Maharashtra State Board Class 7 Mathematics & Statistics Solutions Chapter 14 Algebraic Formulae – Expansion of Squares Practice Set 50 - Digest Answers | Maharashtra Board Solutions

Algebraic Formulae – Expansion of Squares Class 7 Practice Set 50 Answers Solutions Chapter 14

Question 1 Maharashtra Board Solution

Expand: i. (5a + 6b)² ii. iii. (2p – 3q)² iv. v. (ax + by)² vi. (7m – 4)² vii. viii.

Solution & Step-by-Step Answer:

i. (5a + 6b)² Here, A = 5a and B = 6b (5a + 6b)² = (5a)² + 2 × 5a × 6b + (6b)² …. [(A + B)² = A² + 2AB + B²] ∴ (5a + 6b)² = 25a² + 60ab + 36b²

ii.
Here A = and B =

iii. (2p – 3q)²
Here, a = 2p and b = 3q
(2p – 3q)² = (2p)² – 2 × (2p) × (3q) + (3q)²
…. [(a – b)² = a² – 2ab + b²]
∴ (2p – 3q)² = 4p² – 12pq + 9q²

iv.
Here a = x and b =

v. (ax + by)²
Here, A = ax and B = by
(ax + by)² = (ax)² + 2 × ax × by + (by)²
…. [(A + B)² = A² + 2AB + B²]
∴ (ax + by)² = a²x² + 2abxy + b²y²

vi. (7m – 4)²
Here, a = 7m and b = 4
(7m – 4)² = (7m)² – 2 × 7m × 4 + 4²
…. [(a – b)² = a² – 2ab + b²]
∴ (7m – 4)² = 49m² – 56m + 16

vii.
Here a = x and b =

viii.
Here A = a and B =

Question 2 Maharashtra Board Solution

Which of the options given below is the square of the binomial (A) (B) (C) (D)

Solution & Step-by-Step Answer:

(C)

Hint:
= …[(a – b)² = a² – 2ab + b²]
=

Question 3 Maharashtra Board Solution

Of which of the binomials given below is the m²n² + 14mnpq + 49p²q² the expansion? (A) (m + n) (p + q) (B) (mn – pq) (C) (7mn + pq) (D) (mn + 7pq)

Solution & Step-by-Step Answer:

(D) (mn + 7pq)

Hint:
Here, square root of the first term = mn
Square root of the last term = 7pq
∴ Required binomial = (mn + 7pq)²

Question 4 Maharashtra Board Solution

Use an expansion formula to find the values of: i. (997)² ii. (102)² iii. (97)² iv. (1005)²

Solution & Step-by-Step Answer:

i. (997)² = (1000 – 3)² Here, a = 1000 and b = 3 (1000 – 3)² = (1000)² – 2 x 1000 x 3 + 3² …. [(a – b)² = a² – 2ab + b²] = 1000000 – 6000 + 9 = 994009 ∴ (997)² = 994009

ii. (102)² = (100 + 2)²
Here, a = 100 and b = 2
(100 + 2)² = (100)² + 2 x 100 x 2 + 2²
…. [(a + b)² = a² + 2ab + b²]
= 10000 + 400 + 4
= 10404
∴ (102)² = 10404

iii. (97)² = (100 – 3)²
Here, a = 100 and b = 3
(100 – 3)² = (100)² – 2 x 100 x 3 + 3²
…. [(a – b)² = a² – 2ab + b²]
= 10000 – 600 + 9
= 9409
∴ (97)² = 9409

iv. (1005)² = (1000 + 5)²
Here, a = 1000 and b = 5 (1000 + 5)²
= (1000)² + 2 x 1000 x 5 + 5²
…. [(a + b)² = a² + 2ab + b²]
= 1000000+ 10000 + 25
= 1010025
∴ (1005)² = 1010025

Maharashtra Board Class 7 Maths Chapter 14 Algebraic Formulae – Expansion of Squares Practice Set 50 Intext Questions and Activities

Question 1 Maharashtra Board Solution

Use the given values to verify the formulae for squares of binomials. (Textbook pg. no. 92) i. a = -7, b = 8 ii. a = 11,b = 3 iii. a = 2.5,b = 1.2

Solution & Step-by-Step Answer:

i. (a + b)² = (-7 + 8)² = 1² = 1 a² + 2ab + b² = (-7)² + 2 x (-7) x 8 + 8² = 49 – 112 + 64 = 1 ∴(a + b)² = a² + 2ab + b² (a – b)² = (-7 – 8)² = (-15)² = 225 a² – 2ab + b² = (-7)² – 2 x (-7) x 8 + (8)² = 49 + 112 + 64 = 225 ∴(a – b)² = a² – 2ab + b²

ii. (a + b)² = (11 + 3)²
= 14²
= 196
a² + 2ab + b² = 11² + 2 x 11 x 3 + 3²
= 121 + 66 + 9
= 196
∴(a + b)² = a² + 2ab + b²
(a – b)² = (11 – 3)² = 8²
= 64
a² – 2ab + b² = 11² – 2 x 11 x 3 + 3²
= 121 – 66 + 9
= 64
∴(a – b)² = a² – 2ab + b²

iii. (a + b)² = (2.5 + 1.2)²
= 3.7²
= 13.69
a² + 2ab + b² = (2.5)² + 2 x 2.5 x 1.2 + (1.2)²
= 6.25 + 6 + 1.44
= 13.69
∴(a + b)² = a² + 2ab + b²
(a – b)² = (2.5 – 1.2)²
= 1.32
= 1.69
a² – 2ab + b² = (2.5)² – 2 x 2.5 x 1.2 + (1.2)²
= 6.25 – 6 + 1.44
= 1.69
∴(a – b)² = a² – 2ab + b²-

Chapter 14 Algebraic Formulae – Expansion of Squares Practice Set 50 Solutions