In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17 B Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17 B Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17 B Maths book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 17 |

Chapter Name | Perimeter and Areas of Plane Figures |

Exercise | 17 B |

Category | RS 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

Length of the rectangle is 21. 5 meter

Area of the rectangular plot = length × breadth = ( 16 × 21.5 ) m^{2} = 344 m^{2}

The length = 21.5 m and the area = 344 m^{2}

**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 ) m^{2} = 39200 m^{2}

Hence, area of the park = 39200 m^{2}

**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) cm^{2} = 420 cm^{2}

Hence, the other side = 35 cm and the area = 420 cm^{2}

**Question 4:**

Let the breadth of the plot be x meter

Area = Length × Breadth = (28 × x) meter

= 28x m^{2}

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) m^{2} = (15x^{2}) m^{2}

Area of lawn = 3375 m^{2}

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) m^{2}

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) m^{2}

Area of carpet = Length × Breadth

= (2.5 x 0.8) m^{2}

Number of carpets =

= 216

Hence the number of carpet pieces required = 216

**Question 9:**

Area of verandah = (36 × 15) m^{2} = 540 m^{2}

Area of stone = (0.6 × 0.5) m^{2 } [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 m^{2}

l × (28 – l) = 192

28l – l^{2 }= 192

l^{2} – 28l + 192 = 0

l^{2} – 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) m^{2} = 630 m^{2}

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) m^{2} = 390 m^{2}

Area of path without grass = Area of the whole park – area of park with grass

= 630 – 390 = 240 m^{2}

Hence, area of the park to be laid with grass = 240 m^{2}

**Question 12:**

Length of the plot = 125 m

Breadth of the plot = 78 m

Area of plot ABCD = (125 × 78) m^{2} = 9750 m^{2}

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) m^{2} = 11004 m^{2}

Area of path = Area of plot PQRS – Area of plot ABCD

= (11004 – 9750) m^{2}

= 1254 m^{2}

Cost of gravelling = Rs 75 per m^{2}

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) m^{2}

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 – 4x^{2} = 420

178x – 4x^{2} = 420

4x^{2} – 178x + 420 = 0

2x^{2} – 89x + 210 = 0

2x^{2} – 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) m^{2} = 45 x^{2}m^{2}

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) m^{2}

= 4.4 × 3.0 = 13.2 m^{2}

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) m^{2}

Area of floor PQRS without border

= (8 – 2x)(5 – 2x)

= 40 – 16x – 10x + 4x^{2}

= 40 – 26x + 4x^{2}

Area of border = Area of floor ABCD – Area of floor PQRS

= [40 – (40 – 26x + 4x^{2 })] m^{2}

=[40 – 40 + 26x – 4x^{2 }] m^{2}

= (26x – 4x^{2 }) m^{2}

**Question 17:**

Area of road ABCD

= ( 80 × 5 ) m^{2}

= 400 m^{2}

Area of road EFGH

= ( 64 × 5 ) m^{2}

= 320 m^{2}

Area of common road PQRS

= ( 5 × 5 ) m^{2}

= 25 m^{2}

Area of the road to be gravelled

=(400 + 320 – 25) m^{2} = 695 m^{2}

Cost of gravelling the roads

=Rs. (695 × 24) m^{2 }= Rs. 16680

**Question 18:**

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

= 2(14 + 10) × 6.5 = 2 × 24 × 6.5

= 312 m^{2}

Area of two doors = 2 × (2.5 × 1.2) m^{2} = 6 m^{2}

Area of four windows = 4 (1.5 × 1) m^{2 }= 6 m^{2}

Area of four walls to be painted = [Area of 4 walls – Area of two doors – Area of two windows]

= [312 – 6 – 6] m^{2 }= 300 m^{2}

Cost of painting the walls = Rs 38 per m^{2}

Cost of painting 300 m^{2} of walls = Rs 38 x 300

= Rs. 11400

**Question 19:**

Cost of papering the wall at the cost of Rs. 30 m^{2} 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 m^{2} 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 m^{2} (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 m^{2}

Re 1 is the cost of harvesting an area = 10000180 m^{2}

Rs. 1620 is the cost of harvesting an area = (10000180×1620) m^{2}

Area = 90000 m^{2}

Area of square = (side)^{2} = 90000 m^{2}

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 m^{2}

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) cm^{2}

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 ) cm^{2} = 114 cm^{2}

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) cm^{2} = 412.37 cm^{2}

Perimeter ABCD = AD + AB + BC + CD

= 24 cm + 10 cm + 26 cm + 26 cm

= 86 cm

**All Chapter RS Aggarwal Solutions For Class 10 Maths**

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