Class 10: Maths Chapter 15 solutions. Complete Class 10 Maths Chapter 15 Notes.
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RS Aggarwal Solutions for Class 10 Maths Chapter 15–Perimeter and Areas of Plane Figures
RS Aggarwal 10th Maths Chapter 15, Class 10 Maths Chapter 15 solutions
Exercise 15A Solutions
Question 1:

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 = 1000058 m2
If the cost of sowing is Rs. 783, area = (1000058×783) m2
Area of the field = 135000 m2
Let the attitude of the field be x meters
Then, Base = 3x meter
Area of the field =


Hence, the altitude = 300m and the base = 900 m
Question 3:
Let a = 42 cm, b = 34 cm and c = 20 cm

(i) Area of triangle =


(ii) Let base = 42 cm and corresponding height = h cm
Then area of triangle =


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 =


(ii) Let base = 18 cm and altitude = x cm
Then, area of triangle =


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 = (150×530)m=25m
Length of the second side = (150×1230)m=60m
Length of third side = (150×1330)m=65m
Let a = 25 m, b = 60 m, c = 65 m

(s – a) = 50 cm, (s – b) = 15 cm, and (s – c) = 10 cm

Hence, area of the triangle = 750 m2
Question 6:
On dividing 540 m in ratio 25 : 17 : 12, we get
Length of one side = (540×2554)m=250m
Length of second side = (540×1754)m=170m
Length of third side = (540×1254)m=120m
Let a = 250m, b = 170 m and c = 120 m

Then, (s – a) = 29 m, (s – b) = 100 m, and (s – c) = 150m

The cost of ploughing 100 area is = Rs. 18. 80
The cost of ploughing 1 is = Rs. 18.80100
The cost of ploughing 9000 area = Rs. (18.80100×9000)
= 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

Hence, area of the triangle = 60 cm2
Question 8:
Let the sides containing the right angle be x cm and (x × 7) cm

One side = 15 cm and other = (15 × 7) cm = 8 cm

perimeter of triangle (15 + 8 + 17) cm = 40 cm
Question 9:
Let the sides containing the right angle be x and (x × 2) cm

One side = 8 cm, and other (8 × 2) cm = 6 cm
= 10 cm

Therefore, perimeter of the triangle = 8 + 6 + 10 = 24 cm
Question 10:
(i) Here a = 8 cm
Area of the triangle = (3√4×a2) Sq.unit

(ii) Height of the triangle= (3√4×a) Sq.unit

Hence, area = 27.71 cm2 and height = 6.93 cm
Question 11:
Let each side of the equilateral triangle be a cm

Question 12:
Let each side of the equilateral triangle be a cm
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 = 3√4a2

Height of equilateral triangle

Question 14:
Base of right angled triangle = 48 cm
Height of the right angled triangle =


Question 15:
Let the hypotenuse of right angle triangle = 6.5 m
Base = 6 cm

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

Hypotenuse = 2 ×(radius of circumcircle)
= (2 × 8) cm = 16 cm
Base = 16 cm, height = 6 cm
Area of right angled triangle

Hence, area of the triangle= 48 cm2
Question 17:
Let each side a = 13 cm and the base b = 20 cm

Hence, area of the triangle = 83.1 cm2.
Question 18:
Let each equal side be a cm in length.
Then,
Hence, hypotenuse = 28.28 cm and perimeter = 68.28 cm
Question 19:
Let each equal side be a cm and base = 80 cm

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 = 32a
perimeter = (2a + b) cm
Hence each side = 12 cm and Base = 3212 = 18cm
(ii) Area of triangle =


(iii) Height of the triangle =


Question 21:
Let the height be h cm, then a= (h + 2) cm and b = 12 cm

Squaring both sides,

Therefore, a = h + 2 = (8 + 2)cm = 10 cm

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

(ii) The base = 154 cm and let the perpendicular = h cm

Hence, required length of the perpendicular of the triangle is 36 m.
Question 23:

Area of shaded region = Area of ∆ABC – Area of ∆DBC
First we find area of ∆ABC

Second we find area of ∆DBC which is right angled

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.
Area of right isosceles triangle ABC

Hence, area = 50 cm2 and perimeter = 34.14 cm
Exercise 15B Solutions
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
RS Aggarwal Solutions for Class 10 Maths Chapter 15: Download PDF
RS Aggarwal Solutions for Class 10 Maths Chapter 15–Perimeter and Areas of Plane Figures
Chapterwise RS Aggarwal Solutions for Class 10 Maths :
- Chapter 1–Real Numbers
- Chapter 2–Polynomials
- Chapter 3–Linear Equations In Two Variables
- Chapter 4–Quadratic Equations
- Chapter 5–Arithmetic Progression
- Chapter 6–Coordinate Geometry
- Chapter 7–Triangles
- Chapter 8–Circles
- Chapter 9–Constructions
- Chapter 10–Trigonometric Ratios
- Chapter 11–T Ratios Of Some Particular Angles
- Chapter 12–Trigonometric Ratios Of Some Complementary Angles
- Chapter 13–Trigonometric Identities
- Chapter 14–Height and Distance
- Chapter 15–Perimeter and Areas of Plane Figures
- Chapter 16–Areas of Circle, Sector and Segment
- Chapter 17–Volume and Surface Areas of Solids
- Chapter 18–Mean, Median, Mode of Grouped Data
- Chapter 19–Probability
About RS Aggarwal Class 10 Book
Investing in an R.S. Aggarwal book will never be of waste since you can use the book to prepare for various competitive exams as well. RS Aggarwal is one of the most prominent books with an endless number of problems. R.S. Aggarwal’s book very neatly explains every derivation, formula, and question in a very consolidated manner. It has tonnes of examples, practice questions, and solutions even for the NCERT questions.
He was born on January 2, 1946 in a village of Delhi. He graduated from Kirori Mal College, University of Delhi. After completing his M.Sc. in Mathematics in 1969, he joined N.A.S. College, Meerut, as a lecturer. In 1976, he was awarded a fellowship for 3 years and joined the University of Delhi for his Ph.D. Thereafter, he was promoted as a reader in N.A.S. College, Meerut. In 1999, he joined M.M.H. College, Ghaziabad, as a reader and took voluntary retirement in 2003. He has authored more than 75 titles ranging from Nursery to M. Sc. He has also written books for competitive examinations right from the clerical grade to the I.A.S. level.
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