NCERT Solutions for Class 7 Maths Chapter 12 – Algebraic Expression

NCERT Solutions for Class 7 Maths Chapter 12 – Algebraic Expression. Furthermore, here we’ve provided you with the latest solution for Class 7 Maths Chapter 12 – Algebraic Expression. As a result here you’ll find solutions to all the exercises. This NCERT Class 7 solution will help you to score good marks in your exam.

Students can refer to our solution for NCERT Class 7 Maths Chapter 12 – Algebraic Expression. The Chapter 12 Solution of NCERT will help students prepare for the exams and easily crack the exam. Below we’ve provided you with the exercise-wise latest solution.

NCERT Solutions for Class 7 Maths Chapter 12 – Algebraic Expression Exercise Wise Solution

Exercise 12.1 – Page 234 of NCERT
Exercise 12.2 – Page 239 of NCERT
Exercise 12.3 – Page 242 of NCERT
Exercise 12.4 – Page 246 of NCERT

NCERT Solutions for Class 7 Maths Chapter 12 – Algebraic Expression Exercise 12.1 Solution

Here you’ll find NCERT Chapter 12 – Algebraic Expression Exercise 12.1 Solution.
Exercise 12.1: Solutions of Questions on Page Number: 234

Q1: Get the algebraic expressions in the following cases using variables, constants and arithmetic operations.

1. Subtraction of z from y.
2. One-half of the sum of numbers x and y.
3. The number z multiplied by itself.
4. One-fourth of the product of numbers p and q.
5. Numbers x and y both squared and added.
6. Number 5 added to three times the product of number m and n.
7. Product of numbers y and z subtracted from 10.
8. Sum of numbers a and b subtracted from their product.

Answer:

1. y – z
2. 
3. z²
4. 

5. x² + y²
6. 5 + 3 (mn)
7. 10 – yz
8. ab – (a + b)

Q2:

1. Identify the terms and their factors in the following expressions Show the terms and factors by tree diagrams.

(a) x – 3 (b) 1 + x + x²  (c) y – y³

(d) 5xy² + 7x²y (e) – ab + 2b²  – 3a²

2. Identify terms and factors in the expressions given below:
   (a) – 4x + 5 (b) – 4x + 5y (c) 5y + 3y²
   (d) xy + 2x²y² (e) pq + q (f) 1.2 ab – 2.4 b + 3.6 a
   (g) (h) 0.1p² + 0.2 q²

Answer:

(i)

(a)

(b)

(c)

(d)

(e)

(ii)

Row

Expression

Terms

Factors

Q3: Identify the numerical coefficients of terms (other than constants) in the following expressions:

(i) 5 – 3t²  (ii) 1 + t + t²  + t³  (iii) x + 2xy+ 3y

(iv) 100m + 1000n (v) – p²q² + 7pq (vi) 1.2a + 0.8b

(vii) 3.14 r²  (viii) 2 (l + b) (ix) 0.1y + 0.01 y²

Answer:

Row	Expression	Terms	Coefficients
(i)	5 – 3t2	– 3t2	– 3
(ii)	1 + t + t2   + t3	t t2 t3	1 1 1

Q4 :

1. Identify terms which contain x and give the coefficient of x.

(i) y²x + y (ii) 13y²– 8yx (iii) x + y + 2

(iv) 5 + z + zx (v) 1 + x+ xy (vi) 12xy²  + 25

(vii) 7x + xy²

2. Identify terms which contain y² and give the coefficient of y².
   (i) 8 – xy²  (ii) 5y²  + 7x (iii) 2x²y -15xy²  + 7y²

Answer:

(a)

Row	Expression	Terms with x	Coefficient of x
(i)	y2x + y	y2x	y2
(ii)	13y2   – 8yx	– 8yx	– 8y
(iii)	x + y + 2	x	1

(b)

(iv)	5 + z + zx	zx	z
(v)	1 + x + xy	x xy	1 y
(vi)	12xy2   + 25	12xy2	12y2
(vii)	7x+ xy2	7x xy2	7y2

Q5: Classify into monomials, binomials and trinomials.

(i) 4y – 7z (ii) y² (iii) x + y – xy (iv) 100 (v) ab – a – b (vi) 5 – 3t

(vii) 4p²q – 4pq² (viii) 7mn (ix) z²  – 3z + 8

(x) a²  + b²  (xi) z² + z (xii) 1 + x + x²

Answer:

The monomials, binomials, and trinomials have 1, 2, and 3 unlike terms in it respectively.

1. 4y – 7z
   Binomial
2. y²
   Monomial

3. x + y – xy
   Trinomial

4. 100
   Monomial
5. ab – a – b
   Trinomial

6. 5 – 3t
   Binomial
7. 4p²q – 4pq²
   Binomial

8. 7mn
   Monomial
9. z² – 3z + 8
   Trinomial

10. a² + b²
    Binomial

11. z² + z
    Binomial

12. 1 + x + x²
    Trinomial

Q6: State whether a given pair of terms is of like or unlike terms.

(i) 1, 100 (ii) (iii) – 29x, – 29y

(iv) 14xy, 42yx (v) 4m²p, 4mp² (vi) 12xz, 12 x²z²

Answer:

The terms which have the same algebraic factors are called like terms. However, when the terms have different algebraic factors, these are called unlike terms.

(i) 1, 100
Like

2. – 7x,
   Like

3. – 29x, – 29y
   Unlike
4. 14xy, 42yx
   Like
5. 4m²p, 4mp²
   Unlike
6. 12xz, 12x²z²
   Unlike

Q7: Identify like terms in the following:

(a) –xy², – 4yx², 8x², 2xy², 7y, – 11x², – 100x, -11yx, 20x²y, -6x², y, 2xy,3x

(b) 10pq, 7p, 8q, – p²q², – 7qp, – 100q, – 23, 12q²p², – 5p², 41, 2405p, 78qp, 13p²q, qp², 701p²

Answer:

1. –xy², 2xy²
   -4yx², 20x²y
   8x², -11x², -6x²
   7y, y
   -100x, 3x
   -11xy, 2xy

2. 10pq, -7qp, 78qp
   7p, 2405p
   8q, -100q
   –p²q², 12p²q²
   -23, 41
   -5p², 701p²
   13p²q, qp²

NCERT Solutions for Class 7 Maths Chapter 12 – Algebraic Expression Exercise 12.2 Solution

Here you’ll find NCERT Chapter 12 – Algebraic Expression Exercise 12.2 Solution.
Exercise 12.2: Solutions of Questions on Page Number: 239

Q1: Simplify combining like terms:

1. 21b – 32 + 7b – 20b
2. – z²  + 13z²  – 5z + 7z³  – 15z

3. p – (p – q) – q – (q – p)
4. 3a – 2b – ab – (a – b + ab) + 3ab + b – a
5. 5x²y – 5x² + 3y x² – 3y² + x² – y² + 8xy² -3y²
6. (3 y²  + 5y – 4) – (8y – y²  – 4)

Answer:

(i) 21b – 32 + 7b – 20b = 21b + 7b – 20b – 32

= b (21 + 7 – 20) -32

= 8b – 32

(ii) – z²  + 13z²  – 5z + 7z³ – 15z = 7z³ – z²  + 13z² – 5z – 15z

= 7z³ + z² (-1 + 13) + z (-5 – 15)

= 7z³ + 12z² – 20z

(iii) p – (p – q) – q – (q – p) = p – p + q – q – q + p

= p – q

(iv) 3a – 2b – ab – (a – b + ab) + 3ba + b – a

= 3a – 2b – ab – a + b – ab + 3ab + b – a

= 3a – a – a – 2b + b + b – ab – ab + 3ab

= a (3 – 1 – 1) + b (- 2 + 1 + 1) + ab (-1 -1 + 3)

= a + ab

(v) 5x²y – 5x² + 3yx²  – 3y² + x²  – y² + 8xy²  – 3y²

= 5x²y + 3yx² – 5x² + x² – 3y² – y² – 3y² + 8xy²

= x²y (5 + 3) + x²  (-5 + 1) + y²(-3 – 1 – 3) + 8xy²

= 8x²y – 4x²  – 7y² + 8xy²

(vi) (3y²  + 5y – 4) – (8y – y² – 4)

= 3y² + 5y – 4 – 8y + y² + 4

= 3y² + y² + 5y – 8y – 4 + 4

= y² (3 + 1) + y (5 – 8) + 4 (1 – 1)

= 4y²  – 3y

Q2: Add:

(i) 3mn, – 5mn, 8mn, -4mn

(ii) t – 8tz, 3tz – z, z – t

(iii) – 7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3

(iv) a + b – 3, b – a + 3, a – b + 3

(v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy

(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5

(vii) 4x²y, – 3xy², – 5xy², 5x²y

(viii) 3p²q²  – 4pq + 5, – 10p²q², 15 + 9pq + 7p²q²

(ix) ab – 4a, 4b – ab, 4a – 4b

(x)   x² – y²  – 1 , y²  – 1 – x², 1- x²  – y²

Answer:

(i)   3mn + (-5mn) + 8mn + (-4mn) = mn (3 – 5 + 8 – 4)

= 2mn

(ii) (t – 8tz) + (3tz – z) + (z – t) = t – 8tz + 3tz – z + z – t

= t – t – 8tz + 3tz – z + z

= t (1 – 1) + tz (- 8 + 3) + z (- 1 + 1)

= -5tz

(iii) (- 7mn + 5) + (12mn + 2) + (9mn – 8) + (- 2mn – 3)

= – 7mn + 5 + 12mn + 2 + 9mn – 8 – 2mn – 3

= – 7mn + 12mn + 9mn – 2mn + 5 + 2 – 8 – 3

= mn (- 7 + 12 + 9 – 2) + (5 + 2 – 8 – 3)

= 12mn – 4

(iv) (a + b – 3) + (b – a + 3) + (a – b + 3)

= a + b – 3 + b – a + 3 + a – b + 3

= a – a + a + b + b – b – 3 + 3 + 3

= a (1 – 1 + 1) + b (1 + 1 – 1) + 3 (- 1 + 1 + 1)

= a + b + 3

(v) (14x + 10y – 12xy – 13) + (18 – 7x – 10y + 8yx) + 4xy

= 14x + 10y – 12xy – 13 + 18 – 7x – 10y + 8yx + 4xy

= 14x – 7x + 10y – 10y – 12xy + 8yx + 4xy – 13 + 18
= x (14 – 7) + y (10 – 10) + xy (- 12 + 8 + 4) – 13 + 18
= 7x + 5

(vi) (5m – 7n) + (3n – 4m + 2) + (2m – 3mn – 5)

= 5m – 7n + 3n – 4m + 2 + 2m – 3mn – 5

= 5m – 4m + 2m – 7n + 3n – 3mn + 2 – 5

= m (5 – 4 + 2) + n (- 7 + 3) -3mn + 2 – 5

= 3m – 4n – 3mn – 3

(vii) 4x²  y – 3xy²  – 5xy²  + 5x²y = 4x² y + 5x²y – 3xy²  – 5xy²

= x² y (4 + 5) + xy² (- 3 – 5)

= 9x²y – 8xy²

(viii) (3p²q² – 4pq + 5) + (-10 p²q²) + (15 + 9pq + 7p²q²)

= 3p²q² – 4pq + 5 – 10 p²q² + 15 + 9pq + 7p²q²

= 3p²q² – 10 p²q²  + 7p²q²  – 4pq + 9pq + 5 + 15

= p²q²  (3 – 10 + 7) + pq (- 4 + 9) + 5 + 15

= 5pq + 20

(ix) (ab – 4a) + (4b – ab) + (4a – 4b)

= ab – 4a + 4b – ab + 4a – 4b

= ab – ab – 4a + 4a + 4b – 4b

= ab (1 – 1) + a (- 4 + 4) + b(4 – 4)

= 0

(x) (x² – y² – 1) + (y²  – 1 – x²) + (1 – x² – y²)

= x² – y² – 1 + y² – 1 – x² + 1 – x² – y²

= x² – x² – x² – y² + y² – y² – 1 – 1 + 1

= x²(1 – 1 – 1) + y² (-1 + 1 – 1) + (- 1 – 1 + 1)

= – x² – y² – 1

Q3: Subtract:

1. – 5y²  from y²
2. 6xy from – 12xy
3. (a – b) from (a + b)
4. a (b – 5) from b (5 – a)
5. – m²  + 5mn from 4m²  – 3mn + 8
6. – x² + 10x – 5 from 5x – 10
7. 5a² – 7ab + 5b² from 3ab – 2a² -2b²
8. 4pq – 5q² – 3p² from 5p² + 3q² – pq

Answer:

(i)   y² – (-5y²) = y² + 5y² = 6y²

(ii) – 12xy – (6xy) = -18xy

(iii) (a + b) – (a – b) = a + b – a + b = 2b

(iv) b (5 – a) – a (b – 5) = 5b – ab – ab + 5a

= 5a + 5b – 2ab

(v) (4m² – 3mn + 8) – (- m²  + 5mn) = 4m² – 3mn + 8 + m² – 5 mn

= 4m²  + m²  – 3mn – 5 mn + 8

= 5m²  – 8mn + 8

(vi) (5x – 10) – (- x²  + 10x – 5) = 5x – 10 + x²  – 10x + 5

= x² + 5x – 10x – 10 + 5

= x² – 5x – 5

(vii) (3ab – 2a² – 2b²) – (5a²– 7ab + 5b²)

= 3ab – 2a² – 2b²  – 5a²  + 7ab – 5 b²

= 3ab + 7ab – 2a²  – 5a²  – 2b² – 5 b²

= 10ab – 7a² – 7b²

(viii) 4pq – 5q² – 3p² from 5p² + 3q² – pq

(5p² + 3q² – pq) – (4pq – 5q²– 3p²)

= 5p²  + 3q²  – pq – 4pq + 5q²  + 3p²

= 5p² + 3p²  + 3q² + 5q²  – pq – 4pq

= 8p² + 8q² – 5pq

Q4:

1. What should be added to x²  + xy + y²  to obtain 2x²  + 3xy?
2. What should be subtracted from 2a + 8b + 10 to get – 3a + 7b + 16?

Answer :

1. Let a be the required term.

a + (x² + y² + xy) = 2x² + 3xy a = 2x² + 3xy – (x² + y² + xy) a = 2x² + 3xy – x² – y² – xy a

= 2x²  – x²  – y²  + 3xy – xy

= x² – y² + 2xy

2. Let p be the required term.

(2a + 8b + 10) – p = – 3a + 7b + 16

p = 2a + 8b + 10 – (- 3a + 7b + 16)

= 2a + 8b + 10 + 3a – 7b – 16

= 2a + 3a + 8b – 7b + 10- 16

= 5a + b – 6

Q5: What should be taken away from 3x²  – 4y²  + 5xy + 20 to obtain

– x²  – y²  + 6xy + 20?

Answer:

Let p be the required term.

(3x² – 4y² + 5xy + 20) – p = – x² – y² + 6xy + 20 p

= (3x² – 4y² + 5xy + 20) – (- x² – y² + 6xy + 20) = 3x² – 4y² + 5xy + 20 + x² + y² – 6xy – 20 = 3x² +

x² – 4y² + y² + 5xy – 6xy + 20 – 20

= 4x² – 3y² – xy

Q6:

1. From the sum of 3x – y + 11 and – y – 11, subtract 3x – y – 11.
2. From the sum of 4 + 3x and 5 – 4x + 2x², subtract the sum of 3x² – 5x and – x² + 2x + 5.

Answer:

(a) (3x – y + 11) + (- y – 11)

= 3x – y + 11 – y – 11

= 3x – y – y + 11 – 11

= 3x – 2y

(3x – 2y) – (3x – y – 11)

= 3x – 2y – 3x + y + 11

= 3x – 3x – 2y + y + 11

= – y + 11

(b) (4 + 3x) + (5 – 4x + 2x²) = 4 + 3x + 5 – 4x + 2x²

= 3x – 4x + 2x² + 4 + 5

= – x + 2x² + 9

(3x² – 5x) + (- x² + 2x + 5) = 3x² – 5x – x² + 2x + 5

= 3x² – x² – 5x + 2x + 5

= 2x²  – 3x + 5

(- x + 2x²  + 9) – (2x²  – 3x + 5)

= – x + 2x²  + 9 – 2x²  + 3x – 5

= – x + 3x + 2x² – 2x² + 9 – 5

= 2x + 4

NCERT Solutions for Class 7 Maths Chapter 12 – Algebraic Expression Exercise 12.3 Solution

Here you’ll find NCERT Chapter 12 – Algebraic Expression Exercise 12.3 Solution.
Exercise 12.3: Solutions of Questions on Page Number: 242

Q1: If m = 2, find the value of:

(i) m – 2 (ii) 3m – 5 (iii) 9 – 5m

(iv) 3m²  – 2m – 7 (v) 5m/2 – 4

Answer:

(i) m – 2 = 2 – 2 = 0

(ii) 3m – 5 = (3 × 2) – 5 = 6 – 5 = 1

(iii) 9 – 5m = 9 – (5 × 2) = 9 – 10 = – 1

(iv) 3m² – 2m – 7 = 3 × (2 × 2) – (2 × 2) – 7

= 12 – 4 – 7 = 1

(v)

Q2: If p = -2, find the value of:

(i) 4p + 7

(ii) -3p² + 4p + 7

(iii) -2p³ – 3p² + 4p + 7

Answer:

(i) 4p + 7 = 4 x (-2) + 7 = – 8 + 7 = -1

(ii) – 3p²  + 4p + 7 = -3 (-2) x (-2) + 4 x (-2) + 7

= – 12 – 8 + 7 = -13

(iii) -2p³  – 3p² + 4p + 7

= -2 (-2) x (-2) x (-2) – 3 (-2) x (-2) + 4 x (-2) + 7

= 16 – 12 – 8 + 7 = 3

Q3: Find the value of the following expressions, when x = – 1:

(i) 2x – 7 (ii) – x + 2 (iii) x² + 2x + 1 (iv) 2x²  – x – 2

Answer:

(i) 2x – 7

= 2 x (-1) – 7 = -9

(ii) – x + 2 = – (-1) + 2 = 1 + 2 = 3

(iii) x²  + 2x + 1 = (-1) x (-1) + 2 x (-1) + 1

= 1 – 2 + 1 = 0

(iv) 2x²  – x – 2 = 2 (-1) x (-1) – (-1) – 2

= 2 + 1 – 2 = 1

Q4: If a = 2, b = – 2, find the value of:

(i) a²  + b²  (ii) a² + ab + b²  (iii) a²  – b²

Answer:

1. a² + b²

= (2)²  + (-2)²  = 4 + 4 = 8

2. a² + ab + b²

= (2 x 2) + 2 x (-2) + (-2) x (-2)

= 4 – 4 + 4 = 4

3. a² – b² = (2)² – (-2)² = 4 – 4 = 0

Q5: When a = 0, b = – 1, find the value of the given expressions:
(i) 2a + 2b (ii) 2a²  + b²  + 1
(iii) 2a² b + 2ab²  + ab (iv) a² + ab + 2

Answer:

(i) 2a + 2b = 2 x (0) + 2 x (-1) = 0 – 2 = -2

2. 2a² + b²  + 1

= 2 x (0)²  + (-1) x (-1) + 1

= 0 + 1 + 1 = 2

3. 2a²b + 2ab² + ab

= 2 x (0)² x (-1) + 2 x (0) x (-1) x (-1) + 0 x (-1)

= 0 + 0 + 0 = 0

4. a² + ab + 2

= (0)²  + 0 x (-1) + 2

= 0 + 0 + 2 = 2

Q6: Simplify the expressions and find the value if x is equal to 2

(i) x + 7 + 4 (x – 5) (ii) 3 (x + 2) + 5x – 7

(iii) 6x + 5 (x – 2) (iv) 4 (2x -1) + 3x + 11

Answer:

(i) x + 7 + 4 (x – 5) = x + 7 + 4x – 20

= x + 4x + 7 – 20

= 5x – 13

= (5 x 2) – 13

= 10 – 13 = -3

(ii) 3 (x + 2) + 5x – 7 = 3x + 6 + 5x – 7

= 3x + 5x + 6 – 7 = 8x – 1

= (8 x 2) – 1 = 16 – 1 =15

(iii) 6x + 5 (x – 2) = 6x + 5x – 10

= 11x – 10

= (11 x 2) – 10 = 22 – 10 = 12

(iv) 4 (2x – 1) + 3x + 11 = 8x – 4 + 3x + 11

= 11x + 7

= (11 x 2) + 7 =

22 + 7 = 29

Q7: Simplify these expressions and find their values if x = 3, a = – 1, b = – 2.

(i) 3x – 5 – x + 9 (ii) 2 – 8x + 4x + 4

(iii) 3a + 5 – 8a + 1 (iv) 10 – 3b – 4 – 5b

(v) 2a – 2b – 4 – 5 + a

Answer:

(i) 3x – 5 – x + 9 = 3x – x – 5 + 9

= 2x + 4 = (2 x 3) + 4 = 10

(ii) 2 – 8x + 4x + 4 = 2 + 4 – 8x + 4x

= 6 – 4x = 6 – (4 x 3) = 6 – 12 = -6 (iii) 3a + 5 – 8a + 1 = 3a – 8a + 5 + 1

= – 5a + 6 = -5 x (-1) + 6

= 5 + 6 = 11

(iv) 10 – 3b – 4 – 5b = 10 – 4- 3b – 5b

= 6 – 8b = 6 – 8 x (-2)

= 6 + 16 = 22

(v) 2a – 2b – 4 – 5 + a = 2a + a – 2b – 4 – 5

= 3a – 2b – 9s

= 3 x (-1) – 2 (-2) – 9

= – 3 + 4 – 9 = -8

Q8:

1. Ifz = 10, find the value of z³ – 3 (z – 10).
2. If p = – 10, find the value of p² – 2p – 100

Answer:

(i) z³ – 3 (z – 10) = z³  – 3z + 30

= (10 x 10 x 10) – (3 x 10) + 30

= 1000 – 30 + 30 = 1000

(ii) p²  – 2p – 100

= (-10) x (-10) – 2 (-10) – 100

= 100 + 20 – 100 = 20

Q9: What should be the value of a if the value of 2x² + x – a equals to 5, when x = 0?

Answer:

2x² + x – a = 5, when x = 0 (2 x 0) + 0 – a = 5

0 – a = 5 a = -5

Q10: Simplify the expression and find its value when a = 5 and b = -3. 2 (a² + ab) + 3 – ab

Answer:

2 (a²  + ab) + 3 – ab = 2a²  + 2ab + 3 – ab

= 2a²  + 2ab – ab + 3

= 2a² + ab + 3

= 2 x (5 x 5) + 5 x (-3) + 3

= 50 – 15 + 3 = 38

NCERT Solutions for Class 7 Maths Chapter 12 – Algebraic Expression Exercise 12.4 Solution

Here you’ll find NCERT Chapter 12 – Algebraic Expression Exercise 12.4 Solution.
Exercise 12.4: Solutions of Questions on Page Number: 246

Q1: Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic watches or calculators. (a)
(a)

(b)

(c)

If the number of digits formed is taken to be n, the number of segments required to form n digits is given by the algebraic expression appearing on the right of each pattern.

How many segments are required to form 5, 10, 100 digits of the kind –

, ,

Answer:

1. It is given that the number of segments required to form n digits of the kind

is (5n + 1).

Number of segments required to form 5 digits = (5 x 5 + 1)

= 25 + 1 = 26

Number of segments required to form 10 digits = (5 x 10 + 1)

= 50 + 1 = 51

Number of segments required to form 100 digits = (5 x 100 + 1)

= 500 + 1 = 501

2. It is given that the number of segments required to form n digits of the kind is (3n + 1). Number of segments required to form 5 digits = (3 x 5 + 1)

= 15 + 1 = 16

Number of segments required to form 10 digits = (3 x 10 + 1)

= 30 + 1 = 31

Number of segments required to form 100 digits = (3 x 100 + 1)

= 300 + 1 = 301

3. It is given that the number of segments required to form n digits of the kind is (5n + 2). Number of segments required to form 5 digits = (5 x 5 + 2)

= 25 + 2 = 27

Number of segments required to form 10 digits = (5 x 10 + 2)

= 50 + 2 = 52

Number of segments required to form 100 digits = (5 x 100 + 2)

= 500 + 2 = 502

Q2:

Use the given algebraic expression to complete the table of number patterns. The given table can be completed as follows.

NCERT Class 7 Maths All Chapters Solution 

Chapter 1: Integers

Chapter 2: Fractions and Decimals

Chapter 3: Data Handling

Chapter 4: Simple Equations

Chapter 5: Lines and Angles

Chapter 6: The Triangle and its Properties.

Chapter 7: Congruence of Triangles

Chapter 8: Comparing Quantities 

Chapter 9: Rational Numbers

Chapter 10: Practical Geometry

Chapter 11: Perimeter and Area

Chapter 12: Algebraic Expression

Chapter 13: Exponents and Powers

Chapter 14: Symmetry

Chapter 15: Visualising Solid Shapes