## Maharashtra State Board Class 7 Maths Solutions Chapter 14 Algebraic Formulae – Expansion of Squares Practice Set 50

Question 1.

Expand:

i. (5a + 6b)²

ii. \(\left(\frac{\mathrm{a}}{2}+\frac{\mathrm{b}}{3}\right)^{2}\)

iii. (2p – 3q)²

iv. \(\left(x-\frac{2}{x}\right)^{2}\)

v. (ax + by)²

vi. (7m – 4)²

vii. \(\left(x+\frac{1}{2}\right)^{2}\)

viii. \(\left(a-\frac{1}{a}\right)^{2}\)

Solution:

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. \(\left(\frac{\mathrm{a}}{2}+\frac{\mathrm{b}}{3}\right)^{2}\)

Here A = \(\frac { a }{ 2 }\) and B = \(\frac { b }{ 3 }\)

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. \(\left(x-\frac{2}{x}\right)^{2}\)

Here a = x and b = \(\frac { 2 }{ x }\)

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. \(\left(x+\frac{1}{2}\right)^{2}\)

Here a = x and b = \(\frac { 1 }{ 2 }\)

viii. \(\left(a-\frac{1}{a}\right)^{2}\)

Here A = a and B = \(\frac { 1 }{ a }\)

Question 2.

Which of the options given below is the square of the binomial

(A) \(64-\frac{1}{x^{2}}\)

(B) \(64+\frac{1}{x^{2}}\)

(C) \(64-\frac{16}{x}+\frac{1}{x^{2}}\)

(D) \(64+\frac{16}{x}+\frac{1}{x^{2}}\)

Solution:

(C) \(64-\frac{16}{x}+\frac{1}{x^{2}}\)

Hint:

= \(\left(8-\frac{1}{x}\right)^{2}=8^{2}-2 \times 8 \times \frac{1}{x}+\left(\frac{1}{x}\right)^{2}\) …[(a – b)² = a² – 2ab + b²]

= \(64-\frac{16}{x}+\frac{1}{x^{2}}\)

Question 3.

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:

(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.

Use an expansion formula to find the values of:

i. (997)²

ii. (102)²

iii. (97)²

iv. (1005)²

Solution:

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.

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:

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