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OP Malhotra Class 9 ICSE Maths Solutions Chapter 16 Area of Plane Figures
Welcome to NCTB Solutions. Here with this post we are going to help 9th class students for the Solutions of OP Malhotra Class 9 ICSE Math Book, Chapter 16, Area of Plane Figures. Here students can easily find step by step solutions of all the problems for Area of Plane Figures, Exercise 16a, 16b, 16c and 16d Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 16 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.
Area of Plane Figures Exercise 16(a) Solution :
Question no – (1)
Solution :
(i) Rectangle dimensions are = 8 cm by 5 cm
∴ Perimeter of rectangle,
= 2 (l + b)
= 2 (8 + 5)
= (2 × 13) cm
= 26 cm
∴ Area of rectangle,
= (8 × 5) cm2
= 40 cm2
Thus, the perimeter of the rectangle is 26 cm and area is 40 cm2
(ii) Rectangle whose dimensions are = 12 m by 9 m
∴ Perimeter of rectangle,
= 2 (l + b)
= 2 (12 + 9)
= 2 × 21
= 42 cm
∴ Area of rectangle,
= (12 × 9) m2
= 108 m2
Hence, the perimeter of the rectangle will be 42 cm and Area will be 108 m2
(iii) Rectangle whose dimensions are = 3.5 m by 2 m
∴ Perimeter of rectangle,
= 2 (3.5 + 2) m
= (2 × 5.5) m
= 11 m
∴ Area of rectangle,
= (3.5 × 2)
= 7 m2
Hence, the Perimeter of the rectangle is 11 m and Area is 7 m2
Question no – (2)
Solution :
(i) Given, Square whose each side is 3 cm.
First, Perimeter of the square,
= (4 × 3)
= 12 cm
Now, Area of the square,
= 32
= 9 cm2
Thus, the perimeter of the square will be 12 cm and Area will be 9 cm2
(ii) Given, Square whose each side is 5 m
∴ Perimeter of the square,
= (4 × 5)
= 22 cm
∴ Area of the square,
= 52
= 25 cm2
Hence, the perimeter of the square will be 22 cm and Area will be 25 cm2
(iii) Given, Square whose each side is 1.3 m.
∴ Perimeter of the square,
= (4 × 1.3) m
= 5.2 m
∴ Area of the square,
= (1.3)2 m2
= 1.69 m2
Hence, the Perimeter of the square will be 5.2 m and Area will be 1.69 m2
(iv) Square whose each side is 2.4 m
∴ Perimeter of the square,
= (4 × 2.4) m
= 9.6 m
∴ Area of the square,
= (2.4)2
= 5.76 m2
Therefore, the perimeter of the square will be 9.6 m and Area will be 5.76 m2
Question no – (3)
Solution :
(i) Square whose area = 25 m2
Let, length of one side = a
= a2 = 25
= a = 5 m
∴ Perimeter of the square,
= (4 × 5)
= 20 m
Thus, the length of one side will be 5 m and perimeter will be 20 m.
(ii) Square whose area = 144 m2
Let, length of one side = a
= a2 = 144
= a = 12 m
∴ Perimeter of square,
= (4 × 12)
= 48 m
Thus, the length of one side is 12 m and perimeter is 48 m.
(iii) Given, Square whose area = 256 cm2
Let, length of one side = a
= a2 = 256
= a = 16 m
∴ Perimeter of the square,
= (4 × 16)
= 126 dm
Therefore, the length of one side will be 16 m and perimeter will be 126 dm.
(iv) Square whose area = 961 dm2
Let, length of one side = a
= a2 = 961
= a = 19
∴ Perimeter of the square,
= (4 × 19)
= 126 dm
Therefore, the length of one side will be 19 dm and perimeter will be 126 dm.
Question no – (4)
Solution :
As per the question we know,
Perimeter of a rectangle is = 30 cm,
Breadth is = 6 cm
Let, length = l
∴ 2 (l + 6) = 30
= 2l + 12 = 30
= 2l = 30 – 12
= l = 18/2
= l = 9 cm
∴ Area of the rectangle,
= (9 × 6)
= 54 cm2
Therefore, Its length will be 9 cm and are will be 54 cm2
Question no – (5)
Solution :
Area of a rectangular courtyard, 30 m long, is 450 m²
Let, Length = l
= 2 (l + 6) = 30
= l + 6 = 15
= l = 15 – 6
= 9 cm
and, Let, breadth = b
= 30 × b = 450
= b = 450/30
= 15 m
Therefore, its breadth will be 15 m.
Question no – (6)
Solution :
Let, width is = x km
= length = 2x
∴ Perimeter of rectangular filed,
= 2 (2x + x) = 3/5
= 2 × 3x = 3/5
= 30x = 3
= x = 3/30
= 1/20
∴ Length of rectangular filed,
= 2 × 1/10
= 1/5 km
∴ Area of rectangular filed,
= (1/10 × 1/5) km2
= 1/50 km2
Question no – (7)
Solution :
According to the question,
Let, the side of square = a
∴ Side of new square = 20
∴ Area of 1st square = a2
∴ Area of new,
= (2a)2
= 4a2
∴ Ratio,
= 4a2 : a2
= 4 : 1
Therefore, the ratio of the area will be 4 : 1
Question no – (8)
Solution :
Here, Area of square field,
= (89)2 cm2
= 7921 m2
Thus, Area of short of,
= (10,000 – 7921) m2 (we know, 1 hc = 10,000 s.m)
= 2079 m2
Therefore, its area fall short of 2079 m2
Question no – (9)
Solution :
Here, Diagonal,
= √252 + 252
= √625 + 625
= √1250
= 35.35
Hence, the length of its diagonal will be 35.35 cm.
Question no – (10)
Solution :
Here, (10)2 = 62
= 100 = 36 + l2
= l2 = 100 – 36
= l2 = 100 – 36
= 64
= l = 8 cm
∴ Area of the rectangle,
= (8 × 6) cm2
= 48 cm2
Therefore, its area of the rectangle is 48 cm2
Question no – (11)
Solution :
Let, breadth = b
Length = 2b
∴ (2b)2 + b2 = (16 √5)2
= 4b2 + b2 = 256 × 5
= b2 = 256
= b = 16
∴ Length of the rectangle,
= (2 × 16) cm2
= 32 cm2
∴ Area of the rectangle,
= (16 × 32)
= 512 cm2
Question no – (12)
Solution :
Here, Let, side = 2x and 3x
Perimeter,
= (4 × 2x)
= 8x
and, 2nd, perimeter
= (4 × 3x)
= 12x
∴ Ratio,
= 8x : 12x
= 8 : 12
= 2 : 3
∴ Ratio of Area,
= (2x)2 : (3x)2
= 4x2 : 9x2
= 4 : 9
Therefore, the ratio of their perimeters and areas will be 4 : 9
Question no – (13)
Solution :
Here, Area of the floor,
= (30 × 12)m2
= 360 m2
∴ Area of carpet,
= (3 × 2)m2
= 6m2
∴ Number of carpet,
= 360/6
= 60
Therefore, 60 carpet will be required to cover the floor of a hall.
Question no – (14)
Solution :
As per the question, room is 5 m by 4 m
∴ Area of room,
= (5 × 4)m2
= 20 m2
Now, Cost of cementing,
= (20 × 7.50) Rs
= 150 Rs
Therefore, the cost of cementing its floor will be 150 Rs
Question no – (15)
Solution :
Area of the floor,
= (12 × 10)m2
= 120m2
and, Brick Area,
= (20 × 6) cm2
= 120
∴ No of bricks required,
= 120 × 1000/120
= 10,000 bricks
Now, Cost of 10,000 bricks,
= 300 × 10,000/1000
= 3000 Rs
Question no – (16)
Solution :
Let, width = x
∴ L × B = 84 × 84
= 144 × Bb = 84 × 84
= b = 84 × 84/144
= 49 m
Therefore, the width of the rectangular plot will be 49 m.
Question no – (17)
Solution :
In the given question,
A plot 110 metres long and 80 metres broad
Covered with grass leaving 5 metres all around
∴ Area to be laid with grams,
= (110 – 10) (80 – 10)
= (100 × 70) m2
= 7000 m2
Therefore, the area to be laid with grass will be 7000 m2
Question no – (18)
Solution :
Let, the ratio 6x and 5x
= 6x × 5x = 27000
= 30x2 = 27000/30
= 900
= x2 = + 30
= 30
∴ l = 6 × 30 = 180
= b = 5 × 30 = 150
∴ Perimeter,
= 2 (180 + 150) m
= 2 × 330
= 660 m
∴ Total cost,
= (660 × 1.25) Rs
= 825 Rs
Thus, the cost of constructing a fence around the field will be 825 Rs.
Question no – (20)
Solution :
Let, Length be a m
∴ a2 + a2 = 196
= 2a2 = 196
= a2 = 98
∴ Area = 98 m2
∴ Perimeter,
= 4√98
= 39.59
Question no – (21)
Solution :
Area of lawn,
= (10 × 8)m2
= 80 m2
With path area of lawn,
= (12 × 10)m2
= 120m2
∴ Path,
= (120 – 80) m2
= 40 m2
Therefore, the area of path will be 40 m2
Question no – (22)
Solution :
According to the question,
Rectangular field 50 m long and 35 m wide
∴ Area of the shaded portion,
= (50 × 35) – (50 – 5) (35 – 5)
= 1750 – (45 × 30)
= (1750 – 1350) m2
= 400 m2
Thus, the area of the shaded portion will be 400 m2
Question no – (23)
Solution :
Let, Length be = m
∴ a2 + a2 = 196
= 2a2 = 196
= a2 = 98
∴ Area = 98 m2
∴ Perimeter
= 4√98
= 39.59
Question no – (24)
Solution :
Area of paper,
= (40 × 60)m2
= 2400cm2
∴ Area of walls,
= 2 (l + b) h
= 2 (6 + 4) × 3
= 60 m2
= 600000 cm2
Area of door
= 11m2
= 110000 cm2
∴ Remain part,
= (60,000 – 11000)
= 490000 cm2
∴ Piece of paper need,
= 490000 cm2
= 204.16
∴ Costing,
= (204.16 × 1.40) Rs
= 285 Rs
Hence, the cost of the paper for the walls will be 285 Rs.
Question no – (25)
Solution :
Here, Area of wet surface,
= {2(6 + 4) × 1.25 + 6 × 4}m2 ….(according to the question)
= (2 × 10 × 1.25 + 24) m2
= 49 m2
Therefore, the area of the wet surface will be 49 m2
Area of Plane Figures Exercise 16(b) Solution :
Question no – (1)
Solution :
(i) Area of triangle is 6 cm2 and its base is 4 cm
∴ Area
= 1/2 × b × h
= h = Area × 2/b
= 2 × 6/4
= 3 cm
Therefore, it’s height will be 3 cm.
(ii) Triangle sides are = 26 cm, 28 cm, 30 cm
Here, S = 26 + 28 + 30/2
= 84/2
= 42 cm
∴ Area,
= √42(42 – 36) (42 – 28) (42 – 30)
= √42 × 16 × 14 × 12
= 336 cm2
Hence, the area of the triangle is 336 cm2
(iii) Triangle sides are = 48 cm, 73 cm, 55 cm
Here, S = a + b + c/2
= 48 + 73 + 55/2
= 176/2
= 88
∴ Area,
= √88 (88 – 48) (88 – 73) (88 – 55)
= √88 × 40 × 15 × 33 cm2
= √1742400 cm2
= 1320 cm2
Thus, the area of the triangle will be 1320 cm2
(iv) Triangle sides are 21 cm, 20 cm, 13 cm
Here, Side = a + b + c/2
= 21 + 20 + 13/2
= 54/2
= 27
∴ Area,
= √27 (27 – 21) (27 – 20) (27 – 13)
= √27 × 6 × 7 × 14
= √15876
= 126
(v) Triangle whose sides are = 7.5 cm, 18 cm, 19.5 cm
Here, Side = 7.5 + 1.8 + 19.5/2
= 45/2
= 22.5
∴ Area,
= √22.5 (22.5 – 7.5) (22.5 – 18) (22.5 – 19.5)
= √22.5 × 15 × 4.5 × 3
= 67.5 cm2
Hence, the area of the triangle will be 67.5 cm2
Question no – (2)
Solution :
In the question,
Area of triangle is 6 cm2 and its base is 4 cm
∴ Area,
= 1/2 × b × h
= h = Area × 2/b
= 2 × 6/4
= 3 cm
Therefore, it’s height will be 3 cm.
Question no – (3)
Solution :
As we know that,
1 are = 10 m2
1 hectare = 10000 m2
1 decameter = 10 m
Area = 25 × 100 = 2500 m2
h = 20 m
∴ A = 1/2 × b × h
= bh/2 = A
= b = 2A/h
= 2 × 2500/20
= 250 m
Question no – (4)
Solution :
(i) Triangle sides are = 26 cm, 28 cm, 30 cm
Here, S = 26 + 28 + 30/2
= 84/2
= 42 cm
∴ Area,
= √42(42 – 36) (42 – 28) (42 – 30)
= √42 × 16 × 14 × 12
= 336 cm2
Hence, the area of the triangle is 336 cm2
(ii) Triangle sides are = 48 cm, 73 cm, 55 cm
Here, S = a + b + c/2
= 48 + 73 + 55/2
= 176/2
= 88
∴ Area,
= √88 (88 – 48) (88 – 73) (88 – 55)
= √88 × 40 × 15 × 33 cm2
= √1742400 cm2
= 1320 cm2
Thus, the area of the triangle will be 1320 cm2
(iii) Triangle sides are 21 cm, 20 cm, 13 cm
Here, Side = a + b + c/2
= 21 + 20 + 13/2
= 54/2
= 27
∴ Area,
= √27 (27 – 21) (27 – 20) (27 – 13)
= √27 × 6 × 7 × 14
= √15876
= 126
(iv) Triangle whose sides are = 7.5 cm, 18 cm, 19.5 cm
Here, Side = 7.5 + 1.8 + 19.5/2
= 45/2
= 22.5
∴ Area,
= √22.5 (22.5 – 7.5) (22.5 – 18) (22.5 – 19.5)
= √22.5 × 15 × 4.5 × 3
= 67.5 cm2
Hence, the area of the triangle will be 67.5 cm2
Question no – (5)
Solution :
Let, sides of triangle = 25x, 17x, 12x
∴ Perimeter,
= (25x + 17x + 12x)
= 54x
∴ 54x = 540
= x = 540/54
= 10
∴ Sides are = (25 × 10), (17 × 10), (12 × 10)
= 250, = 170, = 120
Now, Side = 250 + 170 + 120/2
= 540/2
= 270
∴ Area = √270 (270 – 250) (270 – 120) (270 – 170)
= √270 × 20 × 150 × 100
= √81000000
= 9000 m2
Therefore, the area of the triangle will be 9000 m2
Question no – (6)
Solution :
Area of square,
= (6 × 6)
= 36 cm2
From fig, Area of △AOF = 1/3 × 36
= 12 cm2
∴ Now, 12 = 1/2 × DF × 6
= DF = 2 × 12/6
= 4 cm
Therefore, the length of FD will be 4 cm.
Question no – (7)
Solution :
Let, here ‘l’ be the length
∴ l × 5 = 20
= l = 20/5 = 4
∴ Area of triangle,
= 1/2 × 5 × 4
= 10 cm2
Therefore, the area of a triangle is 10 cm2
Question no – (8)
Solution :
(i) The area of a triangle with base 4 cm and perp. height 6 cm is 24 cm².
Here, Area = 1/2 × 4 × 6
= 12 cm2
Thus, it is False.
(ii) The area of a triangle with sides measuring a, b, c, is given by √s(s-a)(s-b)(s-c) where s is the perimeter of the triangle.
This statement is False.
(iii) The area of a rectangle and a triangle are equal if they stand on the same base and are between the same parallels.
This statement is False.
Question no – (9)
Solution :
As per the question,
ABC is a triangle in which AB = AC = 4 cm and ∠A = 90°
∴ Area of △ABC,
= 1/2 × 4 × 4
= 8 cm2
Therefore, the area of △ABC will be 8 cm2
Question no – (12)
Solution :
(i) Equilateral triangle whose each side is = 12 cm
∴ Area of equilateral triangle
= √3/4 × a2
= √3/4 × (12)2
= √3/4 × 144
= √3 × 36
= 36√3 cm2
Thus, the area of the equilateral triangle will be 36√3 cm2
(ii) Equilateral triangle whose each side is = 5 cm.
∴ Area = √3/4 × a2
= √3/4 × 5 × 5
= 25√3/4 cm2
Hence, the area of the equilateral triangle will be 25√3/4 cm2
Question no – (13)
Solution :
Let, ‘a’ side of the equilateral
∴ 3a = 24
= a = 24/3 = 8
∴ Area,
= √3/4 × 8 × 8
= 16√3 cm2
Thus, its area will be 16√3 cm2
Question no – (14)
Solution :
Let, ‘a’ be the length of side
∴ Perimeter = √3 × Area
= 3a = √3 × √3/4 × a2
= a2 = 4a
= a = 4 (length)
Question no – (15)
Solution :
Let, ‘a’ be the side,
√3/4 × a2 = 173.2
= a2 = 173.2 × 4/√3
= a = 20 m
∴ Perimeter will be,
= (3 × 20) m
= 60 m
Therefore, its perimeter will be 60 m.
Question no – (16)
Solution :
(i) Each of the equal sides is 8 cm and the base is 9 cm
Here, semi perimeter
= (a + b +c)/2
= (8 + 8 + 9)/2
= 25/2
= 12.5
∴ Area of triangle
= √12.5 (12.5 – 8) (12.5 – 8) (12.5 – 9)
= √12.5 × 4.5 × 4.5 × 3.5
= 29.77 cm2
Thus, the area of the isosceles triangle will be 29.77 cm2
(ii) Each of the equal sides is 10 cm and the base is 12 cm;
Semi perimeter,
= 10 + 10 + 12/2
= 16 cm
Area of isosceles triangle,
= √16 (16 – 10) (16 – 10) (16 – 12)
= √16 × 6 × 6 × 4
= √2304
= 48 cm2
Thus, the area of the isosceles triangle will be 48 cm2
(iii) Sides is 7.4 cm and the base is 6.2 cm.
∴ Area of isosceles triangle,
= 1/2 × b × h
= 1/2 × 6.2 × 7.4
= 22.94 cm2
Hence, the area of the isosceles triangle will be 22.94 cm2
(iv) Given in the question,
Perimeter is = 11 cm
Base is = 4 cm
Area of isosceles triangle
= (1/2 × 4 × 11)
= 22
Therefore, the area of the isosceles triangle will be 22 cm2
Question no – (17)
Solution :
(i) Here, Base = 24,
Area = 192 sqm
∴ Area = 1/2 × b × h
192 = 1/2 × 24 × h
= 24h = 192 × 2
= h = 192 × 2/24
= 16 cm
∴ Perimeter,
= (24 + 24 + 16) cm
= 64 cm
Thus, perimeter of the isosceles triangle will be 64 cm.
Question no – (18)
Solution :
Let, length of equal side = x and base = y
∴ y = 2/3 × 2x
= 4x/3
∴ Now, x + x + 4x/3 = 42
= 3x + 3x + 4x = 126
= 10x = 126
= x = 126/10 = 12.6
∴ Base = (4/3 × 12.6) = 16.8
Now, By theorem,
h2 + (8.4)2 = (12.6)2
= h2 = (12.6)2 – (8.4)2
= 88.2
= h √88.2
∴ Area,
= 1/2 × 16 . 8 × √88.2
= 78.9 cm2
Next Chapter Solution :
👉 Chapter 17 👈