 162 cm

Question 20:
Perimeter of an isosceles triangle = 42 cm
(i) Let each side be a cm, then base =  \frac { 3 }{ 2 } { a }  
perimeter = (2a + b) cm
RS Aggarwal Class 10 Book pdf download Chapter 17 Perimeter and Areas of Plane Figures ex 17a
Hence each side = 12 cm and Base = \frac { 3 }{ 2 } { 12 }   = 18cm
(ii) Area of triangle = RS Aggarwal Class 10 Book pdf download Chapter 17 Perimeter and Areas of Plane Figures ex 17a

Class 10 Maths RS Aggarwal Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17a
(iii) Height of the triangle = Class 10 Maths RS Aggarwal Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17a

Class 10 Maths RS Aggarwal Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17a

Question 21:
Let the height be h cm, then a= (h + 2) cm and b = 12 cm
Maths RS Aggarwal Class 10 Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17a
Squaring both sides,
Maths RS Aggarwal Class 10 Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17a
Therefore, a = h + 2 = (8 + 2)cm = 10 cm
Maths RS Aggarwal Class 10 Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17a
Hence, area of the triangle = 48 cm2.

Question 22:
Perimeter of triangle = 324 cm
(i) Length of third side = (324 – 85 – 154) m = 85 m
Let a = 85 m, b = 154 m, c = 85 m
RS Aggarwal Solutions Class 10 2018 Chapter 17 Perimeter and Areas of Plane Figures ex 17a
(ii) The base = 154 cm and let the perpendicular = h cm
RS Aggarwal Solutions Class 10 2018 Chapter 17 Perimeter and Areas of Plane Figures ex 17a
Hence, required length of the perpendicular of the triangle is 36 m.

Question 23:
RS Aggarwal Solutions Class 10 2018 Chapter 17 Perimeter and Areas of Plane Figures ex 17a
Area of shaded region = Area of ∆ABC – Area of ∆DBC
First we find area of ∆ABC
RS Aggarwal Solutions Class 10 2018 Chapter 17 Perimeter and Areas of Plane Figures ex 17a
Second we find area of ∆DBC which is right angled
rs-aggarwal-class-10-solutions-perimeter-and-areas-of-plane-figures-17a-q23-3
Area of shaded region = Area of ∆ABC – Area of ∆DBC
= (43.30 – 24) = 19. 30
Area of shaded region = 19.3

Question 24:
Let ∆ABC is a isosceles triangle. Let AC, BC be the equal sides
Then AC = BC = 10cm. Let AB be the base of ∆ABC right angle at C.
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures ex 17a
Area of right isosceles triangle ABC
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures ex 17a
Hence, area = 50 cm2 and perimeter = 34.14 cm

Exercise 17B

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 ex 17b
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 ex 17b
By Pythagoras theorem, we have
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures ex 17b
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 ex 17b
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 ex 17b
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 Class 10 Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17b
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 Class 10 Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17b
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 Class 10 Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17b
= 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 Class 10 Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17b
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
Solution of RS Aggarwal Class 10 Chapter 17 Perimeter and Areas of Plane Figures ex 17b
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
Solution of RS Aggarwal Class 10 Chapter 17 Perimeter and Areas of Plane Figures ex 17b
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.
Solution of RS Aggarwal Class 10 Chapter 17 Perimeter and Areas of Plane Figures ex 17b
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 2017 Chapter 17 Perimeter and Areas of Plane Figures ex 17b
RS Aggarwal Solutions Class 10 2017 Chapter 17 Perimeter and Areas of Plane Figures ex 17b
(98x – 49) = 1911
98x = 1911 + 49
RS Aggarwal Solutions Class 10 2017 Chapter 17 Perimeter and Areas of Plane Figures ex 17b
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 = (\frac { 13.2 }{ 0.80 })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 Class 10 Maths Book Solutions pdf Chapter 17 Perimeter and Areas of Plane Figures ex 17b
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 Class 10 Maths Book Solutions pdf Chapter 17 Perimeter and Areas of Plane Figures ex 17b

Question 17:
Area of road ABCD
= ( 80 × 5 ) m2
= 400 m2
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 17 Perimeter and Areas of Plane Figures ex 17b
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 Class 10 Maths Book Solutions pdf Chapter 17 Perimeter and Areas of Plane Figures ex 17b
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 Class 10 Book pdf download Chapter 17 Perimeter and Areas of Plane Figures ex 17b

Question 20:
(i) Area of the square =  \frac { 1 }{ 2 } { (diagonal) }^{ 2 }   Sq.unit
RS Aggarwal Class 10 Book pdf download Chapter 17 Perimeter and Areas of Plane Figures ex 17b
(ii) Side of the square = \sqrt { 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 =  \frac { 1 }{ 2 } { (diagonal) }^{ 2 }   Sq.unit
Let diagonal of square be x
RS Aggarwal Class 10 Book pdf download Chapter 17 Perimeter and Areas of Plane Figures ex 17b
Length of diagonal = 16 m
Side of square =  \sqrt { 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 =  \frac { 1 }{ 2 } { (diagonal) }^{ 2 }   Sq.unit = 80000 m2 (given)
Class 10 Maths RS Aggarwal Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17b
Time taken to cross the field along the diagonal
Class 10 Maths RS Aggarwal Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17b
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 = \frac { 10000 }{ 180 }   m2
Rs. 1620 is the cost of harvesting an area = (\frac { 10000 }{ 180 } \times 1620)    m2
Area = 90000 m2
Area of square = (side)2 = 90000 m2
side = \sqrt { 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 =  \frac { 28000 }{ 14 }   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\times \frac { 54 }{ 100 } )  = Rs. 135000

Question 25:
Largest possible size of square tile = HCF of 525 cm and 378 cm
= 21 cm
Number of tiles =  \frac { Area\quad of\quad rectangle }{ Area\quad of\quad square\quad tiles }    
=  \frac { (525\times 378) }{ (21\times 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
Maths RS Aggarwal Class 10 Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17b
For area of ∆DBC
a = 29 cm, b = 21 cm, c = 20 cm
Maths RS Aggarwal Class 10 Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17b

Question 27:
Maths RS Aggarwal Class 10 Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17b
Area of quad. ABCD = Area of ∆ABC + Area of ∆ACD
Maths RS Aggarwal Class 10 Solutions Chapter 17 Perimeter and Areas of Plane Figures ex 17b
Now, we find area of a ∆ACD
RS Aggarwal Solutions Class 10 2018 Chapter 17 Perimeter and Areas of Plane Figures ex 17b
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 2018 Chapter 17 Perimeter and Areas of Plane Figures ex 17b
For area of equilateral ∆DBC, we have
a = 26 cm
RS Aggarwal Solutions Class 10 2018 Chapter 17 Perimeter and Areas of Plane Figures ex 17b
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

We hope the given Solutions of RS Aggarwal Solutions Class 10 Chapter 17 are helpful to complete your math homework. If you have any doubts, please comment below. We will try to provide online math tutoring for you.


RS Aggarwal Solutions Class 10

  1. Real Numbers
  2. Polynomials
  3. Linear equations in two variables
  4. Triangles
  5. Trigonometric Ratios
  6. T-Ratios of Some Particular Angles
  7. Trigonometric Identities
  8. Trigonometric Ratios of Complementary Angles
  9. Mean, Median, Mode of Grouped Data, of Grouped Data Ex.9A Ex.9B Ex.9C and 9D Ex.9E
  10. Quadratic Equations Ex.10A Ex.10B Ex.10C Ex.10D
  11. Arithmetic Progressions
  12. Circles
  13. Constructions
  14. Height and Distance
  15. Probability
  16. Coordinate Geometry
  17. Perimeter and Areas of Plane Figures
  18. Areas of Circle, Sector and Segment
  19. Volume and Surface Areas of Solids

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