Mathematics Part i Solutions Solutions for Class 8 Math Chapter 5 Algebra are provided here with simple step-by-step explanations. These solutions for Algebra are extremely popular among class 8 students for Math Algebra Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part i Solutions Book of class 8 Math Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part i Solutions Solutions. All Mathematics Part i Solutions Solutions for class 8 Math are prepared by experts and are 100% accurate.

#### Page No 73:

#### Question 1:

Can’t you find these products on your own?

(i) (*p* + *q*) (2*m* + 3*n*)

(ii) (4*x* + 3*y*) (2*a* + 3*b*)

(iii) (4*a* + 2*b*) (5*c* + 3*d*)

(iv) (*m* + *n*) (5*a* + *b*)

(v) (2*x* + 3*y*) (*x* + 2*y*)

(vi) (3*a* + 2*b*) (*x* + 2*y*)

#### Answer:

By the general principle, we know that:

(*x *+ *y*) (*u* + *v*) = *xu* + *xv* + *yu* + *yv*

(i)

(*p* + *q*) (2*m* + 3*n*) = (*p* × 2*m*) + (*p* × 3*n*) + (*q* × 2*m*) + (*q* × 3*n*)

= 2*pm *+ 3*pn* + 2*qm* + 3*qn*

(ii)

(4*x* + 3*y*) (2*a* + 3*b*) = (4*x* × 2*a*) + (4*x* × 3*b*) + (3*y* × 2*a*) + (3*y* × 3*b*)

= 8*ax* + 12*bx* + 6*ay* + 9*by*

(iii)

(4*a* + 2*b*) (5*c* + 3*d*) = (4*a* × 5*c*) + (4*a* × 3*d*) + (2*b* × 5*c*) + (2*b* × 3*d*)

= 20*ac* + 12*ad* + 10*bc* + 6*bd*

(iv)

(*m* + *n*) (5*a* + *b*) = (*m* × 5*a*) + (*m* × *b*) + (*n* × 5*a*) + (*n *× *b*)

= 5*am *+ *bm* + 5*an* + *bn*

(v)

(2*x* + 3*y*) (*x* + 2*y*) = (2*x* × *x*) + (2*x* × 2*y*) + (3*y* × *x*) + (3*y* × 2*y*)

= 2*x*^{2} + 4*xy* + 3*xy* + 6*y*^{2}

(vi)

(3*a* + 2*b*) (*x* + 2*y*) = (3*a* × *x*) + (3*a* × 2*y*) + (2*b* × *x*) + (2*b* × 2*y*)

= 3*ax* + 6*ay* + 2*bx* + 4*by*

#### Page No 74:

#### Question 1:

Find these products on your own:

(i) (*x* + 3*y*) (2*a* − *b*)

(ii) (3*x* + 5*y*) (3*m* − 2*n*)

(iii) (2*r* − 3*s*) (*t* − *u*)

(iv) (*a* − *b*) (4*x* − 3*y*)

(v) (3*a* − 5*b*) (2*c* − *d*)

(vi) (2*p* + 5*q*) (3*r* − 4*s*)

#### Answer:

By the general principle, we know that:

(*x *+ *y*) (*u* − *v*) = *xu* − *xv* + *yu* − *yv*

(*x *− *y*) (*u* − *v*) = *xu* − *xv* − *yu* + *yv*

(*x *− *y*) (*u* + *v*) = *xu* + *xv* − *yu* − *yv*

(i)

(*x* + 3*y*) (2*a* − *b*) = (*x *× 2*a*) − (*x *× *b*) + (3*y* × 2*a*) − (3*y* × *b*)

= 2*ax* − *bx* + 6*ay* − 3*by*

(ii)

(3*x* + 5*y*) (3*m* − 2*n*) = (3*x* × 3*m*) − (3*x* × 2*n*) + (5*y* × 3*m*) − (5*y* × 2*n*)

= 9*mx* − 6*nx* + 15*my* − 10*ny*

(iii)

(2*r* − 3*s*) (*t* − *u*) = (2*r* × *t*) − (2*r* × *u*) − (3*s* × *t*) + (3*s* × *u*)

= 2*rt* − 2*ru* − 3*st* + 3*su*

(iv)

(*a* − *b*) (4*x* − 3*y*) = (*a* × 4*x*) − (*a* × 3*y*) − (*b* × 4*x*) + (*b *× 3*y*)

= 4*ax* − 3*ay* − 4*bx* + 3*by*

(v)

(3*a* − 5*b*) (2*c* − *d*) = (3*a* × 2*c*) − (5*b* × 2*c*) − (3*a* × *d*) + (5*b* × *d*)

= 6*ac* − 10*bc* − 3*ad* + 5*bd*

(vi)

(2*p* + 5*q*) (3*r* − 4*s*) = (2*p* × 3*r*) − (2*p* × 4*s*) + (5*q* × 3*r*) − (5*q* × 4*s*)

= 6*pr* − 8*ps* + 15*qr* −20*qs*

#### Page No 76:

#### Question 1:

Find the squares of the following algebraic expressions

(i) 2*x* + 3*y*

(ii) *x* + 2

(iii) 2*x* + 1

#### Answer:

We know that (*x* + *y*)^{2} = *x*^{2} + *y*^{2} + 2*xy*

(i)

Square of 2*x* + 3*y *= (2*x* + 3*y*)^{2}

= (2*x*)^{2} + (3*y*)^{2} + 2 × 2*x *× 3*y*

= 4*x*^{2 }+ 9*y*^{2} + 12*xy*

(ii)

Square of *x* + 2* = *(*x* + 2)^{2}

= (*x*)^{2} + (ii)^{2} + 2 × 2 × *x*

= *x*^{2} + 4 + 4*x*

(iii)

Square of 2*x* + 1* *= (2*x* + 1)^{2}

= (2*x*)^{2} + (i)^{2 }+ 2 × 2*x* × 1

= 4*x*^{2} + 1+ 4*x*

#### Page No 76:

#### Question 2:

Compute the squares of these numbers mentally.

(i) 102

(ii) 202

(iii) 1001

(iv) 2002

(v) 205

(vi) 10.3

#### Answer:

We know that (*x* + *y*)^{2} = *x*^{2} + *y*^{2} + 2*xy*

(i)

102* *= 100 + 2

⇒ (102)^{2} = (100 + 2)^{2}

= (100)^{2} + 2 × 100* *× 2 + 2^{2}

= 10000 + 400 + 4

=10404

(ii)

202* *= 200 + 2

⇒ (202)^{2} = (200 +2)^{2}

= (200)^{2} + 2 × 200* *× 2 + 2^{2}

= 40000 + 800 + 4

= 40804

(iii)

1001* *= 1000 + 1

⇒ (1001)^{2} = (1000 + 1)^{2}

= (1000)^{2} + 2 × 1000* *× 1 + 1^{2}

= 1000000 + 2000 + 1

= 1002001

(iv)

2002* *= 2000 + 2

⇒ (2002)^{2} = (2000 + 2)^{2}

= (2000)^{2} + 2 × 2000* *× 2 + 2^{2}

= 4000000 + 8000 + 4

= 4008004

(v)

205* *= 200 + 5

⇒ (205)^{2} = (200 + 5)^{2}

= (200)^{2} + 2 × 200* *× 5 + 5^{2}

= 40000 + 2000 + 25

= 42025

(vi)

10.3* *= 10 + 0.3

⇒ (10.3)^{2} = (10 + 0.3)^{2}

= (10)^{2} + 2 × 10* *× 0.3 + (0.3)^{2}

= 100 + 6 + 0.09

=106.09

#### Page No 76:

#### Question 3:

Prove that *x*^{2} + 6*x* + 9 is a perfect square for all natural numbers *x*. What can you say about its square root?

#### Answer:

To show that the expression *x*^{2} + 6*x* + 9 is a perfect square, we need to show that it can be written as the square of a linear expression.

*x*^{2} + 6*x* + 9 = *x*^{2} + 2 × 3 × *x *+ 3^{2}

= (*x* + 3)^{2 }{(*x* + *y*)^{2} = *x*^{2} + *y*^{2} + 2*xy*}

∴ *x*^{2} + 6*x* + 9 = (*x* + 3)^{2}

Hence, *x*^{2} + 6*x* + 9 is a perfect square.

Square root of *x*^{2} + 6*x* + 9 =

=

= *x* + 3

#### Page No 76:

#### Question 4:

Prove that in the sequence 1, 2, 3, ….. of natural numbers, 1 added to the product of any two alternatives numbers perfect square.

#### Answer:

Let us assume a number *a* in the sequence of natural numbers 1, 2, 3,…

Alternate number after *a* = *a* + 2

Product of the two alternate numbers = (*a* + 2) × *a*

= *a*^{2 }+ 2*a*

If one is added to this expression, it will become *a*^{2 }+ 2*a* + 1.

Now, *a*^{2 }+ 2*a* + 1 = *a*^{2} + 2 × *a* × 1 + 1^{2}

= (*a* + 1)^{2 }{(*x* + *y*)^{2} = *x*^{2} + *y*^{2} + 2*xy*}

Thus, when 1 is added to the product of any two alternate natural numbers, the result obtained is a perfect square.

#### Page No 77:

#### Question 1:

Find the squares of the algebraic expressions given below:

(i) 2*x* − 3*y*

(ii) *x* − 2

(iii) 2*x* − 1

#### Answer:

We know that (*x* − *y*)^{2} = *x*^{2} + *y*^{2} − 2*xy*

(i)

Square of 2*x* − 3*y *= (2*x* − 3*y*)^{2}

= (2*x*)^{2} + (3*y*)^{2 }− 2 × 2*x* × 3*y*

= 4*x*^{2} + 9*y*^{2 }− 12*xy*

(ii)

Square of *x* − 2* *= (*x* − 2)^{2}

=* x*^{2} + 2^{2} − 2 × *x* × 2

= *x*^{2} + 4 − 4*x*

(iii)

Square of 2*x* − 1* *= (2*x* − 1)^{2}

= (2*x*)^{2} + (i)^{2} − 2 × 2*x* × 1

= 4*x*^{2} + 1 − 4*x*

#### Page No 77:

#### Question 2:

Mentally compute the squares of the numbers given below:

(i) 98

(ii) 198

(iii) 999

(iv) 1998

(v) 195

(vi) 9.7

#### Answer:

We know that (*x* − *y*)^{2} = *x*^{2} + *y*^{2} − 2*xy*

(i)

98* *= 100 − 2

⇒ (98)^{2} = (100 − 2)^{2}

= (100)^{2} + 2^{2 }− 2 × 100* *× 2

= 10000 + 4 − 400

= 9604

(ii)

198* *= 200 − 2

⇒ (198)^{2} = (200 − 2)^{2}

= (200)^{2} + 2^{2 }− 2 × 200* *× 2

= 40000 + 4 − 800

= 39204

(iii)

999 = 1000 −1

⇒ (999)^{2} = (1000 −1)^{2}

= (1000)^{2} + 1^{2 }− 2 × 1000* *× 1

= 1000000 + 1 − 2000

= 998001

(iv)

1998* *= 2000 − 2

⇒ (1998)^{2} = (2000 − 2)^{2}

= (2000)^{2} + 2^{2 }− 2 × 2000* *× 2

= 4000000 + 4 − 8000

= 3992004

(v)

195* *= 200 − 5

⇒ (195)^{2} = (200 − 5)^{2}

= (200)^{2} + 5^{2} − 2 × 200* *× 5

= 40000 + 25 − 2000

= 38025

(vi)

9.7* *= 10 − 0.3

⇒ (9.7)^{2} = (10 − 0.3)^{2}

= (10)^{2} + (0.3)^{2 }− 2 × 10* *× 0.3

= 100 + 0.09 − 6

= 94.09

#### Page No 77:

#### Question 3:

Prove that *x*^{2} − 6*x* + 9 is a perfect square for all natural numbers *x*. What can you say about its square root?

#### Answer:

To show that the expression *x*^{2} − 6*x* + 9 is a perfect square, we need to show that it can be written as the square of a linear expression.

⇒ *x*^{2} − 6*x* + 9 = *x*^{2} − 2 × 3 × *x *+ 9

= (*x* − 3)^{2 }{(*x* − *y*)^{2} = *x*^{2} + *y*^{2} − 2*xy*}

∴ *x*^{2} − 6*x* + 9 = (*x* − 3)^{2}

Hence, *x*^{2} − 6*x* + 9 is a perfect square.

Square root of *x*^{2} − 6*x* + 9 =

=

= *x* − 3

#### Page No 78:

#### Question 1:

Can’t you easily find the products below like this?

(i) 51 × 49

(ii) 98 × 102

(iii) 10.2 × 9.8

(iv) 7.3 × 6.7

#### Answer:

We know that (*x* − *y*)(*x* + *y*) = *x*^{2} − *y*^{2}

(i)

The number 51 can be written as (50 + 1) and 49 can be written as (50 − 1).

∴ 51 × 49 = (50 + 1) (50 − 1)

= 50^{2} − 1^{2}

= 2500 − 1

= 2499

(ii)

The number 98 can be written as (100 − 2) and 102 can be written as (100 + 2).

∴ 98 × 102 = (100 − 2) (100 + 2)

= (100)^{2} − 2^{2}

= 10000 − 4

= 9996

(iii)

The number 10.2 can be written as (10 + 0.2) and 9.8 can be written as (10 − 0.2).

∴ 10.2 × 9.8 = (10 + 0.2) (10 − 0.2)

= (10)^{2} − (0.2)^{2}

= 100 − 0.04

= 99.96

(iv)

The number 7.3 can be written as (7 + 0.3) and 6.7 can be written as (7 − 0.3).

7.3 × 6.7 = (7 + 0.3) × (7 − 0.3)

= 7^{2} − (0.3)^{2}

= 49 − 0.09

= 48.91

#### Page No 79:

#### Question 1:

Compute the following in your head:

(i) 67^{2} − 33^{2}

(ii) 123^{2} − 122^{2}

(iii)

(iv) 0.27^{2} − 0.23^{2}

#### Answer:

We know that *x*^{2} − *y*^{2} = (*x* + *y*) (*x *− *y*)

(i)

67^{2 }− 33^{2} = (67 + 33) (67 − 33)

= 100 × 34

= 3400

(ii)

123^{2 }− 122^{2} = (123 + 122) (123 − 122)

= 245 × 1

= 245

(iii)

(iv)

(0.27)^{2 }− (0.23)^{2} = (0.27 + 0.23) (0.27 − 0.23)

= 0.5 × 0.04

= 0.02

#### Page No 79:

#### Question 2:

The diagonal of a rectangle is 65 centimetres long and one of its sides is 63 centimeres long. What is the length of the other side?

#### Answer:

Given: A rectangle ABCD with length of one side, DC = 63 cm and length of the diagonal, BD = 65 cm.

We know that each angle of a rectangle measures 90°.

In ΔBCD, ∠BCD = 90°

Using Pythagoras theorem, we get:

BD^{2} = BC^{2} + CD^{2}

⇒ (65)^{2} = BC^{2} + (63)^{2}

⇒ BC^{2 }= (65)^{2} − (63)^{2}

= (65 − 63) (65 + 63) {*x*^{2} − *y*^{2} = (*x* − *y*) (*x *+ *y*)}

= 2 × 128

= 256

⇒ BC = 16

Thus, the length of the other side of the rectangle is 16 cm.

#### Page No 80:

#### Question 1:

Prove that the difference of the squares of two consecutive natural is equal to their sum.

#### Answer:

Let *n* be a natural number.

Its consecutive natural number = *n* + 1

Sum of these consecutive natural numbers = *n* + (*n* +1) = 2*n* +1

Difference of the squares of two consecutive natural numbers

= (*n* + 1)^{2 }− *n*^{2}

= (*n *+ 1 − *n*) (*n* + 1 +* n*) {*x*^{2} − *y*^{2} = (*x* − *y*) (*x *+ *y*)}

= 2*n* + 1

Thus, the difference of the squares of two consecutive natural numbers is equal to their sum.

#### Page No 80:

#### Question 2:

In how many different ways you can write 15 as the difference of the squares of two natural numbers?

#### Answer:

The factors of 15 are 1, 3, 5 and 15.

We can write 15 as:

15 = 1 × 15

= (8 − 7) (8 + 7) {*x*^{2} − *y*^{2} = (*x* − *y*) (*x *+ *y*)}

= 8^{2} − 7^{2}

So, 15 can be written as the difference of the squares of 8 and 7.

15 = 3 × 5

= (4 − 1) (4 + 1) {*x*^{2} − *y*^{2} = (*x* − *y*) (*x *+ *y*)}

= 4^{2} − 1^{2}

So, 15 can be written as the difference of the squares of 4 and 1.

Thus, there are two ways to write 15 as the difference of the squares of two natural numbers.

View NCERT Solutions for all chapters of Class 8