### Algebraic Expressions and Identities - Worksheets

CBSE Worksheet-1

Class 08 - Mathematics (Algebraic Expressions and Identities)

Class 08 - Mathematics (Algebraic Expressions and Identities)

General Instructions: All questions are compulsory. Q.1 to Q.2 carries one mark each. Q.3 to Q.7 carries two marks each. Q.8 and Q.9 carries three marks each. Q.10 to Q.12 carries four marks each.

- Evaluate : 5t + 4 if t = 3
- Write two terms which are like 4mn2
- State true or false:
- The degree of polynomial of 3x3 + 7x2y2 - 7x2 +7x is 3.
- Multiply 2 and a and add 5 to the result can be expressed as 2a + 5.
- Two more than a can be expressed as 2a.
- The numerical coefficient of 8xy3 is 8.

- Fill Up the following
- Any algebraic expression, which has one or more terms is called a _________.
- Factor of x2 – 7x + 12 is ______.
- The degree of constant term is ______.
- An_______ is an equality which is true for all values of variable in it.

- Match the following:
Column A Column B 1. (x + 1) (x+1) (a) a2–2ab + b2 2. (m - 3 ) (m + 3) (b) x2 +(a+b)x + ab 3. (y - 2) ( y - 2) (c) a2 - b2 4. (x + 3) (x +5) (d) a2 + 2ab + b2 - Find the areas of rectangle with the following monomials as their lengths and breadths respectively :(3mn, 4np)
- Obtain the product of xy, yz, zx
- Find the product using a suitable identity : (3xy – 5)2
- Evaluate: (x – 2)(x+ 2) (x2 + 4)
- The base and the altitude of a triangle are (3x – 4y) and (6x + 5y) respectively. Find its area.
- The perimeter of triangle is 8y2 – 9y + 4 and its two sides are 3y2 – 5y and 4y2 + 12. Find its third side.
- Prove that (x – y) (x+y) + (y- z) (y+ z) + (z – x) (z + x) = 0

CBSE Worksheet-1

Class 08 - Mathematics (Algebraic Expressions and Identities)

Solution

Class 08 - Mathematics (Algebraic Expressions and Identities)

Solution

- 5t + 4 = 5 × 3 + 4

= 15+ 4

= 19 - Two terms which are like 4mn2 are mn2 and – 9 mn2 .
- False
- True
- False
- True

- polynomial
- (x – 3) (x – 4)
- zero
- identity

- 1. (d), 2.(c), 3. (a), 4. (b)
- Area of the rectangle

= Length × Breadth

= (3mn) × (4np)

= (3 × 4) × (mn) × (np)

= 12 × m × (n × n) × p

= 12mn2p. - Required product

= (xy) × (yz) × (zx)

= (x × x) × (y × y) × (z × z)

= x2 × y2 × z2

= x2y2z2 - (a – b)2 = a2 – 2ab + b2

(3xy – 5)2 = (3xy)2 –2 × 3xy × 5 + (5)2

= 9x2y2 – 30xy + 25 - (x – 2)(x+ 2) (x2 + 4)

= (x2 – 22) (x2 + 4)

= (x2 – 4) (x2 + 4)

= (x2)2 – 42

= x4 – 16 - Base = (3x – 4y)

Altitude = (6x + 5y)

Area of triangle = ½ ×base × altitude

= × (3x – 4y) (6x + 5y)

= [3x (6x + 5y) – 4y (6x + 5y)]

= [18x2 + 15xy – 24xy – 20y2]

= [18x2– 9xy – 20y2] sq. unit

= 9x2– xy – 10y2] sq. unit - Perimeter of triangle = Sum of all sides

Third side = Perimeter of triangle – Sum of two sides

= (8y2 – 9y + 4) – (3y2– 5y + 4y2 + 12)

= (8y2 – 9y + 4) – (7y2 – 5y + 12)

= 8y2 – 9y + 4 – 7y2 + 5y – 12

= y2 – 4y – 8 - LHS = (x – y) (x+y) + (y- z) (y+ z) + (z – x) (z + x)

= x2 + xy – yx – y2 + y2 + yz – zy – z2 + z2 + zx – xz – x2

= 0