zaheerpashamd

M.Sc, B.Ed ; Teacher in Mathematics, SDVR ZPHS B. Gangaram, Sathuaplly (Md) Khammam (dt) Telangana, India

10th Maths Real Numbers Exercise 1.3 Solutions

Exercise 1.3 Solutions – Class X Mathematics

Exercise 1.3 Solutions – Class X Mathematics

These solutions are based on the Telangana State Class X Mathematics textbook, focusing on decimal expansions and rational numbers. Mathematical expressions are rendered using MathJax.

1. Write the following rational numbers in their decimal form and also state which are terminating and which are non-terminating, repeating decimal.

(i) \( \frac{3}{8} \)

\( \frac{3}{8} = 3 \div 8 = 0.375 \) (long division)
Denominator \( 8 = 2^3 \), only 2 as factor, so terminating.

Decimal = 0.375, Terminating

(ii) \( \frac{229}{400} \)

\( \frac{229}{400} = 229 \div 400 = 0.5725 \) (long division)
Denominator \( 400 = 2^4 \times 5^2 \), only 2 and 5, so terminating.

Decimal = 0.5725, Terminating

(iii) \( \frac{4}{5} \)

\( \frac{4}{5} = 4 \div 5 = 0.8 \) (long division)
Denominator \( 5 = 5^1 \), only 5, so terminating.

Decimal = 0.8, Terminating

(iv) \( \frac{2}{11} \)

\( \frac{2}{11} = 2 \div 11 = 0.181818\ldots \) (long division)
Repeating digit: 18
Denominator \( 11 \) has factor other than 2 or 5, so non-terminating repeating.

Decimal = 0.\overline{18}, Non-terminating repeating

(v) \( \frac{8}{125} \)

\( \frac{8}{125} = 8 \div 125 = 0.064 \) (long division)
Denominator \( 125 = 5^3 \), only 5, so terminating.

Decimal = 0.064, Terminating

2. Without performing division, state whether the following rational numbers will have a terminating decimal form or a non-terminating, repeating decimal form.

A rational number \( \frac{p}{q} \) (in lowest form) has a terminating decimal if \( q = 2^m \times 5^n \), where \( m \) and \( n \) are non-negative integers.

(i) \( \frac{13}{3125} \)

Denominator \( 3125 = 5^5 \)
Only 5 as factor, so terminating.

Terminating

(ii) \( \frac{15}{16} \)

Denominator \( 16 = 2^4 \)
Only 2 as factor, so terminating.

Terminating

(iii) \( \frac{23}{2^3 \cdot 5^2} \)

Denominator \( 2^3 \times 5^2 = 8 \times 25 = 200 \)
Only 2 and 5 as factors, so terminating.

Terminating

(iv) \( \frac{7218}{3^2 \cdot 5^2} \)

Denominator \( 3^2 \times 5^2 = 9 \times 25 = 225 \)
Has 3 as factor (not just 2 or 5), so non-terminating repeating.

Non-terminating repeating

(v) \( \frac{143}{110} \)

Denominator \( 110 = 2 \times 5 \times 11 \)
Has 11 as factor (not just 2 or 5), so non-terminating repeating.

Non-terminating repeating

(vi) \( \frac{23}{2^3 \cdot 5^2} \)

Denominator \( 2^3 \times 5^2 = 8 \times 25 = 200 \)
Only 2 and 5 as factors, so terminating.

Terminating

(vii) \( \frac{129}{2^2 \cdot 5^2 \cdot 7^2} \)

Denominator \( 2^2 \times 5^2 \times 7^2 = 4 \times 25 \times 49 = 4900 \)
Has 7 as factor (not just 2 or 5), so non-terminating repeating.

Non-terminating repeating

(viii) \( \frac{9}{15} \)

Denominator \( 15 = 3 \times 5 \)
Has 3 as factor (not just 2 or 5), so non-terminating repeating.

Non-terminating repeating

(ix) \( \frac{36}{100} \)

Denominator \( 100 = 2^2 \times 5^2 \)
Only 2 and 5 as factors, so terminating.

Terminating

(x) \( \frac{77}{210} \)

Denominator \( 210 = 2 \times 3 \times 5 \times 7 \)
Has 3 and 7 as factors (not just 2 or 5), so non-terminating repeating.

Non-terminating repeating

3. Write the following rationals in decimal form using Theorem 1.4.

(i) \( \frac{13}{25} \)

Denominator \( 25 = 5^2 \)
Express with power of 10: \( \frac{13}{25} = \frac{13 \times 4}{25 \times 4} = \frac{52}{100} = 0.52 \)
Terminating decimal.

Decimal = 0.52

(ii) \( \frac{15}{16} \)

Denominator \( 16 = 2^4 \)
Express with power of 10: \( \frac{15}{16} = \frac{15 \times 625}{16 \times 625} = \frac{9375}{10000} = 0.9375 \)
Terminating decimal.

Decimal = 0.9375

(iii) \( \frac{23}{2^3 \cdot 5^2} \)

Denominator \( 2^3 \times 5^2 = 8 \times 25 = 200 \)
Express with power of 10: \( \frac{23}{200} = \frac{23 \times 5}{200 \times 5} = \frac{115}{1000} = 0.115 \)
Terminating decimal.

Decimal = 0.115

(iv) \( \frac{7218}{3^2 \cdot 5^2} \)

Denominator \( 3^2 \times 5^2 = 9 \times 25 = 225 \)
Has 3, cannot be expressed as \( 2^m \times 5^n \) alone, so non-terminating repeating.
Approximate: \( 7218 \div 225 \approx 32.08 \) (repeating).

Decimal = 32.08\overline{…}, Non-terminating repeating

(v) \( \frac{143}{110} \)

Denominator \( 110 = 2 \times 5 \times 11 \)
Has 11, cannot be expressed as \( 2^m \times 5^n \) alone, so non-terminating repeating.
Approximate: \( 143 \div 110 \approx 1.3 \) (repeating).

Decimal = 1.3\overline{…}, Non-terminating repeating

4. Express the following decimals in the form of \( \frac{p}{q} \), and write the prime factors of \( q \). What do you observe?

(i) 43.123

Let \( x = 43.123 \)
\( 1000x = 43123.123 \)
\( 1000x – x = 43123.123 – 43.123 \)
\( 999x = 43080 \)
\( x = \frac{43080}{999} \)
Prime factors of \( 999 = 3^3 \times 37 \)
Observation: Denominator has factors other than 2 and 5, indicating non-terminating repeating.

Form = \(\frac{43080}{999}\), Prime factors of \( q = 3^3 \times 37\)

(ii) 0.1201201

Let \( x = 0.1201201 \)
\( 1000000x = 120120.1 \)
\( 1000x = 120.1201 \)
\( 1000000x – 1000x = 120120.1 – 120.1201 \)
\( 999000x = 120000 \)
\( x = \frac{120000}{999000} = \frac{4}{33} \) (simplify)
Prime factors of \( 33 = 3 \times 11 \)
Observation: Denominator has factors other than 2 and 5, indicating non-terminating repeating.

Form = \(\frac{4}{33}\), Prime factors of \( q = 3 \times 11\)

(iii) 43.12

Let \( x = 43.12 \)
\( 100x = 4312.12 \)
\( 100x – x = 4312.12 – 43.12 \)
\( 99x = 4269 \)
\( x = \frac{4269}{99} = \frac{1423}{33} \) (simplify)
Prime factors of \( 33 = 3 \times 11 \)
Observation: Denominator has factors other than 2 and 5, indicating non-terminating repeating.

Form = \(\frac{1423}{33}\), Prime factors of \( q = 3 \times 11\)

(iv) 0.63

Let \( x = 0.63 \)
\( 100x = 63.63 \)
\( 100x – x = 63.63 – 0.63 \)
\( 99x = 63 \)
\( x = \frac{63}{99} = \frac{7}{11} \) (simplify)
Prime factors of \( 11 = 11 \)
Observation: Denominator has factors other than 2 and 5, indicating non-terminating repeating.

Form = \(\frac{7}{11}\), Prime factors of \( q = 11\)

10th Maths Real Numbers Exercise 1.2 Solutions

Exercise 1.2 Solutions – Class X Mathematics

Exercise 1.2 Solutions – Class X Mathematics

These solutions are based on the Telangana State Class X Mathematics textbook, focusing on the Fundamental Theorem of Arithmetic. Mathematical expressions are rendered using MathJax.

1. Express each of the following numbers as a product of its prime factors.

(i) 140

Divide by smallest prime: \( 140 \div 2 = 70 \)
\( 70 \div 2 = 35 \)
\( 35 \div 5 = 7 \)
\( 7 \div 7 = 1 \)
Prime factors: \( 2 \times 2 \times 5 \times 7 = 2^2 \times 5 \times 7 \)

Product = \( 2^2 \times 5 \times 7 \)

(ii) 156

\( 156 \div 2 = 78 \)
\( 78 \div 2 = 39 \)
\( 39 \div 3 = 13 \)
\( 13 \div 13 = 1 \)
Prime factors: \( 2 \times 2 \times 3 \times 13 = 2^2 \times 3 \times 13 \)

Product = \( 2^2 \times 3 \times 13 \)

(iii) 3825

\( 3825 \div 5 = 765 \)
\( 765 \div 5 = 153 \)
\( 153 \div 3 = 51 \)
\( 51 \div 3 = 17 \)
\( 17 \div 17 = 1 \)
Prime factors: \( 5 \times 5 \times 3 \times 3 \times 17 = 5^2 \times 3^2 \times 17 \)

Product = \( 5^2 \times 3^2 \times 17 \)

(iv) 5005

\( 5005 \div 5 = 1001 \)
\( 1001 \div 7 = 143 \)
\( 143 \div 11 = 13 \)
\( 13 \div 13 = 1 \)
Prime factors: \( 5 \times 7 \times 11 \times 13 \)

Product = \( 5 \times 7 \times 11 \times 13 \)

(v) 7429

\( 7429 \div 17 = 437 \) (17 is a prime factor)
\( 437 \div 19 = 23 \) (19 and 23 are prime)
\( 23 \div 23 = 1 \)
Prime factors: \( 17 \times 19 \times 23 \)

Product = \( 17 \times 19 \times 23 \)

2. Find the LCM and HCF of the following integers by the prime factorization method.

(i) 12, 15 and 21

Prime Factorization:
  • \( 12 = 2^2 \times 3 \)
  • \( 15 = 3 \times 5 \)
  • \( 21 = 3 \times 7 \)
HCF: Lowest power of common factors = \( 3 \)
LCM: Highest power of all factors = \( 2^2 \times 3 \times 5 \times 7 = 420 \)

HCF = 3, LCM = 420

(ii) 17, 23, and 29

Prime Factorization:
  • \( 17 = 17 \)
  • \( 23 = 23 \)
  • \( 29 = 29 \)
HCF: No common factors, so \( 1 \)
LCM: \( 17 \times 23 \times 29 = 11339 \)

HCF = 1, LCM = 11339

(iii) 8, 9, and 25

Prime Factorization:
  • \( 8 = 2^3 \)
  • \( 9 = 3^2 \)
  • \( 25 = 5^2 \)
HCF: No common factors, so \( 1 \)
LCM: \( 2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 1800 \)

HCF = 1, LCM = 1800

(iv) 72 and 108

Prime Factorization:
  • \( 72 = 2^3 \times 3^2 \)
  • \( 108 = 2^2 \times 3^3 \)
HCF: Lowest power of common factors = \( 2^2 \times 3^2 = 36 \)
LCM: Highest power of all factors = \( 2^3 \times 3^3 = 216 \)

HCF = 36, LCM = 216

(v) 306 and 657

Prime Factorization:
  • \( 306 = 2 \times 3 \times 3 \times 17 = 2 \times 3^2 \times 17 \)
  • \( 657 = 3 \times 3 \times 73 = 3^2 \times 73 \)
HCF: Lowest power of common factors = \( 3^2 = 9 \)
LCM: Highest power of all factors = \( 2 \times 3^2 \times 17 \times 73 = 22338 \)

HCF = 9, LCM = 22338

3. Check whether \(6^n\) can end with the digit 0 for any natural number \(n\).

For a number to end with 0, it must be divisible by 10, i.e., have factors \( 2 \) and \( 5 \).
\( 6^n = (2 \times 3)^n = 2^n \times 3^n \)
Contains \( 2 \) but no \( 5 \), so not divisible by 10.
Examples: \( 6^1 = 6 \), \( 6^2 = 36 \), \( 6^3 = 216 \) (none end with 0).

Conclusion: \( 6^n \) cannot end with 0 for any natural number \( n \).

4. Explain why \(7 \times 11 \times 13 + 13\) and \(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5\) are composite numbers.

First number: \( 7 \times 11 \times 13 + 13 \)
Factor out 13: \( 13 \times (7 \times 11 + 1) = 13 \times (77 + 1) = 13 \times 78 \)
78 is composite (\( 78 = 2 \times 3 \times 13 \)), so \( 13 \times 78 \) is composite.
Second number: \( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 \)
\( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \)
\( 5040 + 5 = 5045 \)
Check divisibility: \( 5045 \div 5 = 1009 \) (ends with 5, divisible by 5), and 1009 is prime but \( 5045 = 5 \times 1009 \), a product of primes > 1.

Conclusion: Both are composite.

5. How will you show that \((17 \times 11 \times 2) + (17 \times 11 \times 5)\) is a composite number? Explain.

Factor out common terms: \( (17 \times 11 \times 2) + (17 \times 11 \times 5) = 17 \times 11 \times (2 + 5) \)
\( 2 + 5 = 7 \)
So, \( 17 \times 11 \times 7 \)
Product of three primes \( > 1 \), hence composite.
Calculate: \( 17 \times 11 = 187 \), \( 187 \times 7 = 1309 \)
1309 is composite (e.g., divisible by 7: \( 1309 \div 7 = 187 \)).

Conclusion: It is composite.

6. What is the last digit of \(6^{100}\)?

Last digit depends on the units digit of \(6^n\).
\( 6^1 = 6 \) (ends with 6)
\( 6^2 = 36 \) (ends with 6)
\( 6^3 = 216 \) (ends with 6)
Pattern: Units digit is always 6 for any \( n \).
So, \( 6^{100} \) ends with 6.

Last digit = 6

10th Maths Real Numbers Exercise 1.1 Solutions

Exercise 1.1 Solutions – Class X Mathematics

Exercise 1.1 Solutions – Class X Mathematics

These solutions are based on the Telangana State Class X Mathematics textbook, focusing on Euclid’s algorithm and the division algorithm. Mathematical expressions are rendered using MathJax.

1. Use Euclid’s algorithm to find the HCF of:

Euclid’s algorithm: For two positive integers \(a\) and \(b\) (where \(a > b\)), divide \(a\) by \(b\), take the remainder, and repeat until the remainder is 0. The last non-zero remainder is the HCF.

(i) 900 and 270

Step 1: \( 900 = 270 \times 3 + 90 \) (Quotient = 3, Remainder = 90)
Step 2: \( 270 = 90 \times 3 + 0 \) (Quotient = 3, Remainder = 0)
Since remainder is 0, HCF is the last divisor: 90.

HCF(900, 270) = 90

(ii) 196 and 38220

Step 1: \( 38220 = 196 \times 194 + 164 \) (Quotient = 194, Remainder = 164)
Step 2: \( 196 = 164 \times 1 + 32 \) (Quotient = 1, Remainder = 32)
Step 3: \( 164 = 32 \times 5 + 4 \) (Quotient = 5, Remainder = 4)
Step 4: \( 32 = 4 \times 8 + 0 \) (Quotient = 8, Remainder = 0)
Since remainder is 0, HCF is the last divisor: 4.

HCF(196, 38220) = 4

(iii) 1651 and 2032

Step 1: \( 2032 = 1651 \times 1 + 381 \) (Quotient = 1, Remainder = 381)
Step 2: \( 1651 = 381 \times 4 + 127 \) (Quotient = 4, Remainder = 127)
Step 3: \( 381 = 127 \times 3 + 0 \) (Quotient = 3, Remainder = 0)
Since remainder is 0, HCF is the last divisor: 127.

HCF(1651, 2032) = 127

2. Use division algorithm to show that any positive odd integer is of the form \(6q + 1\), \(6q + 3\), or \(6q + 5\), where \(q\) is some integer.

Division algorithm: \( a = bq + r \), where \( 0 \leq r < b \). Let \( a \) be a positive odd integer, \( b = 6 \), so \( r = 0, 1, 2, 3, 4, 5 \).
Check remainders:
  • \( r = 0 \): \( a = 6q \) (even, e.g., 6, 12)
  • \( r = 1 \): \( a = 6q + 1 \) (odd, e.g., 1, 7)
  • \( r = 2 \): \( a = 6q + 2 \) (even, e.g., 2, 8)
  • \( r = 3 \): \( a = 6q + 3 \) (odd, e.g., 3, 9)
  • \( r = 4 \): \( a = 6q + 4 \) (even, e.g., 4, 10)
  • \( r = 5 \): \( a = 6q + 5 \) (odd, e.g., 5, 11)
Since \( a \) is odd, \( r = 1, 3, 5 \). Thus, \( a = 6q + 1 \), \( 6q + 3 \), or \( 6q + 5 \).

Conclusion: Every positive odd integer is of the form \(6q + 1\), \(6q + 3\), or \(6q + 5\).

3. Use division algorithm to show that the square of any positive integer is of the form \(3p\) or \(3p + 1\).

Let \( a \) be a positive integer, \( a = 3q + r \), where \( r = 0, 1, 2 \).
Compute \( a^2 \):
  • \( r = 0 \): \( a = 3q \), \( a^2 = (3q)^2 = 9q^2 = 3(3q^2) \), so \( a^2 = 3p \) (\( p = 3q^2 \)).
  • \( r = 1 \): \( a = 3q + 1 \), \( a^2 = (3q + 1)^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1 \), so \( a^2 = 3p + 1 \) (\( p = 3q^2 + 2q \)).
  • \( r = 2 \): \( a = 3q + 2 \), \( a^2 = (3q + 2)^2 = 9q^2 + 12q + 4 = 3(3q^2 + 4q + 1) + 1 \), so \( a^2 = 3p + 1 \) (\( p = 3q^2 + 4q + 1 \)).
Thus, \( a^2 = 3p \) or \( 3p + 1 \).

Conclusion: The square of any positive integer is of the form \(3p\) or \(3p + 1\).

4. Use division algorithm to show that the cube of any positive integer is of the form \(9m\), \(9m + 1\), or \(9m + 8\).

Let \( a = 3q + r \), where \( r = 0, 1, 2 \).
Compute \( a^3 \):
  • \( r = 0 \): \( a = 3q \), \( a^3 = (3q)^3 = 27q^3 = 9(3q^3) \), so \( a^3 = 9m \) (\( m = 3q^3 \)).
  • \( r = 1 \): \( a = 3q + 1 \), \( a^3 = (3q + 1)^3 = 27q^3 + 27q^2 + 9q + 1 = 9(3q^3 + 3q^2 + q) + 1 \), so \( a^3 = 9m + 1 \) (\( m = 3q^3 + 3q^2 + q \)).
  • \( r = 2 \): \( a = 3q + 2 \), \( a^3 = (3q + 2)^3 = 27q^3 + 54q^2 + 36q + 8 = 9(3q^3 + 6q^2 + 4q) + 8 \), so \( a^3 = 9m + 8 \) (\( m = 3q^3 + 6q^2 + 4q \)).
Thus, \( a^3 = 9m \), \( 9m + 1 \), or \( 9m + 8 \).

Conclusion: The cube of any positive integer is of the form \(9m\), \(9m + 1\), or \(9m + 8\).

5. Show that one and only one out of \(n\), \(n + 2\), or \(n + 4\) is divisible by 3, where \(n\) is any positive integer.

Let \( n = 3q + r \), where \( r = 0, 1, 2 \).
Check remainders:
  • \( r = 0 \):
    • \( n = 3q \), remainder = 0 (divisible by 3).
    • \( n + 2 = 3q + 2 \), remainder = 2.
    • \( n + 4 = 3q + 4 = 3(q + 1) + 1 \), remainder = 1.
    • Only \( n \) is divisible by 3.
  • \( r = 1 \):
    • \( n = 3q + 1 \), remainder = 1.
    • \( n + 2 = 3q + 3 = 3(q + 1) \), remainder = 0 (divisible by 3).
    • \( n + 4 = 3q + 5 = 3(q + 1) + 2 \), remainder = 2.
    • Only \( n + 2 \) is divisible by 3.
  • \( r = 2 \):
    • \( n = 3q + 2 \), remainder = 2.
    • \( n + 2 = 3q + 4 = 3(q + 1) + 1 \), remainder = 1.
    • \( n + 4 = 3q + 6 = 3(q + 2) \), remainder = 0 (divisible by 3).
    • Only \( n + 4 \) is divisible by 3.
In each case, exactly one number is divisible by 3.

Conclusion: One and only one of \(n\), \(n + 2\), or \(n + 4\) is divisible by 3.

10th Maths Pair of Linear Equations In Two Variables MC Questions

Class 10 Maths: Pair of Linear Equations in Two Variables MCQ Quiz

Class 10 Mathematics: Pair of Linear Equations in Two Variables MCQ Quiz

Chapter 4 - 50 Important Questions with Detailed Explanations

Linear Equations Graphical Method Algebraic Methods Consistency Real-life Problems Equation Pairs
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10th Maths Polynomials MC Questions

Class 10 Maths: Polynomials MCQ Quiz

Class 10 Mathematics: Polynomials MCQ Quiz

Chapter 3 - 50 Important Questions with Detailed Explanations

Polynomial Definition Degree of Polynomial Zeroes of Polynomial Geometric Meaning Coefficients & Zeroes Division Algorithm
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10th Maths Sets MC Questions

Class 10 Maths: Sets MCQ Quiz

Class 10 Mathematics: Sets MCQ Quiz

Chapter 2 - 50 Important Questions with Detailed Explanations

Set Definition Roster & Set-Builder Form Subsets Venn Diagrams Union & Intersection Cardinality
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10th Maths Real Numbers MC Questions

Class 10 Maths: Real Numbers & Logarithms MCQ Quiz

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Chapter 1 - 57 Important Questions with Detailed Explanations

Euclid's Division Lemma Fundamental Theorem of Arithmetic HCF & LCM Rational & Irrational Numbers Logarithms
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Class 1 Maths Practice : Addition & Subtraction






1st Grade Math Practice


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6th Class Maths Year Plan

6th Class Maths Year Plan

YEAR PLAN  2023-2024
CLASS:-6TH EM
SUBJECT:MATHEMATICS
PERIODS REQUIRED:170
TEACHING PERIODS:75
PRACTICE PERIODS:95
MD.ZAHEER PASHA, SA (Maths), ZPSS KANDUKURU Vemsoor (md)
https://mathsstudent.com
LEARNING OUTCOMES
The learner—
Solves problems involving large numbers by applying appropriate operations (addition, subtraction, multiplication and division)
• Recognises and appreciates (through patterns) the broad classification of numbers as even, odd, prime, co-prime, etc.
• Applies HCF or LCM in a particular situation • Solves problem involving addition and subtraction of integers.
• Uses fractions and decimals in different situations which involve money, length, temperature etc. For example, 7½ metres of cloth. Distance between two places is 112.5 km etc.
• Solves problems on daily life situations involving addition and subtraction of fractions / decimals
• Uses variable with different operations to generalise a given situation. e.g., Perimeter of a rectangle with sides x units and 3 units is 2(x+3) units.
• Compares quantities using ratios in different situations. e.g., the ratio of girls to boys in a particular class in 3:2
• Uses unitary method in solving various word problems. For example, if the cost of a dozen notebooks is given she finds the cost of 7 notebooks by first finding the cost of 1 notebook.
• Describes geometrical ideas like line, line segment, open and closed figures, angle, triangle, quadrilateral, circle, etc., with the help of examples in surroundings.
• Demonstrates an understanding of angles by identifying examples of angles in the surroundings classifying angles according to their measure estimating the measure of angles demonstrates an understanding of line symmetry by identifying symmetrical 2-dimensional (2-D) shapes which are symmetrical along one or more lines creating symmetrical 2-D shapes.
• Classifies triangles into different groups/ types on the basis of their angles and sides. For example- scalene, isosceles or equilateral on the basis of sides, etc.
• Classifies quadrilaterals into different groups/types on the basis of their sides / angles. https://mathsstudent.com/        

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7th Class Maths Year Plan

7th Class Maths Year Plan

YEAR PLAN  2023-2024
CLASS:-7TH EM
SUBJECT:MATHEMATICS
PERIODS REQUIRED:170
TEACHING PERIODS:74
PRACTICE PERIODS:96
LEARNING OUTCOMES
The learner—
• Multiplies/divides two integers
• Interprets the division and multiplication of fractions
. • For example interprets as of . Also is interpreted as how many make ?
• Uses algorithms to multiply and divide fractions/decimals.
• Solves problems related to daily life situations involving rational numbers
• Uses exponential form of numbers to simplify problems involving multiplication and division of large numbers
. • Represents daily life situations in the form of a simple equation and solves it
• Adds/subtracts algebraic expressions • Distinguishes quantities that are in proportion. For example, tells that 15, 45, 40, 120 are in proportion as is the same as
• Solves problems related to conversion of percentage to fraction and decimal and vice versa
• Caculates profit/loss percent and rate percent in simple interest
• Classifies pairs of angles based on their properties as linear, supplementary,complementary, adjacent and vertically opposite and finds value of the one when the other is given.
• Verifies the properties of various pairs of angles formed when a transversal cuts Two lines
• Finds unknown angle of a triangle when its two angles are known
• Explains congruency of triangles on the basis of the information given about them like (SSS, SAS, ASA, RHS)
• Using ruler and a pair of compasses constructs, a line parallel to a given line from a point outside it and triangles finds out approximate area of closed shapes by using unit square grid/ graph sheet
• Calculates areas of the regions enclosed in a rectangle and a square
• Finds various representative values for simple data from her/his daily life contexts like mean, median and mode
• Recognises variability in real life situation such as, variations in the height of students in her class and uncertainty in happening of events like throwing a coin.
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9th Class Maths Year Plan

9th Class Maths Year Plan

CLASS:-9TH EM
SUBJECT:MATHEMATICS
PERIODS REQUIRED:170
TEACHING PERIODS:70
PRACTICE PERIODS:100
LEARNING OUTCOMES
The learner —
• Applies logical reasoning in classifying real numbers, proving their properties and using them in different situations.
• Identifies/classifies polynomials among algebraic expressions and factorises them by
applying appropriate algebraic identities.
• Relates the algebraic and graphical representations of a linear equation in one or two variables and applies the concept to daily life situations.
• Identifies similarities and differences among different geometrical shapes.
• Derives proofs of mathematical statements particularly related to geometrical concepts, like parallel lines, triangles, quadrilaterals, circles, etc., by applying axiomatic approach and solves problems using them.
• Finds areas of all types of triangles by using appropriate formulae and apply them in real life situations.
• Constructs different geometrical shapes like bisectors of line segments, angles and triangles under given conditions and provides reasons for the processes of such constructions.
• Develops strategies to locate points in a Cartesian plane.
• Identifies and classifies the daily life situations in which mean, median and mode can be used.
• Analyses data by representing it in different forms like, tabular form (grouped or
ungrouped), bar graph, histogram (with equal and varying width and length), and frequency polygon.
• Calculates empirical probability through experiments and describes its use in words.
• Derives formulae for surface areas and volumes of different solid objects like,
cubes, cuboids, right circular cylinders/ cones, spheres and hemispheres and applies
them to objects found in the surroundings.
• Solves problems that are not in the familiar context of the child using above learning. These problems should include the situations to which the child is not exposed earlier        

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UNNATHI Maths Teaching Periods and Steps     

UNNATHI Maths Teaching Periods and Steps 

 Teaching Period-1 (TP-1)( Introduction Period)

  1. పూర్వ భావనల అవగాహన (Under Stand Pre-requisites)
  2. పాఠం యొక్క ప్రాముఖ్యత (Relevance of lesson)
  3. భావనల క్రమం(Concept Map) (Flow Chart)
  4. సారాంశం (Summary)   

Teaching Period-2 (TP-2) ( Chapter Teaching)

  1. పూర్వ భావనలు ( Chapter Pre-requisites)
  2. కీలక భావనలు(Key Concepts)/  కీలక పదాలు(Key Words)
  3. భావనల అవగాహన ( Conceptual under standing)/అభ్యసన కృత్యాలు
  4. సమస్యా సాధన (Problem solving)
  5. మాదిరి సమస్య సాధన (Model Problem Solving)
  6. పునఃశ్చరణ(Sum Up)/సారాంశం

Practice Period (PP)

  1. కీలక పదాలు లేదా భావనలు (Key Words/Concepts)
  2. ఒకే రూప వాక్యాలు/ఒకే రకమైన వాక్యాలు చదవడం
  3. విద్యార్ధులచే సమస్యా సాధన (Problem solving )
  4. రాసిన వాక్యాలు/సాధించిన సమస్యలు సరిచేయుట /న్యూమరసి స్కిల్స్ సాధించడం

 Tangents and Secants to a Circle

Tangents and Secants to a Circle

  • The line PQ and the circle have no common point. In this case PQ is a nonintersecting line with respect to the circle.
  • The line PQ intersects the circle at two points A and B. It forms a chord AB on the circle with two common points. In this case the line PQ is a secant of the circle.
  • There is only one point A, common to the line PQ and the circle. This line is called a tangent to the circle.

The common point of the tangent and the circle is called the point of contact and the tangent is said to touch the circle at the common point.

  • We can draw infinite tangents to the circle.
  • We can draw two tangents to the circle from a point away from it.
  • The tangent at any point of a circle is perpendicular to the radius through the point of contact.

EXERCISE-9.I

  1. Fill in the blanks
    • i) A tangent to a circle intersects it in ….one………… point (s).
    • ii) A line intersecting a circle in two points is called a………secant……
    • iii) The number of tangents drawn at the end points of the diameter is……. two…….
    • iv) The common point of a tangent to a circle and the circle is called ……. point of contact…..
    • v) We can draw…. infinite….. tangents to a given circle.
  2. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 13 cm. Find length of PQ.

𝑃𝑄2 + 𝑂𝑃2 = 𝑂𝑄2

𝑃𝑄2 + 52 = 132

𝑃𝑄2 + 25 = 169

𝑃𝑄2 = 169 − 25 = 144 = 122

∴ 𝑃𝑄 = 12 𝑐𝑚

3. Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.

∠𝑂𝐴𝑃 = 900 𝑎𝑛d ∠𝑂𝐵𝑆 = 900

⇒ ∠𝐵𝐴𝑃 = 900 𝑎𝑛𝑑 ∠𝐴𝐵𝑆 = 900

⇒ ∠𝐵𝐴𝑃 = ∠𝐴𝐵𝑆

⇒ 𝐴𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑒 𝑖𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝑎𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝑒𝑞𝑢𝑎𝑙

⇒ 𝑃𝑄 ∥ 𝑅𝑆

Prob1:Prove The centre of a circle lies on the bisector of the angle between two tangents drawn from a point outside it.