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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) cm^{2}

= 40 cm^{2}

Thus, the perimeter of the rectangle is 26 cm and area is 40 cm^{2}

**(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) m^{2}

= 108 m^{2}

Hence, the perimeter of the rectangle will be 42 cm and Area will be 108 m^{2}

**(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 m^{2}

Hence, the Perimeter of the rectangle is 11 m and Area is 7 m^{2}

**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 cm^{2}

Thus, the perimeter of the square will be 12 cm and Area will be 9 cm^{2}

**(ii) Given, Square whose each side is 5 m**

**∴** Perimeter of the square,

= (4 × 5)

= 22 cm

**∴**** Area of the square,**

= 5^{2}

= 25 cm^{2}

Hence, the perimeter of the square will be 22 cm and Area will be 25 cm^{2}

**(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} m^{2}

= 1.69 m^{2}

Hence, the Perimeter of the square will be 5.2 m and Area will be 1.69 m^{2}

**(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 m^{2}

Therefore, the perimeter of the square will be 9.6 m and Area will be 5.76 m^{2}

**Question no – (3) **

**Solution : **

**(i) Square whose area = 25 m**^{2}

Let, length of one side = a

= a^{2} = 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 m**^{2}

Let, length of one side = a

= a^{2} = 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 cm**^{2}

Let, length of one side = a

= a^{2} = 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 dm**^{2}

Let, length of one side = a

= a^{2} = 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 cm^{2}

Therefore, Its length will be 9 cm and are will be 54 cm^{2}

**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) km^{2}

= 1/50 km^{2}

**Question no – (7) **

**Solution : **

According to the question,

Let, the side of square = a

**∴ **Side of new square = 20

**∴ **Area of 1^{st} square = a^{2}

**∴ Area of new, **

= (2a)^{2}

= 4a^{2}

**∴**** Ratio,**

= 4a^{2 }: a^{2}

= 4 : 1

Therefore, the ratio of the area will be 4 : 1

**Question no – (8) **

**Solution : **

Here, Area of square field,

= (89)^{2} cm^{2}

= 7921 m^{2}

**Thus, Area of short of, **

= (10,000 – 7921) m^{2 }(we know, 1 hc = 10,000 s.m)

= 2079 m^{2}

Therefore, its area fall short of 2079 m^{2}

**Question no – (9) **

**Solution : **

Here, Diagonal,

= √25^{2} + 25^{2}

= √625 + 625

= √1250

= 35.35

Hence, the length of its diagonal will be 35.35 cm.

**Question no – (10) **

**Solution : **

Here, (10)^{2} = 6^{2}

= 100 = 36 + l^{2}

= l^{2} = 100 – 36

= l^{2} = 100 – 36

= 64

= l = 8 cm

**∴**** Area of the rectangle, **

= (8 × 6) cm^{2}

= 48 cm^{2}

Therefore, its area of the rectangle is 48 cm^{2}

**Question no – (11) **

**Solution : **

Let, breadth = b

Length = 2b

**∴ **(2b)^{2} + b^{2} = (16 √5)^{2}

= 4b^{2} + b^{2} = 256 × 5

= b^{2} = 256

= b = 16

**∴**** Length of the rectangle, **

= (2 × 16) cm^{2}

= 32 cm^{2}

**∴**** Area of the rectangle, **

= (16 × 32)

= 512 cm^{2}

**Question no – (12) **

**Solution : **

Here, Let, side = 2x and 3x

**Perimeter,**

= (4 × 2x)

= 8x

and, 2^{nd}, perimeter

= (4 × 3x)

= 12x

**∴**** Ratio,**

= 8x : 12x

= 8 : 12

= 2 : 3

**∴**** Ratio of Area,**

= (2x)^{2} : (3x)^{2}

= 4x^{2} : 9x^{2}

= 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)m^{2}

= 360 m^{2}

**∴**** Area of carpet,**

= (3 × 2)m^{2}

= 6m^{2}

**∴**** 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)m^{2}

= 20 m^{2}

**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)m^{2}

= 120m^{2}

**and, Brick Area,**

= (20 × 6) cm^{2}

= 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) m^{2}

= 7000 m^{2}

Therefore, the area to be laid with grass will be 7000 m^{2}

**Question no – (18) **

**Solution :**

Let, the ratio 6x and 5x

= 6x × 5x = 27000

= 30x^{2} = 27000/30

= 900

= x^{2} = + 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

**∴** a^{2} + a^{2} = 196

= 2a^{2} = 196

= a^{2} = 98

**∴** Area = 98 m^{2}

**∴**** Perimeter,**

= 4√98

= 39.59

**Question no – (21) **

**Solution : **

**Area of lawn,**

= (10 × 8)m^{2}

= 80 m^{2}

**With path area of lawn,**

= (12 × 10)m^{2}

= 120m^{2}

**∴**** Path,**

= (120 – 80) m^{2}

= 40 m^{2}

Therefore, the area of path will be 40 m^{2}

**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) m^{2}

= 400 m^{2}

Thus, the area of the shaded portion will be 400 m^{2}

**Question no – (23) **

**Solution : **

Let, Length be = m

**∴** a^{2} + a^{2} = 196

= 2a^{2} = 196

= a^{2} = 98

**∴** Area = 98 m^{2}

**∴** Perimeter

= 4√98

= 39.59

**Question no – (24) **

**Solution : **

**Area of paper, **

= (40 × 60)m^{2}

= 2400cm^{2}

**∴**** Area of walls, **

= 2 (l + b) h

= 2 (6 + 4) × 3

= 60 m^{2}

= 600000 cm^{2}

**Area of door**

= 11m^{2}

= 110000 cm^{2}

**∴**** Remain part,**

= (60,000 – 11000)

= 490000 cm^{2}

**∴**** Piece of paper need,**

= 490000 cm^{2}

= 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}m^{2 }….(according to the question)

= (2 × 10 × 1.25 + 24) m^{2}

= 49 m^{2}

Therefore, the area of the wet surface will be 49 m^{2}

**Area of Plane Figures Exercise 16(b) Solution :**

**Question no – (1) **

**Solution : **

**(i) Area of triangle is 6 cm**^{2}** 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 cm^{2}

Hence, the area of the triangle is 336 cm^{2}

**(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 cm^{2}

= √1742400 cm^{2}

= 1320 cm^{2}

Thus, the area of the triangle will be 1320 cm^{2}

**(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 cm^{2}

Hence, the area of the triangle will be 67.5 cm^{2}

**Question no – (2) **

**Solution : **

In the question,

Area of triangle is 6 cm^{2} 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 m^{2}

1 hectare = 10000 m^{2}

1 decameter = 10 m

Area = 25 × 100 = 2500 m^{2}

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 cm^{2}

Hence, the area of the triangle is **336 cm**^{2}

**(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 cm^{2}

= √1742400 cm^{2}

= 1320 cm^{2}

Thus, the area of the triangle will be 1320 cm^{2}

**(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 cm^{2}

Hence, the area of the triangle will be 67.5 cm^{2}

**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 m^{2}

Therefore, the area of the triangle will be 9000 m^{2}

**Question no – (6) **

**Solution : **

Area of square,

= (6 × 6)

= 36 cm^{2}

From fig, Area of △AOF = 1/3 × 36

= 12 cm^{2}

**∴ **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 cm^{2}

Therefore, the area of a triangle is 10 cm^{2}

**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 cm^{2}

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 cm^{2}

Therefore, the area of △ABC will be 8 cm^{2}

**Question no – (12) **

**Solution : **

**(i) Equilateral triangle whose each side is = 12 cm**

**∴** Area of equilateral triangle

= √3/4 × a^{2}

= √3/4 × (12)^{2}

= √3/4 × 144

= √3 × 36

= 36√3 cm^{2}

Thus, the area of the equilateral triangle will be 36√3 cm^{2}

**(ii) Equilateral triangle whose each side is = 5 cm.**

**∴** Area = √3/4 × a^{2}

= √3/4 × 5 × 5

= 25√3/4 cm^{2}

Hence, the area of the equilateral triangle will be 25√3/4 cm^{2}

**Question no – (13) **

**Solution : **

Let, ‘a’ side of the equilateral

**∴ **3a = 24

= a = 24/3 = 8

**∴** Area,

= √3/4 × 8 × 8

= 16√3 cm^{2}

Thus, its area will be 16√3 cm^{2}

**Question no – (14) **

**Solution : **

Let, ‘a’ be the length of side

**∴** Perimeter = √3 × Area

= 3a = √3 × √3/4 × a^{2}

= a^{2} = 4a

= a = 4 (length)

**Question no – (15) **

**Solution : **

Let, ‘a’ be the side,

√3/4 × a^{2} = 173.2

= a^{2} = 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 cm^{2}

Thus, the area of the isosceles triangle will be 29.77 cm^{2}

**(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 cm^{2}

Thus, the area of the isosceles triangle will be 48 cm^{2}

**(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 cm^{2}

Hence, the area of the isosceles triangle will be 22.94 cm^{2}

**(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 cm^{2}

**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,**

h^{2} + (8.4)^{2} = (12.6)^{2}

= h^{2} = (12.6)^{2} – (8.4)^{2}

= 88.2

= h √88.2

**∴**** Area,**

= 1/2 × 16 . 8 × √88.2

= 78.9 cm^{2}

**Next Chapter Solution : **

👉 Chapter 17 👈