Worksheet on polygon angle sum theorem

Note: Sum of all angles of an n-sided polygon is given by =(n-2)×180  Problem: 01 Find the angle measure of x  Problem: 02 What is the sum of all angles of  (i) a hexagon  (ii) an octagon  (iii) a regular 14 sided polygon  Problem: 03 In the following figures, find the value of x. (i) … Read more

Worksheet on square root using factorization method

In this worksheet, you’ll get various problems to solve based on square root using the factorization method. But, if you don’t know, how we find square root using the factorization method, check these posts  Learn to find square root using the factorization method part 1  Finding square root using factorization method with more problems  Q.1 … Read more

Finding square root using ones tens method

Procedure for calculating square root using one’s tens method  Step: 01 The given number should be a perfect square. Step: 02 We all know, perfect square ending with 1,4,6 or 9 have two possible one digits. As you can see in the given table : Step: 03 Select the last two-digit of the given number … Read more

Finding square root using prime factorization method part 2

  Q.1 Find the square root of 625 using the prime factorization method  Solution:  625=5×5×5×5  Make pairs 625=5×5×5×5  √625=√5×5×5×5  √625=5×5 √625=25  Q.2 Find the square root of 144  using the prime factorization method  Solution:  144=2×2×2×2×3×3 Make pairs  144=2×2×2×2×3×3 √144=√2×2×2×2×3×3 √144=2×3×2 √144=12 Q.3 Find the square root of 400 using the prime factorization method  Solution:  400=2×2×2×2×5×5 … Read more

Finding Square root using prime factorization method

Procedure to find square root using prime factorization method  Write the prime factors of the given number  Make pairs of the equal factors  Write one factor corresponding to each pair  Multiply the obtained factors Q.1 Find the square root of 256 using the prime factorization method. Solution: 256=2×2×2×2×2×2×2×2 Make pairs  256=2×2×2×2×2×2×2×2 √256=√2×2×2×2×2×2×2×2 √256=2×2×2×2 √256=16  Q.2 … Read more

Worksheet on algebraic identities with answer

Here are some questions solution of algebraic identities 

Q.1 Write the following in the expanded form 

(i) (3x+4y)²

(ii) (2x-3y)(2x+3y) 

(iii) (y-6)(y-9) 

Q.2 Using algebraic identity evaluate : 

(i) (98)²

(ii) 48×52

(iii) 101 ×103 

(iv) 153×153 -47×47 

Q.3 If x-\frac{1}{x}=5 ,find the value of the 

(i) x²+\frac{1}{x²}

(ii) x+\frac{1}{x}

Q.4 Find z ,if 9z=79²-61²

Q.5 Show that 

(i) (2x-7y)²+28xy=4x²+49y²

(ii) (a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0 

Solution: 

Problem: 01

Write the following in the expanded form 

(i) (3x+4y)²

(ii) (2x-3y)(2x+3y) 

(iii) (y-6)(y-9) 

Solution:(i) 

Using the identity :  (x+y)²=x²+y²+2xy 

(3x+4y)²=(3x)²+(4y)²+2.(3x).(4y) 

(3x+4y)²=9x²+16y²+24xy 

Solution:(ii) 

Using the identity : (x-y)(x+y)=x²-y²

(2x-3y)(2x+3y) =(2x)²-(3y)²

(2x-3y)(2x+3y) =4x²-9y²

Solution:(iii) 

Using the identity : (x+a)(x+b)=x²+(a+b)x+ab 

(y-6)(y-9) 

Here a=-6 and b=-9 

(y-6)(y-9) =y²+{-6+(-9)}y+(-6)(-9) 

(y-6)(y-9) =y²+{-6-9}y+54 

(y-6)(y-9) =y²-15y+54 

Problem: 02

Using algebraic identity evaluate : 

(i) (98)²

(ii) 48×52

(iii) 101 ×103 

(iv) 153×153 -47×47 

Solution:(i) 

(98)² can be written as (100-2)² which is of form (a-b)²

Using the identity : (a-b)²=a²-2ab+b²

(100-2)²=(100)²-2.100.2.+(2)²

(100-2)²=10000-400+4 

(100-2)²=9604 

Solution(ii) 

48×52 can be written as (50-2)×(50+2) which is of form (a-b)(a+b) 

Using the identity : (a-b)(a+b)=a²-b²

(50-2)×(50+2)=(50)²-(2)²

(50-2)×(50+2)=2500-4 

(50-2)×(50+2)=2496 

Solution(iii)

101 ×103 can be written as (100+1)(100+3) which is of form (x+a)(x+b) 

Using the identity : (x+a)(x+b) =x²+(a+b)x+ab 

 We have a=1 and b=3 

(100+1)(100+3)=(100)²+(1+3)(100)+1.3

(100+1)(100+3)=10000+(4)(100)+3 

(100+1)(100+3)=10000+400+3 

(100+1)(100+3)=10403 

Solution(iv) 

153×153 -47×47  is of form a.a-b.b=a²-b² 

Using identity : a²-b²=(a+b)(a-b) 

153×153 -47×47 =(153)²-(47)²

(153)²-(47)²=(153+47)(153-47) 

(153)²-(47)²=106×200

(153)²-(47)²=21200 

Problem:03

If x-\frac{1}{x}=5 ,find the value of the 

(i) x²+\frac{1}{x²}

(ii) x+\frac{1}{x}

Solution(i) 

We have  , x-\frac{1}{x}=5

Squaring on both sides 

(x-\frac{1}{x})²=(5)²

Using identity : (x-y)²=x²-2xy+y² 

(x)²-2.x.\frac{1}{x}+(\frac{1}{x²})=25 

x²-2.+(\frac{1}{x²})=25 

x²+\frac{1}{x²}-2=25 

x²+\frac{1}{x²}=27 

Thus ,value of x²+\frac{1}{x²}=27                  ………………(i) 

Solution(ii) 

Find the value of (x+\frac{1}{x}

(x+\frac{1}{x})²=x²+2.x.\frac{1}{x}+\frac{1}{x²}

(x+\frac{1}{x})²=x²+2+\frac{1}{x²}                  ………………….(ii) 

We have value of x²+\frac{1}{x²}=27  from eq(i) 

Substitute in eq(ii) 

(x+\frac{1}{x})²=27+2 

(x+\frac{1}{x})²=29 

x+\frac{1}{x}=√29 

Problem: 04 Find z ,if 9z=79²-61²

Solution: 

We have , 9z=79²-61²

Using the identity :a²-b²=(a+b)(a-b) 

So, 79²-61²=(79+61)(79-61) =(140)(18) 

9z= 140.18 

z=\frac{140.18}{9}

z= 280 

Problem: 05

Show that 

(i) (2x-7y)²+28xy=4x²+49y²

(ii) (a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0 

Solution(i) :

(2x-7y)²+28xy 

Exapand Using identity : (x-y)²=x²-2xy+y²

=(2x)²-2.(2x).(7y)+(7y)²+28xy 

= 4x²-28xy+49y²+28xy 

=4x²+49y²+0 

=4x²+49y²

Hence proved (2x-7y)²+28xy=4x²+49y²

Solution(ii) 

(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0 

Using identity : (x-y)(x+y)=x²-y²

(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)

(a-b)(a+b)=a²-b²,(b-c)(b+c)=b²-c² and (c-a)(c+a)=c²-a²

=a²-b²+b²-c²+c²-a²

=0

Hence proved  ,(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0 

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Worksheet on Multiplication of binomials

Q.1 Simplify  (i) (a²+b²)(a+b)  (ii) (3x²-y²)(10x²-3y²) (iii) 3x(-6x+11y)-(5x+12y)(2x+9y)  (iv) 7x(2x-3)-9x(2x²)-5x(2x+3)  Q.2 Simplify this  (7x-9y²)(2x²-3y)-(5x+2y)(9x-10y)-2x(3x+5y)  Q.3 Find the following products  (i) (10a-11b)(5a+1)  (ii) (1 – x3) × (x2 + 2) Answer Problem: 01 Simplify  (i) (a²+b²)(a+b)  (ii) (3x²-y²)(10x²-3y²) (iii) 3x(-6x+11y)-(5x+12y)(2x+9y)  (iv) 7x(2x-3)-9x(2x²)-5x(2x+3)  Solution:(i)  (a²+b²)(a+b) =a²(a+b)+b²(a+b)  (a²+b²)(a+b) =a³+a²b+b²a+b³ (a²+b²)(a+b) =a³+b³+ab(a+b)  Solution :(ii)  (3x²-y²)(10x²-3y²)=3x²(10x²-3y²)-y²(10x²-3y²) (3x²-y²)(10x²-3y²)=30-9x²y²-10x²y²+3 (3x²-y²)(10x²-3y²)=30+3-19x²y² Solution:(iii)  3x(-6x+11y)-(5x+12y)(2x+9y) =(-18x²+33xy)-{5x(2x+9y)+12x(2x+9y)} … Read more