Category: CBSE

Extra questions for Class 9 Maths Chapter 3 Coordinate Geometry are given here. These questions will help the students to build a better understanding of the Coordinate Geometry concept in Maths. These Class 9 Chapter 3 questions are important for CBSE Class 9 Maths exams. https://meritbatch.com/coordinate-geometry-ncert-extra-questions-for-class-9-maths-chapter-3/

Class 9 Coordinate Geometry Extra Questions

Extra Questions for Class 9 Maths Chapter 3 Coordinate Geometry

Coordinate Geometry Class 9 Extra Questions


Coordinate Geometry Class 9 Worksheet With Solutions

Coordinate Geometry Class 9 Questions With Solutions


Ch 3 Maths Class 9 Extra Questions

Class 9 Maths Ch 3 Extra Questions




Class 9 Maths Chapter 3 Extra Questions With Solutions

Coordinate Geometry Class 9 Worksheet

Class 9 Maths Chapter 3 Test Paper

Coordinate Geometry Class 9 Extra Questions With Solutions

Extra Questions For Class 9 Maths Chapter 3

NCERT Solutions for Class 9 Maths

Extra questions for Class 9 Maths Chapter 4 Linear Equations in Two Variables are given here. These questions will help the students to build a better understanding of the Linear Equations in Two Variables concept in Maths. These Class 9 Chapter 4 questions are important for CBSE Class 9 Maths exams. https://meritbatch.com/linear-equations-in-two-variables-ncert-extra-questions-for-class-9-maths/

Class 9 Linear Equations in Two Variables Extra Questions

Extra Questions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

NCERT Solutions for Class 9 Maths

Extra questions for Class 9 Maths Chapter 2 Polynomials are given here. These questions will help the students to build a better understanding of the Polynomials concept in Maths. These Class 9 Chapter 2 questions are important for CBSE Class 9 Maths exams. https://meritbatch.com/polynomials-ncert-extra-questions-for-class-9-maths/

Class 9 Polynomials Extra Questions

Extra Questions for Class 9 Maths Chapter 2 Polynomials

Polynomials Class 9 Extra Questions

Polynomial Calculator is an online tool that helps to calculate the result of addition, subtraction, multiplication, and division of two polynomials.

Polynomial Class 9 Extra Questions

Class 9 Maths Chapter 2 Extra Questions

Is this a polynomial calculator allows you to divide two polynomials to find the quotient and the remainder of the division.

Polynomials Class 9 Important Questions

Polynomial Questions Class 9

Class 9 Maths Polynomials Extra Questions

Polynomials Class 9 Questions


Polynomials Class 9 Extra Questions With Answers Pdf

Extra Questions For Class 9 Maths Polynomials

NCERT Solutions for Class 9 Maths

Extra questions for Class 9 Maths Chapter 6 Lines and Angles are given here. These questions will help the students to build a better understanding of the Lines and Angles concept in Maths. These Class 9 Chapter 6 questions are important for CBSE Class 9 Maths exams. https://meritbatch.com/lines-and-angles-ncert-extra-questions-for-class-9-maths/

Class 9 Lines and Angles Extra Questions

Extra Questions for Class 9 Maths Chapter 6 Lines and Angles
Lines And Angles Class 9 Extra Questions

Class 9 Lines And Angles Extra Questions

Lines And Angles Class 9 Worksheet With Answers

Lines And Angles Class 9 Important Questions

Lines And Angles Class 9 Questions

Extra Questions Of Lines And Angles Class 9

Class 9 Maths Ch 6 Extra Questions

Lines And Angles Class 9 Questions With Solutions


Questions On Lines And Angles For Class 9

Lines And Angles Class 9 Questions And Answers

Ch 6 Maths Class 9 Extra Questions

Extra Questions For Class 9 Maths Chapter 6

NCERT Solutions for Class 9 Maths

Extra questions for Class 9 Maths Chapter 7 Triangles are given here. These questions will help the students to build a better understanding of the Triangles concept in Maths. These Class 9 Chapter 7 questions are important for CBSE Class 9 Maths exams. https://meritbatch.com/triangles-ncert-extra-questions-for-class-9-maths-chapter-7/

 Class 9 Triangles Extra Questions

Extra Questions for Class 9 Maths Chapter 7 Triangles

Triangles Class 9 Extra Questions

NCERT Solutions for Class 9 Maths

Playing with Numbers Class 8 Extra Questions Maths Chapter 16 are created from the examination perspective and hence will be a great help for students’ exam preparation. These questions are available in PDF format and students can download them for free from Extra Questions for Class 8 Maths. https://meritbatch.com/playing-with-numbers-ncert-extra-questions-for-class-8-maths-chapter-16/

Class 8 Playing With Numbers Extra Questions

Extra Questions for Class 8 Maths Chapter 16 Playing with Numbers

Playing with Numbers Class 8 Extra Questions Very Short Answer Type

Playing With Numbers Class 8 Extra Questions Question 1.
Write the following numbers in generalised form.
(a) ab
(b) 85
(c) 132
(d) 1000
Solution:
(a) ab = 10 × a + 1 × b = 10a + b
(b) 85 = 10 × 8 + 1 × 5 = 10 × 8 + 5
(c) 132 = 100 × 1 + 10 × 3 + 1 × 2 = 100 × 1 + 10 × 3 + 2
(d) 1000 = 1000 × 1

Class 8 Maths Chapter 16 Extra Questions Question 2.
Write the following in usual form.
(a) 3 × 100 + 0 × 10 + 6
(b) 5 × 1000 + 3 × 100 + 2 × 10 + 1
Solution:
(a) 3 × 100 + 0 × 10 + 6 = 300 + 0 + 6 = 306
(b) 5 × 1000 + 3 × 100 + 2 × 10 + 1 = 5000 + 300 + 20 + 1 = 5321

Playing With Numbers Class 8 Worksheet Question 3.
Which of the following numbers are divisible by 3?
(i) 106
(ii) 726
(iii) 915
(iv) 1008
Solution:
(i) Sum of the digits of 106 = 1 + 0 + 6 = 7 which is not divisible by 3.
Hence 106 is not divisible by 3.
(ii) Sum of the digits of 726 = 7 + 2 + 6 = 15 which is divisible by 3.
Hence 726 is divisible by 3.
(iii) Sum of the digits of 915 = 9 + 1 + 5 = 15 which is divisible by 3.
Hence 915 is divisible by 3.
(iv) Sum of the digits of 1008 = 1 + 0 + 0 + 8 = 9 which is divisible by 3.
Hence 1008 is divisible by 3.

Playing With Numbers Class 8 Extra Questions With Answers Question 4.
Prove that the sum of the given numbers and the numbers obtained by reversing their digits is divisible by 11.
(a) 89
(b) ab
(c) 69
(d) 54
Solution:
(a) Given number = 89
Number obtained by reversing the order of digits = 98
Sum = 89 + 98 = 187 ÷ 11 = 17
Hence, the required number is 11.

(b) Given number = ab = 10a + b
Number obtained by reversing the digits = 10b + a
Sum = (10a + b) + (10b + a)
= 10a + b + 10b + a
= 11a + 11b
= 11(a + b) ÷ 11
= a + b

(c) Given number = 69
Number obtained by reversing the digits = 96
Sum = 69 + 96 = 165 ÷ 11 = 15
Hence, the required number is 11.

(d) Given number = 54
Number obtained by reversing the digits = 45
Sum = 54 + 45 = 99 ÷ 11 = 9
Hence, the required number is 11.

Playing With Numbers Class 8 Worksheets With Answers Question 5.
Prove that the difference of the given numbers and the numbers obtained by reversing their digits is divisible by 9.
(i) 59
(ii) xy
(iii) xyz
(iv) 203
Solution:
(i) Given number = 59
Number obtained by reversing the digits = 95
Difference = 95 – 59 = 36 ÷ 9 = 4
Hence, the required number is 9.

(ii) Given number = xy = 10x + y
Number obtained by reversing the digits = 10y + x
Difference = (10x + y) – (10y + x)
= 10x + y – 10y – x
= 9x – 9y
= 9(x – y) ÷ 9
= x – y
Hence, the required number is 9.

(iii) Given number = xyz = 100x + 10y + z
Number obtained by reversing the digits = 100z + 10y + x
Difference = (100x + 10y + z) – (100z + 10y + x)
= 100x + 10y + z – 100z – 10y – x
= 99x – 99z
= 99(x – z)
= 99(x – 2) ÷ 9
= 11 (x – z)
Hence, the required number is 9.

(iv) Given number = 203
Number obtained by reversing the digits = 302
Difference = 302 – 203 = 99 ÷ 9 = 11
Hence, the required number is 9.

Class 8 Maths Ch 16 Extra Questions Question 6.
If a, b, c are three digits of a three-digit number, prove that abc + cab + bca is a multiple of 37.
Solution:
We have abc + cab + bca
abc = 100a + 10b + c
cab = 100c + 10a + b
bca = 100b + 10c + a
Adding abc + cab + bca = 111a +111b + 111c
= 111 (a + b+ c)
= 37 × 3 (a + b + c) which is a multiple of 37.
Hence proved.

Playing with Numbers Class 8 Extra Questions Short Answer Type

Ch 16 Maths Class 8 Extra Questions Question 7.
Complete the magic square given below so that the sum of the numbers in each row or in each column or along each diagonal is 15.

Solution:
(i) A = 15 – (8 + 1) = 15 – 9 = 6
(ii) F = 15 – (8 + 5) = 15 – 13 = 2
(iii) C = 15 – (A + F) = 15 – (6 + 2) = 15 – 8 = 7
(iv) E = 15 – (1 + 5) = 15 – 6 = 9
(v) D = 15 – (E + F) = 15 – (9 + 2) = 15 – 11 = 4
(vi) B = 15 – (8 + 4) = 15 – 12 = 3
Hence the required square is

Playing With Numbers Extra Questions Question 8.
Find the values of P and Q from the given addition problem

Solution:
Here, 3 + Q = 7
⇒ Q = 7 – 3 = 4
Now taking second column, we get
4 + 7 = 11 i.e. 1 is carried over to third column
⇒ 1 + P + 2 = 9
⇒ 3 + P = 9
P = 9 – 3 = 6
Hence the value of P = 6 and Q = 4

Class 8 Playing With Numbers Extra Questions Question 9.
Find the values of p, q and r in the following multiplication problem.

Solution:
6 × 4 = 24, Here 2 is carried over second column
⇒ 6 × p + 2 – 3 × 10 = 2 [∵ 21 – 3 × 6 = 3]
⇒ 6p – 30 = 0
⇒ p = 5
Now the multiplication problem becomes,

Here 2 + r = 4
⇒ r = 2
q × 354 = 1062
⇒ q = 3
Hence, p = 5, q = 3, r = 2

Questions On Playing With Numbers For Class 8 Question 10.
Observe the following patterns:
1 × 9 – 1 = 8
21 × 9 – 1 = 188
321 × 9 – 1 = 2888
4321 × 9 – 1 = 38888
Find the value of 87654321 × 9 – 1
Solution:
From the pattern, we observe that there are as many eights in the result as the first digit from the right which is to be multiplied by 9 and reduced by 1.
87654321 × 9 – 1 = 788888888

Playing with Numbers Class 8 Extra Questions High Order Thinking Skills (HOTS) Type

Chapter 16 Maths Class 8 Extra Questions Question 11.
Complete the cross number puzzle with the given column.


Solution:

Hence the complete square is

Question 12.
The product of two 2-digit numbers is 1431. The product of their tens digits is 10 and the product of their units digits is 21. Find the numbers. Solution:
Let the required two 2-digit numbers be 10a + b and 10p + q as per the condition, we have
a × p = 10 and b × q = 21
a = 2 and p = 5 or a = 5 and p = 2
Similarly b × q = 21
b = 3 and q = 7 or b = 7 and q = 3
10p + q = 57 or 10p + q = 53
and 10a + b = 23 or 10a + b = 27
Since the units digit of product 1431 is 1.
Numbers are 57 and 23 or 53 and 27.
Now 57 × 23 = 1311 and 53 × 27 = 1431 which is given.
Hence, the required numbers are 53 and 27.

NCERT Solutions for Class 8 Maths

Visualising Solid Shapes Class 8 Extra Questions Maths Chapter 10 are created from the examination perspective and hence will be a great help for students’ exam preparation. These questions are available in PDF format and students can download them for free from Extra Questions for Class 8 Maths. https://meritbatch.com/visualising-solid-shapes-ncert-extra-questions-for-class-8-maths-chapter-10/

Class 8 Visualising Solid Shapes Extra Questions

Extra Questions for Class 8 Maths Chapter 10 Visualising Solid Shapes

Visualising Solid Shapes Class 8 Extra Questions Very Short Answer Type

Visualising Solid Shapes Class 8 Extra Questions Question 1.
Draw any four 3-dimensional figures.
Solution:

Visualising Solid Shapes Class 8 Worksheets Question 2.
Verify Euler’s formula for a right triangular prism.

Solution:
Number of vertices (V) = 6
Number of faces (F) = 5
and number of edges (E) = 9
Euler’s formula:
V + F – E = 2
⇒ 6 + 5 – 9 = 2
⇒ 2 = 2
Hence, the formula is verified.

Visualising Solid Shapes Class 8 Worksheets Pdf Question 3.
Find the number of vertices of hexagonal prisms.
Solution:
Number of vertices = 2 × Number of sides = 2 × 6 = 12

Visualising Solid Shapes Extra Questions Question 4.
Verify whether a polyhedron can have 10 faces, 20 edges and 15 vertices.
Solution:
We have
Number of faces F = 10
Number of edges E = 20
and number of vertices V = 15
Euler’s formula:
V + F – E = 2
⇒ 15 + 10 – 20 = 2
⇒ 5 ≠ 2
Hence, it is not possible to have a polyhedron satisfying the above data.

Visualizing Solid Shapes Class 8 Extra Questions Question 5.
If F = 18 and V = 10, then find the value of E in Euler’s formula.
Solution:
We know that
V + F – E = 2
⇒ 10 + 18 – E = 2
⇒ 28 – E = 2
⇒ E = 28 – 2 = 26
Hence, the required value of E = 26

Class 8 Maths Ch 10 Extra Questions Question 6.
Draw the front, side and top views of the following 3-D figures.

Solution:

Visualising Solid Shapes Class 8 Important Questions Question 7.
Draw the nets of the following polyhedrons.
(i) Cuboid
(ii) Triangular prism with a base equilateral triangle.
(iii) Square pyramid.
Solution:
(i) The net pattern of cuboid

(ii) The net pattern of a triangular prism

(iii) Net pattern of square pyramid

Ch 10 Maths Class 8 Extra Questions Question 8.
The given net is made up of two equilateral triangles and three rectangles.

(i) Name the solid it represents.
(ii) Find the number of faces, edges and vertices.
Solution:
(i) The given figure represents the net prims of the triangular prism
(ii) Number of faces = 5
Number of edges = 9
Number of vertices = 6

Visualising Solid Shapes Class 8 Extra Questions Short Answer Type

Extra Questions For Class 8 Maths Visualising Solid Shapes Question 9.
Match the following:

Solution:
(a) → (vi)
(b) → (v)
(c) → (ii)
(d) → (i)
(e) → (iii)
(f) → (v)

Ncert Class 8 Maths Chapter 10 Extra Questions Question 10.
Using Euler’s formula, fill in the blanks:

	Faces	Vertices	Edges
(a)	6	8	—
(b)	—	10	15
(c)	4	—	6
(d)	5	6	—
(e)	8	12	—
(f)	7	7	—

Solution:
(a) F + V – E = 2
⇒ 6 + 8 – E = 2
⇒ 14 – E = 2
⇒ E = 14 – 2 = 12

(b) F + V – E = 2
⇒ F + 10 – 15 = 2
⇒ F – 5 = 2
⇒ F = 2 + 5 = 7

(c) F + V – E = 2
⇒ 4 + V – 6 = 2
⇒ V – 2 = 2
⇒ V = 2 + 2 = 4

(d) F +V – E = 2
⇒ 5 + 6 – E = 2
⇒ 11 – E = 2
⇒ E = 11 – 2 = 9

(e) F + V – E = 2
⇒ 8 + 12 – E = 2
⇒ 20 – E = 2
⇒ E = 20 – 2 = 18

(f) F + V – E = 2
⇒ 7 + 7 – E = 2
⇒ 14 – E = 2
⇒ E = 14 – 2 = 12
Hence (a) → 12, (b) → 7, (c) → 4, (d) → 9, (e) → 18, (f) → 12

Questions On Visualising Solid Shapes Class 8 Question 11.
Name the solids that have:
(i) 4 faces
(ii) 8 triangular faces
(iii) 6 faces
(iv) 1 curved surface
(v) 5 faces and 5 vertices
(vi) 6 rectangular faces and 2 hexagonal faces
Solution:
(i) Tetrahedron
(ii) Regular octahedron
(iii) Cube and cuboid
(iv) Cylinder
(v) Square and a rectangular pyramid
(vi) Hexagonal prism

Extra Questions For Class 8 Maths Chapter 10 Question 12.
Complete the table:

Solid	F	V	E	F + V	E + 2
Cuboid	—	—	—	—	—
Triangular pyramid	—	—	—	—	—
Triangular prism	—	—	—	—	—
Pyramid with square base	—	—	—	—	—
Prism with square base	—	—	—	—	—

Solution:

Solid	F	V	E	F + V	E + 2
Cuboid	6	8	12	14	14
Triangular pyramid	4	4	6	8	8
Triangular prism	5	6	9	11	11
Pyramid with square base	5	5	8	10	10
Prism with square base	6	8	12	14	14

Class 8 Maths Visualising Solid Shapes Extra Questions Question 13.
Use isometric dot paper to sketch a rectangular prism with length 4 units, height 2 units and width 3 units. (NCERT Exemplar)
Solution:
Steps:
1. Draw a parallelogram with sides 4 units and 3 units. This is the top of the prism (Fig. 1).

2. Start at one vertex. Draw a line passing through two dots. Repeat for the other three vertices. Draw the hidden edges as a dashed line (Fig. 2).

3. Connect the ends of the lines to complete the prism (Fig. 3).

NCERT Solutions for Class 8 Maths

Algebraic Expressions and Identities Class 8 Extra Questions Maths Chapter 9 are created from the examination perspective and hence will be a great help for students’ exam preparation. These questions are available in PDF format and students can download them for free from Extra Questions for Class 8 Maths. https://meritbatch.com/algebraic-expressions-and-identities-ncert-extra-questions-for-class-8-maths/

Class 8 Algebraic Expressions and Identities Extra Questions

Extra Questions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities

Algebraic Expressions and Identities Class 8 Extra Questions Very Short Answer Type

Algebraic Expressions And Identities Class 8 Worksheets With Solutions Question 1.
Write two examples of each of
(i) Monomials
(ii) Binomials
(iii) Trinomials
Solution:
(i) Monomials:
(a) 3x
(b) 5xy²
(ii) Binomials:
(a) p + q
(b) -5a + 2b
(iii) Trinomials:
(a) a + b + c
(b) x² + x + 2

Algebraic Expressions Class 8 Extra Questions Question 2.
Identify the like expressions.
5x, -14x, 3x² + 1, x², -9x², xy, -3xy
Solution:
Like terms: 5x and -14x, x² and -9x², xy and -3xy

Class 8 Maths Chapter 9 Extra Questions Question 3.
Identify the terms and their coefficients for each of the following expressions:
(i) 3x²y – 5x
(ii) xyz – 2y
(iii) -x – x²
Solution:

Class 8 Algebraic Expressions Extra Questions Question 4.
Add: -3a²b², –\(\frac { 5 }{ 2 }\) a²b², 4a²b², \(\frac { 2 }{ 3 }\) a²b²
Solution:

Algebraic Expressions And Identities Class 8 Extra Questions With Answers Question 5.
Add: 8x² + 7xy – 6y², 4x² – 3xy + 2y² and -4x² + xy – y²
Solution:

Class 8 Maths Algebraic Expressions Extra Questions Question 6.
Subtract: (4x + 5) from (-3x + 7)
Solution:
(-3x + 7) – (4x + 5) = -3x + 7 – 4x – 5 = -3x – 4x + 7 – 5 = -7x + 2

Questions On Algebraic Expressions For Class 8 Question 7.
Subtract: 3x² – 5x + 7 from 5x² – 7x + 9
Solution:
(5x² – 7x + 9) – (3x² – 5x + 7)
= 5x² – 7x + 9 – 3x² + 5x – 7
= 5x² – 3x² + 5x – 7x + 9 – 7
= 2x² – 2x + 2

Algebraic Expressions Class 8 Worksheets With Answers Question 8.
Multiply the following expressions:
(a) 3xy² × (-5x²y)
(b) \(\frac { 1 }{ 2 }\) x²yz × \(\frac { 2 }{ 3 }\) xy²z × \(\frac { 1 }{ 5 }\) x²yz
Solution:

Ch 9 Maths Class 8 Extra Questions Question 9.
Find the area of the rectangle whose length and breadths are 3x²y m and 5xy² m respectively.
Solution:
Length = 3x²y m, breadth = 5xy² m
Area of rectangle = Length × Breadth = (3x²y × 5xy²) sq m = (3 × 5) × x²y × xy² sq m = 15x³y³ sq m

Algebraic Expressions Extra Questions Class 8 Question 10.
Multiply x² + 7x – 8 by -2y.
Solution:

Algebraic Expressions and Identities Class 8 Extra Questions Short Answer Type

Extra Questions On Algebraic Expressions For Class 8 Question 11.
Simplify the following:
(i) a² (b² – c²) + b² (c² – a²) + c² (a² – b²)
(ii) x²(x – 3y²) – xy(y² – 2xy) – x(y³ – 5x²)
Solution:
(i) a² (b² – c²) + b² (c² – a²) + c² (a² – b²)
= a²b² – a²c²) + b²c² – b²a²) + c²a² – c²b²)
= 0
(ii) x²(x – 3y²) – xy(y² – 2xy) – x(y³ – 5x²)
= x³ – 3x²y² – xy³ + 2x²y² – xy³ + 5x³
= x³ + 5x³ – 3x²y² + 2x²y² – xy³ – xy³
= 6x³ – x²y² – 2xy³

Ncert Class 8 Maths Chapter 9 Extra Questions Question 12.
Multiply (3x² + 5y²) by (5x² – 3y²)
Solution:
(3x² + 5y²) × (5x² – 3y²)
= 3x²(5x² – 3y²) + 5y²(5x² – 3y²)
= 15x⁴ – 9x²y² + 25x²y² – 15y⁴
= 15x⁴ + 16x²y² – 15y⁴

Chapter 9 Class 8 Maths Extra Questions Question 13.
Multiply (6x² – 5x + 3) by (3x² + 7x – 3)
Solution:
(6x² – 5x + 3) × (3x² + 7x – 3)
= 6x²(3x² + 7x – 3) – 5x(3x² + 7x – 3) + 3(3x² + 7x – 3)
= 18x⁴ + 42x³ – 18x² – 15x³ – 35x² + 15x + 9x² + 21x – 9
= 18x⁴ + 42x³ – 15x³ – 18x² – 35x² + 9x² + 15x + 21x – 9
= 18x⁴ + 27x³ – 44x² + 36x – 9

Algebraic Expressions Class 8 Questions Question 14.
Simplify:
2x²(x + 2) – 3x (x² – 3) – 5x(x + 5)
Solution:
2x²(x + 2) – 3x(x² – 3) – 5x(x + 5)
= 2x³ + 4x² – 3x³ + 9x – 5x² – 25x
= 2x³ – 3x³ – 5x² + 4x² + 9x – 25x
= -x³ – x² – 16x

Extra Questions Of Algebraic Expressions For Class 8 Question 15.
Multiply x² + 2y by x³ – 2xy + y³ and find the value of the product for x = 1 and y = -1.
Solution:
(x² + 2y) × (x³ – 2xy + y³)
= x²(x³ – 2xy + y³) + 2y(x³ – 2xy + y³)
= x⁵ – 2x³y + x²y³ + 2x³y – 4xy² + 2y⁴
= x⁵ + x²y³ – 4xy² + 2y⁴
Put x = 1 and y = -1
= (1)⁵ + (1)² (-1)³ – 4(1)(-1)² + 2(-1)⁴
= 1 + (1) (-1) – 4(1)(1) + 2(1)
= 1 – 1 – 4 + 2
= -2

Class 8 Chapter 9 Maths Extra Questions Question 16.
Using suitable identity find:
(i) 48² (NCERT Exemplar)
(ii) 96²
(iii) 231² – 131²
(iv) 97 × 103
(v) 181² – 19² = 162 × 200 (NCERT Exemplar)
Solution:

Class 8 Ch 9 Maths Extra Questions Question 17.

Solution:

Question 18.
Verify that (11pq + 4q)² – (11pq – 4q)² = 176pq² (NCERT Exemplar)
Solution:
LHS = (11pq + 4q)² – (11pq – 4q)² = (11pq + 4q + 11pq – 4q) × (11pq + 4q – 11pq + 4q)
[using a² -b² = (a – b) (a + b), here a = 11pq + 4q and b = 11 pq – 4q]
= (22pq) (8q)
= 176 pq²
= RHS.
Hence Verified.

Question 19.
Find the value of \(\frac { { 38 }^{ 2 }-{ 22 }^{ 2 } }{ 16 }\), using a suitable identity. (NCERT Exemplar)
Solution:

Question 20.
Find the value of x, if 10000x = (9982)² – (18)² (NCERT Exemplar)
Solution:
RHS = (9982)² – (18)² = (9982 + 18)(9982 – 18)
[Since a² -b² = (a + b) (a – b)]
= (10000) × (9964)
LHS = (10000) × x
Comparing L.H.S. and RHS, we get
10000x = 10000 × 9964
x = 9964

NCERT Solutions for Class 8 Maths