## Maharashtra State Board Class 7 Maths Solutions Chapter 14 Algebraic Formulae – Expansion of Squares Practice Set 51

Question 1.

Use the formula to multiply the following:

i. (x + y)(x – y)

ii. (3x – 5)(3x + 5)

iii. (a + 6)(a – 6)

iv. \(\left(\frac{x}{5}+6\right)\left(\frac{x}{5}-6\right)\)

Solution:

i. Here, a = x, b = y

(x + y)(x – y) = x² – y²

…. [(a + b)(a – b) = a² – b²]

ii. Here, a = 3x, b = 5

(3x – 5) (3x + 5) = (3x)² – 5²

…. [(a + b)(a – b) = a² – b²]

= 9x² – 25

iii. Here, A = a, B = 6

(a + 6) (a – 6) = a² – 6²

…. [(A + B)(A – B) = A² – B²]

= a² – 36

iv. Here, a = \(\frac { x }{ 5 }\), b = 6

\(\left(\frac{x}{5}+6\right)\left(\frac{x}{5}-6\right)=\left(\frac{x}{5}\right)^{2}-(6)^{2}\)

…. [(a + b)(a – b) = a² – b²]

= \(\frac{x^{2}}{25}-36\)

Question 2.

Use the formula to find the values:

i. 502 × 498

ii. 97 × 103

iii. 54 × 46

iv. 98 × 102

Solution:

i. 502 × 498 = (500 + 2) (500 – 2)

Here, a = 500, b = 2

∴ (500 + 2) (500 – 2) = 500² – 2²

…. [(a + b)(a – b) = a² – b²]

= 250000 – 4

= 249996

∴ 502 × 498 = 249996

ii. 97 × 103 = (100 – 3) (100 + 3)

Here, a = 100, b = 3

∴ (100 – 3) (100 + 3) = 100² – 3²

…. [(a + b)(a – b) = a² – b²]

= 10000 – 9

= 9991

∴ 97 × 103 = 9991

iii. 54 × 46 = (50 + 4) (50 – 4)

Here, a = 50, b = 4

∴ (50 + 4) (50 – 4) = 50² – 4²

…. [(a + b)(a – b) = a² – b²]

= 2500 – 16 = 2484

∴ 54 × 46 = 2484

iv. 98 × 102 = (100 – 2) (100 + 2)

Here, a = 100, b = 2

∴ (100 – 2) (100 + 2) = 100² – 2²

…. [(a + b)(a – b) = a² – b²]

= 10000 – 4

= 9996

∴ 98 × 102 = 9996