Mathematics Solutions Solutions for Class 7 Maths Chapter 15 Algebraic Formulae Expansion Of Squares are provided here with simple step-by-step explanations. These solutions for Algebraic Formulae Expansion Of Squares are extremely popular among Class 7 students for Maths Algebraic Formulae Expansion Of Squares Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 7 Maths Chapter 15 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 7 Maths are prepared by experts and are 100% accurate.

#### Question 1:

Expand.

(i) ( 5+ 6b)2    (ii)    (iii)    (iv)

(v)     (vi)      (vii)      (viii)

It is known that, (a + b)2 = a2 + 2ab + b2 and (a − b)2 = a2 − 2ab + b2.

#### Question 2:

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

(i)     (ii)   (iii)      (iv)

The given binomial is $\left(8-\frac{1}{x}\right)$.

Hence, the correct answer is option (iii).

#### Question 3:

Of which of the binomials given below is m2n2 + 14mnpq + 49p2q2 the expansion?

(i) ( + n) (p + q) (ii) (mn- pq) (iii) (7mn + pq) (iv) (mn + 7pq)

Let us check each of the given options.
(i) (m + n)(p + q)
= m(p + q) + n(p + q)
mp + mq + np + nq
So, it is not a correct option.

(ii) (mn − pq)2
= (mn)2 − 2 × (mn) × (pq) + (pq)2    [∵ (a − b)2 = a2 − 2ab + b2]
= m2n2 − 2mnpq + p2q2
So, it is not a correct option.

(iii) (7mn + pq)2
= (7mn)2 + 2 × (7mn) × (pq) + (pq)2    [∵ (a + b)2 = a2 + 2ab + b2]
= 49m2n2 + 14mnpq + p2q2
So, it is not a correct option.

(iv) (mn + 7pq)2
= (mn)2 + 2 × (mn) × (7pq) + (7pq)2    [∵ (a + b)2 = a2 + 2ab + b2]
= m2n2 + 14mnpq + 49p2q2
So, it is a correct option.

Hence, the correct answer is option (iv).

#### Question 4:

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

It is known that, (a + b)2 = a2 + 2ab + b2 and (a − b)2 = a2 − 2ab + b2

(i) (997)2
= (1000 − 3)2
= (1000)2 − 2 × 1000 × 3 + (3)2
= 1000000 − 6000 + 9
= 994009

(ii) (102)2
= (100 + 2)2
= (100)2 + 2 × 100 × 2 + (2)2
= 10000 + 400 + 4
= 10404

(iii) (97)2
= (100 − 3)2
= (100)2 − 2 × 100 × 3 + (3)2
= 10000 − 600 + 9
= 9409

(iv) (1005)2
= (1000 + 5)2
= (1000)2 + 2 × 1000 × 5 + (5)2
= 1000000 + 10000 + 25
= 1010025

#### 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)

It is known that, (a + b) (a − b) = a2 − b2.

#### Question 2:

Use the formula to find the values.
(i) 502 × 498 (ii) 97 × 103 (iii) 54 × 46 (iv) 98 × 102

It is known that, (a + b) (a − b) = a2 − b2.

(i) 502 × 498
= (500 + 2) × (500 − 2)
= (500)2 −  (2)2
= 250000 − 4
= 249996

(ii) 97 × 103
= (100 − 3) × (100 + 3)
= (100)2 − (3)2
= 10000 − 9
= 9991

(iii) 54 × 46
= (50 + 4) × (50 − 4)
= (50)2 − (4)2
= 2500 − 16
= 2484

(iv) 98 × 102
= (100 − 2) × (100 + 2)
= (100)2 − (2)2
= 10000 − 4
= 9996

#### Question 1:

Factorise the following expressions and write them in the product form.

(i) 201 a3 b2 , (ii) 91 xyt2 , (iii) 24 a2 b2 , (iv) tr2 s3

(i) 201a3b2
= 3 × 67 × a × a × a × b × b

(ii) 91xyt2
= 7 × 13 × x × y × t × t

(iii) 24a2b2
= 2 × 2 × 2 × 3 × a × a × b × b

(iv) tr2s3
= t × × r × s × s × s

#### Question 1:

Factorise the following expressions.
(i) p 2$-$q 2         (ii) 4x2 $-$ 25y2  (iii) y2 $-$ 4        (iv) p$-$ $\frac{1}{25}$
(v) 9x$-\frac{1}{16}{y}^{2}$ (vi) x2 $-$ $\frac{1}{{x}^{2}}$     (vii) a2b $-$ab   (viii) 4 x2y $-$x2
(ix) $\frac{1}{2}{y}^{2}$   (x)

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