Bhagya

Students can download Maths Chapter 2 Real Numbers Ex 2.6 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.6

Question 1.
Simplify the following using addition and subtraction properties of surds:
(i) 5\(\sqrt{3}\) + 18\(\sqrt{3}\) – 2\(\sqrt{3}\)
(ii) 4\(\sqrt[3]{5}\) + 2\(\sqrt[3]{5}\) – 3\(\sqrt[3]{5}\)
(iii) 3\(\sqrt{75}\) + 5\(\sqrt{48}\) – \(\sqrt{243}\)
(iv) 5\(\sqrt[3]{40}\) + 2\(\sqrt[3]{625}\) – 3\(\sqrt[3]{320}\)
Solution:
(i) 5\(\sqrt{3}\) + 18\(\sqrt{3}\) – 2\(\sqrt{3}\) = (5 + 18 – 2)\(\sqrt{3}\)
= (23 – 2) \(\sqrt{3}\) = 21\(\sqrt{3}\)

(ii) 4\(\sqrt[3]{5}\) + 2\(\sqrt[3]{5}\) – 3\(\sqrt[3]{5}\) = (4 + 2 – 3) \(\sqrt[3]{5}\)
= (6 – 3) \(\sqrt[3]{5}\) = 3\(\sqrt[3]{5}\)

(iii) 3\(\sqrt{75}\) + 5\(\sqrt{48}\) – \(\sqrt{243}\) = \(3\sqrt{5^{2}×3} + 5\sqrt{2^{4}×3} – \sqrt{3^{5}}\)

= 3 × 5\(\sqrt{3}\) + 5 × 2²\(\sqrt{3}\) – 3²\(\sqrt{3}\) = 15\(\sqrt{3}\) + 20\(\sqrt{3}\) – 9\(\sqrt{3}\)
= (15 + 20 – 9)\(\sqrt{3}\)
= (35 – 9)\(\sqrt{3}\)
= 26 \(\sqrt{3}\)

(iv) 5\(\sqrt[3]{40}\) + 2\(\sqrt[3]{625}\) – 3\(\sqrt[3]{320}\)

= 5\(\sqrt[3]{2^{3}×5} + 2\sqrt[3]{5^{3}×5} – 3\sqrt[3]{2^{3}×2^{3}×5}\)
5 × 2\(\sqrt[3]{5} + 2 × 5\sqrt[3]{5} – 3 × 2 × 2\sqrt[3]{5} \)
= 10\(\sqrt[3]{5} + 10\sqrt[3]{5} – 12\sqrt[3]{5} \)
= 20\(\sqrt[3]{5} – 12\sqrt[3]{5}\)
= 8\(\sqrt[3]{5}\)

Question 2.
Simplify the following using multiplication and division properties of surds:
(i) \(\sqrt{3}\) × \(\sqrt{5}\) × \(\sqrt{2}\)
(ii) \(\sqrt{35}\) ÷ \(\sqrt{7}\)
(iii) \(\sqrt[3]{27}\) × \(\sqrt[3]{8}\) × \(\sqrt[3]{125}\)
(iv) (7\(\sqrt{a}\) – 5\(\sqrt{b}\)) (7\(\sqrt{a}\) + 5\(\sqrt{b}\))
(v) (\(\sqrt{\frac{225}{729}} – \sqrt{\frac{25}{144}}\)) ÷ \(\sqrt{\frac{16}{81}}\)
Solution:
(i) \(\sqrt{3}\) × \(\sqrt{5}\) × \(\sqrt{2}\) = \(\sqrt{3×5×2} = \sqrt{30}\)

(ii) \(\sqrt{37} ÷ \sqrt{7} = \frac{\sqrt{35}}{\sqrt{7}} = \sqrt{\frac{35}{7}} = \sqrt{5}\)

(iii) \(\sqrt[3]{27}\) × \(\sqrt[3]{8}\) × \(\sqrt[3]{125}\) = \(\sqrt[3]{27×8×125}\)
= \(\sqrt[3]{3^{3}×2^{3}×5^{3}}\) = 3 × 2 × 5 = 30

(iv) (7\(\sqrt{a}\) – 5\(\sqrt{b}\)) (7\(\sqrt{a}\) + 5\(\sqrt{b}\))
[using a2 – b2 = (a + b) (a – b)]
(7\(\sqrt{a}\) – 5\(\sqrt{b}\)) (7\(\sqrt{a}\) + 5\(\sqrt{b}\)) = \((7\sqrt{a})^{2} – (5\sqrt{b})^{2}\) = 49a – 25b

(v) (\(\sqrt{\frac{225}{729}} – \sqrt{\frac{25}{144}}\)) ÷ \(\sqrt{\frac{16}{81}}\)


= \(\frac{5}{36}\) × \(\frac{9}{4}\)
= \(\frac{5×1}{4×4}\)
= \(\frac{5}{16}\)

Question 3.
If \(\sqrt{2}\) = 1.414, \(\sqrt{3}\) = 1.732, \(\sqrt{5}\) = 2.236, \(\sqrt{10}\) = 3.162, then find the values of the following correct to 3 places of decimals.
(i) \(\sqrt{40}\) – \(\sqrt{20}\)
(ii) \(\sqrt{300}\) + \(\sqrt{90}\) – \(\sqrt{8}\)
Solution:
(i) \(\sqrt{40}\) – \(\sqrt{20}\) = \(\sqrt{4×10} – \sqrt{4×5} = 2\sqrt{10} – 2\sqrt{5}\)
= 2 × 3.162 – 2 × 2.236 = 6.324 – 4.472 = 1.852

(ii) \(\sqrt{300}\) + \(\sqrt{90}\) – \(\sqrt{8}\) = \(\sqrt{3×100} + \sqrt{9×10} – \sqrt{4×2}\)
= 10\(\sqrt{3}\) + 3\(\sqrt{10}\) – 2\(\sqrt{2}\)
= 10 × 1.732 + 3 × 3.162 – 2 × 1.414
= 17.32 + 9.486 – 2.828
= 26.806 – 2.828
= 23.978

Question 4.
Arrange surds in descending order
(i) \(\sqrt[3]{5}\), \(\sqrt[9]{4}\), \(\sqrt[6]{3}\)
Solution:
LCM of 3, 9 and 6 is 18

\(\sqrt[3]{5}\) = \(\sqrt[3×6]{5^{6}}\) = \(\sqrt[18]{15625}\)
\(\sqrt[9]{4}\) = \(\sqrt[2×9]{4^{2}}\) = \(\sqrt[18]{16}\)
\(\sqrt[6]{3}\) = \(\sqrt[3×6]{3^{3}}\) = \(\sqrt[18]{27}\)
\(\sqrt[18]{15625}\) > \(\sqrt[18]{27}\) > \(\sqrt[18]{16}\)
\(\sqrt[3]{5}\) > \(\sqrt[6]{3}\) > \(\sqrt[9]{4}\)

(ii) \(\sqrt[2]{\sqrt[3]{5}}\), \(\sqrt[3]{\sqrt[4]{7}}\), \(\sqrt{\sqrt{3}}\)
Solution:
\(\sqrt[2]{\sqrt[3]{5}}\) = \(\sqrt[6]{5}\); \(\sqrt[3]{\sqrt[4]{7}}\) = \(\sqrt[12]{7}\); \(\sqrt{\sqrt{3}}\) = \(\sqrt[4]{3}\)
LCM of 6, 12 and 4 is 12

\(\sqrt[2]{\sqrt[3]{5}}\) = \(\sqrt[6]{5}\) = \(\sqrt[12]{5^{2}}\) = \(\sqrt[12]{25}\)
\(\sqrt[3]{\sqrt[4]{7}}\) = \(\sqrt[12]{7}\) = \(\sqrt[12]{7}\)
\(\sqrt{\sqrt{3}}\) = \(\sqrt[4]{3}\) = \(\sqrt[12]{3^{3}}\) = \(\sqrt[12]{27}\)
\(\sqrt[12]{27}\) > \(\sqrt[12]{25}\) > \(\sqrt[12]{7}\)
\(\sqrt{\sqrt{3}}\) > \(\sqrt[2]{\sqrt[3]{5}}\) > \(\sqrt[3]{\sqrt[4]{7}}\)

Question 5.
Can you get a pure surd when you find:
(i) the sum of two surds
(ii) the difference of two surds
(iii) the product of two surds
(iv) the quotient of two surds
Justify each answer with an example.
Solution:
(i) Yes we can get a surd.
Example:
(a) 3\(\sqrt{2}\) + 5\(\sqrt{2}\) = (3 + 5)\(\sqrt{2}\) = 8\(\sqrt{2}\)
(b) 3\(\sqrt{6}\) + 2\(\sqrt{6}\) = (3 + 2)\(\sqrt{6}\) = 5\(\sqrt{6}\)

(ii) Yes we can get a surd.
Example:
(a) \(\sqrt{75}\) – \(\sqrt{48}\) = \(\sqrt{25×3}\) – \(\sqrt{16×3}\) = (5 – 4) \(\sqrt{3}\) = \(\sqrt{3}\)
(b) \(\sqrt{98}\) – \(\sqrt{72}\) = \(\sqrt{49×2}\) – \(\sqrt{36×2}\) = (7 – 6) \(\sqrt{2}\) = \(\sqrt{2}\)

(iii) Yes we can get a surd.
Example:
(a) \(\sqrt{8}\) × \(\sqrt{6}\) = \(\sqrt{8×6}\) = \(\sqrt{48}\)
(b) \(\sqrt{11}\) × \(\sqrt{3}\) = \(\sqrt{11×3}\) = \(\sqrt{33}\)

(iv) Yes we can get a surd.
Example:
(a) \(\sqrt{55}\) ÷ \(\sqrt{5}\) = \(\frac{\sqrt{11×5}}{\sqrt{5}} = \sqrt{11}\)
(b) \(\sqrt{65}\) ÷ \(\sqrt{5}\) = \(\frac{\sqrt{13×5}}{\sqrt{13}} = \sqrt{5}\)

Question 6.
Can you get a rational number when you compute:
(i) the sum of two surds
(ii) the difference of two surds
(iii) the product of two surds
(iv) the quotient of two surds
Justify each answer with an example.
Solution:
(i) Yes, the sum of two surds will give a rational number.
Example:
(a) (2 + \(\sqrt{3}\)) + (2 – \(\sqrt{3}\)) = 4
(b) (\(\sqrt{5}\) + 4) + (7 – \(\sqrt{5}\)) = 11

(ii) Yes, the difference of two surds will give a rational number.
Example:
(a) (5 + \(\sqrt{7}\)) – (- 5 + \(\sqrt{7}\)) = 10
(b) (\(\sqrt{11}\) + 5) – (-3 + \(\sqrt{11}\)) = 8

(iii) Yes, the product of two surds will give a rational number.
Example:
(a) \(\sqrt{125}\) × \(\sqrt{45}\) = \(\sqrt{25×5}\) × \(\sqrt{9×5}\) = 5\(\sqrt{5}\) × 3\(\sqrt{5}\) = 15 × 5 = 75
(b) \(\sqrt{150}\) × \(\sqrt{6}\) = \(\sqrt{25×6}\) × \(\sqrt{6}\) = 5\(\sqrt{6}\) × \(\sqrt{6}\) = 5 × 6 = 30

(iv) Yes. The quotient of two surds will give a rational number.
Example:
(a) \(\sqrt{32}\) ÷ \(\sqrt{8}\) = \(\frac{\sqrt{8×4}}{\sqrt{8}} = \sqrt{4}\) = 2
(b) \(\sqrt{50}\) ÷ \(\sqrt{2}\) = \(\frac{\sqrt{25×2}}{\sqrt{2}} = \sqrt{25}\) = 5

Students can download Maths Chapter 2 Numbers and Sequences Ex 2.4 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.4

Question 1.
Find the next three terms of the following sequence.
(i) 8, 24, 72,…
(ii) 5, 1, -3, …
(iii) \(\frac { 1 }{ 4 } \), \(\frac { 2 }{ 9 } \), \(\frac { 3 }{ 16 } \)
(i) 216, 648, 1944 (This sequence is multiple of 3)
Next three terms are 216, 648, 1944
(ii) Next three terms are -7, -11, -15
(adding -4 with each term)
(iii) Next three terms are \(\frac { 4 }{ 25 } \),\(\frac { 5 }{ 36 } \) and \(\frac { 6 }{ 49 } \)
[using \(\frac{n}{(n+1)^{2}}\)]

Question 2.
Find the first four terms of the sequences whose nth terms are given by
(i) a_n = n³ – 2
(ii) a_n = (-1)ⁿ⁺¹ n(n+1)
(iii) a_n = 2n² – 6
Solution:
t_n = a_n = n³ -2
(i) a₁ = 1³ – 2 = 1 – 2 – 1
a₂ = 2³ – 2 = 8 – 2 = 6
a₃ = 3³ – 2 = 27 – 2 = 25
a₄ = 4³ – 2 = 64 – 2 = 62
∴ The first four terms are -1, 6, 25, 62, ……….

(ii) a_n = (-1)ⁿ⁺¹ n(n + 1)
a₁ = (-1)¹⁺¹ (1) (1 +1)
= (-1)² (1) (2) = 2
a₂ = (-1)²⁺¹ (2) (2 + 1)
= (-1)³ (2) (3)= -6
a₃ = (-1)³⁺¹ (3) (3 + 1)
= (-1)⁴ (3) (4) = 12
a₄ = (-1)⁴⁺¹ (4) (4 + 1)
= (-1)⁵ (4) (5) = -20
∴ The first four terms are 2, -6, 12, -20,…

(iii) a_n = 2n² – 6
a₁ = 2(1)² – 6 = 2 – 6 = -4
a₂ = 2(2)² – 6 = 8 – 6 = 2
a₃ = 2(3)² – 6 = 18 – 6 = 12
a₄ = 2(4)² – 6 = 32 – 6 = 26
∴ The first four terms are -4, 2, 12, 26, …

Question 3.
Find the n^th term of the following sequences
(i) 2, 5, 10, 17, ……
Answer:
(1² + 1);(2² + 1),(3² + 1),(4² + 1)….
n^th term is n² + 1
a_n = n² + 1

(ii) 0,\(\frac { 1 }{ 2 } \),\(\frac { 2 }{ 3 } \) ……
Answer:
(\(\frac { 1-1 }{ 1 } \)), (\(\frac { 2-1 }{ 2 } \)), (\(\frac { 3-1 }{ 3 } \)) …..
n^th term is \(\frac { n-1 }{ n } \)
a_n = \(\frac { n-1 }{ n } \)

(iii) 3,8,13,18,…….
Answer:
[5(1) -2], [5(2) – 2], [5(3) – 2], [5(4) – 2] ….
The n^th term is 5n – 2
a_n = 5n – 2

Question 4.
Find the indicated terms of the sequences whose n^th terms are given by

(i) a_n = \(\frac { 5n }{ n+2 } \) ; a₆ and a₁₃
Answer:
a_n = \(\frac { 5n }{ n+2 } \)
a₆ = \(\frac { 5(6) }{ 6+2 } \) = \(\frac { 30 }{ 8 } \) = \(\frac { 15 }{ 4 } \)
a₁₃ = \(\frac { 5(13) }{ 13+2 } \) = \(\frac{5 \times 13}{15}\) = \(\frac { 13 }{ 3 } \)
a₆ = \(\frac { 15 }{ 4 } \), a₁₃ = \(\frac { 13 }{ 3 } \)

(ii) a_n = – (n² – 4); a₄ and a₁₁
Answer:
a_n = -(n² – 4)
a₄ = -(4² – 4)
= – (16 – 4)
= -12
a₁₁ = -(11² – 4)
= – (121 – 4)
= – 117
a₄ = -12 and a₁₁ = -117

Question 5.
Find a₈ and a₁₅ whose n^th term is a_n

Answer:

Question 6.
If a₁ = 1, a₂ = 1 and a_n = 2a_n-1 + a_n-2, n > 3, n ∈ N, then find the first six terms of the sequence.
Solution:
a₁ = 1, a₂ = 1, a_n = 2a_n-1 + a_n-2
a₃ = 2a_(3-1) + a_(3-2)
= 2a₂ + a₁
= 2 × 1 + 1 = 3
a₄ = 2a_(4-1) + a_(4-2)
= 2a₃ + a₂
= 2 × 3 + 1 = 7
a₅ = 2a_(5-1) + a_(5-2)
= 2a₄ + a₃
= 2 × 7 + 3 = 17
a₆ = 2a_(6-1) + a_(6-2)
= 2a₅ + a₄
= 2 × 17 + 7
= 34 + 7
= 41
∴ The first six terms of the sequence are 1, 1, 3, 7, 17, 41 ………..

Students can download Maths Chapter 2 Numbers and Sequences Ex 2.8 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.8

Question 1.
Find the sum of first n terms of the G.P.
(i) 5, -3, \(\frac { 9 }{ 5 } \),-\(\frac { 27 }{ 25 } \), ……
(ii) 256,64,16,…….
Answer:
(i) 5,-3,\(\frac { 9 }{ 5 } \),\(\frac { 27 }{ 55 } \), ….. n terms

(ii) 256,64,16,…….
Answer:
Here a = 256, r = \(\frac { 64 }{ 256 } \) = \(\frac { 1 }{ 4 } \)

Question 2.
Find the sum of first six terms of the G.P. 5,15,45,…
Answer:
Here a = 5, r = \(\frac { 15 }{ 3 } \) = 3, n = 6

Sum of first 6 terms = 1820

Question 3.
Find the first term of the G.P. whose common ratio 5 and whose sum to first 6 terms is 46872.
Answer:
Common ratio (r) = 5
S₆ = 46872

The first term of the G.P. is 12.

Question 4.
Find the sum to infinity of (i) 9 + 3 + 1 + ….(ii) 21 + 14 + \(\frac { 28 }{ 3 } \) ……
Answer:
(i) 9 + 3 + 1 + ….
a = 9, r = \(\frac { 3 }{ 9 } \) = \(\frac { 1 }{ 3 } \)
Sum of infinity term = \(\frac { a }{ 1 – r } \) = \(\frac{9}{1-\frac{1}{3}}\)

Question 5.
If the first term of an infinite G.P. is 8 and its sum to infinity is \(\frac { 32 }{ 3 } \) then find the common ratio.
Answer:
Here a = 8, S∞ = \(\frac { 32 }{ 3 } \)
\(\frac { a }{ 1-r } \) = \(\frac { 32 }{ 3 } \)
\(\frac { 8 }{ 1-r } \) = \(\frac { 32 }{ 3 } \)
32 – 32 r = 24 ⇒ 32 r = 8
r = \(\frac { 8 }{ 32 } \) = \(\frac { 1 }{ 4 } \)
Common ration = \(\frac { 1 }{ 4 } \)

Question 6.
Find the sum to n terms of the series
(i) 0.4 + 0.44 + 0.444 + …… to n terms
Answer:


(ii) 3 + 33 + 333 + ………… to n terms
Answer:
S_n = 3 + 33 + 333 + …. to n terms
= 3[1 + 11 + 111 + …. to n terms]
= \(\frac { 3 }{ 9 } \) [9 + 99 + 999 + …. n terms]
= \(\frac { 1 }{ 3 } \) [(10 – 1) + (100 – 1) + (1000 – 1) + …… n terms]
= \(\frac { 1 }{ 3 } \) [10 + 100 + 1000 + ….. n terms – (1 + 1 + 1 ….. n terms)]

Question 7.
Find the sum of the Geometric series 3 + 6 + 12 + …….. + 1536
Answer:
3 + 6 + 12 …. +1536
a = 3, r = \(\frac { 6 }{ 3 } \) = 2
t_n = 1536


∴ Sum of the series is 3069

Question 8.
Kumar writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with the instruction that they continue the process similarly. Assuming that the process is unaltered and it costs ₹2 to mail ong letter, find the amount spent on postage when 8th set of letters is mailed.
Answer:
When kumar writes a letter to his friend.Friend writes a letter to another person.
It form a G.P
The G.P is 4, 16, 64,………
Here a = 4, r = 4
The last term is 4 (4)^8-1 = 4(4)⁷

Question 9.
Find the rational form of the number 0.123 .
Answer:
Let x = \(\overline { 0.123 } \)
= 0.123123123….
= 0.123 + 0.000123 + 000000123 + ….
This is an infinite G.P

Question 10.
If S_n = (x + y) + (x² + xy + y²) + (x³ + x²y + xy² + y³) + ………… n terms then prove that
Answer:
S_n = (x + y) + (x² + xy + y²) + (x³ + x²y + xy² + y³) + …….. n
Multiply by x
x S_n = x(x + y) + x(x² + xy + y²) + x(x³ + x²y + xy² + y³) + ……….. n
= x² + xy + x³ + x²y + xy² + x⁴ + x³y + x²y² + xy³ + …… n terms ……(1)
Multiply by y
yS_n = y(x + y) + y(x² + xy + y²) + y(x³ + x²y + xy² + y³) + ….. n
= xy + y² + x²y + xy² + y³ + x³y + x²y² + xy³ + y⁴ + ….. n terms
Subtract (1) and (2)
x S_n – y S_n = x² + xy + x³ + x²y + xy² + x⁴ + x³y + x²y² + xy³ + …….
– xy + y² + x²y + xy² + y³ + x³y + x²y² + xy³ + y⁴ + ……
(x – y) S_n = (x² + x³ + x⁴ + ……) – (y² + y³ + y⁴ + ……)
[ a = x²; r = x and a = y²; r = y, S_n = \(\frac{a\left(r^{n}-1\right)}{r-1}\)]

Hence it is proved.

Students can download Maths Chapter 2 Real Numbers Ex 2.9 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.9

Question 1.
If n is a natural number then √n is……….
(a) always a natural number
(b) always an irrational number
(c) always a rational number
(d) may be rational or irrational
Solution:
(d) may be rational or irrational

Question 2.
Which of the following is not true?
(a) Every rational number is a real number
(b) Every integer is a rational number
(c) Every real number is an irrational number
(d) Every natural number is a whole number
Solution:
(c) Every real number is an irrational number

Question 3.
Which one of the following, regarding sum of two irrational numbers, is true?
(a) always an irrational number
(b) may be a rational or irrational number
(c) always a rational number
(d) always an integer
Solution:
(b) may be a rational or irrational number

Question 4.
Which one of the following has a terminating decimal expansion?
(a) \(\frac{5}{64}\)
(b) \(\frac{8}{9}\)
(c) \(\frac{14}{15}\)
(d) \(\frac{1}{12}\)
Solution:
(a) \(\frac{5}{64}\)
Hint:
\(\frac{5}{64}\) = \(\frac{5}{2^{6}}\)

Question 5.
Which one of the following is an irrational number?
(a) \(\sqrt{25}\)
(b) \(\sqrt{\frac{9}{4}}\)
(c) \(\frac{7}{11}\)
(d) π
Solution:
(d) π
Hint:
We take frequently π as \(\frac{22}{7}\) (which gives the value of 3.1428571428571…….) to be its correct value, but in reality these are only approximations

Question 6.
An irrational number between 2 and 2.5 is………
(a) \(\sqrt{11}\)
(b) √5
(c) \(\sqrt{2.5}\)
(d) √8
Solution:
(b) √5
Hint:
√5 = 2.236, it lies between 2 and 2.5

Question 7.
The smallest rational number by which \(\frac{1}{3}\) should be multiplied so that its decimal expansion terminates with one place of decimal is ………
(a) \(\frac{1}{10}\)
(b) \(\frac{3}{10}\)
(c) 3
(d) 30
Solution:
(b) \(\frac{3}{10}\)
Hint:
\(\frac{1}{3}\) × \(\frac{3}{10}\) = \(\frac{1}{10}\) = \(\frac{1}{2×5}\) it has terminating decimal expansion.

Question 8.
If \(\frac{1}{7}\) = 0.\(\overline { 142857 }\) then the value of \(\frac{5}{7}\) is ………..
(a) 0.\(\overline { 142857 }\)
(b) 0.\(\overline { 714285 }\)
(c) 0.\(\overline { 571428 }\)
(d) 0.714285
Solution:
(b) 0.\(\overline { 714285 }\)

Question 9.
Find the odd one out of the following.
(a) \(\sqrt{32}×\sqrt{2}\)
(b) \(\frac{\sqrt{27}}{\sqrt{3}}\)
(c) \(\sqrt{72}×\sqrt{8}\)
(d) \(\frac{\sqrt{54}}{\sqrt{18}}\)
Solution:
(b) \(\frac{\sqrt{27}}{\sqrt{3}}\)
Hint:
\(\frac{\sqrt{27}}{\sqrt{3}}\) = \(\frac{\sqrt{27}}{\sqrt{3}}\) = √9 = 3. It is an odd number

Question 10.
0.\(\overline { 34 }\) + 0.3\(\overline { 4 }\) = ……….
(a) 0.6\(\overline { 87 }\)
(b) 0.\(\overline { 68 }\)
(c) 0.6\(\overline { 8 }\)
(d) 0.68\(\overline { 7 }\)
Solution:
(a) 0.6\(\overline { 87 }\)
Hint:
0.34343434
0.34444444
0.68787878

Question 11.
Which of the following statement is false?
(a) The square root of 25 is 5 or -5
(b) \(-\sqrt{25}\) = -5
(c) \(\sqrt{25}\) = 5
(d) \(\sqrt{25}\) = ±5
Solution:
(d) \(\sqrt{25}\) = ±5

Question 12.
Which one of the following is not a rational number?
(a) \(\sqrt{\frac{8}{18}}\)
(b) \(\frac{7}{3}\)
(c) \(\sqrt{0.01}\)
(d) \(\sqrt{13}\)
Solution:
(d) \(\sqrt{13}\)

Question 13.
\(\sqrt{27}\) + \(\sqrt{12}\) = ……….
(a) \(\sqrt{39}\)
(b) 5√6
(c) 5√3
(d) 3√5
Solution:
(d) 3√5
Hint:
\(\sqrt{27}\) + \(\sqrt{12}\) = \(\sqrt{9×3}\) + \(\sqrt{3×4}\) = 3√3 + 2√3 = 5√3

Question 14.
If \(\sqrt{80}\) = k√5, then k = ………
(a) 2
(b) 4
(c) 8
(d) 16
Solution:
(b) 4
Hint:
\(\sqrt{80}\) = k√5
\(\sqrt{16×5}\) = k√5
4√5 = k√5
∴ k = 4

Question 15.
4√7 × 2√3 = ……….
(a) 6\(\sqrt{10}\)
(b) 8\(\sqrt{21}\)
(c) 8\(\sqrt{10}\)
(d) 6\(\sqrt{21}\)
Solution:
(b) 8\(\sqrt{21}\)
Hint:
4√7 × 2√3 = 4 × 2\(\sqrt{7×3}\) = 8\(\sqrt{21}\)

Question 16.
When written with a rational denominator, the expression \(\frac {2\sqrt{3}}{3\sqrt{2}}\) can be simplified as……..
(a) \(\frac {\sqrt{2}}{3}\)
(b) \(\frac {\sqrt{3}}{2}\)
(c) \(\frac {\sqrt{6}}{3}\)
(d) \(\frac {2}{3}\)
Solution:
(c) \(\frac {\sqrt{6}}{3}\)

Question 17.
When (2√5 – √2)² is simplified, we get………
(a) 4√5 + 2√2
(b) 22 – 4\(\sqrt{10}\)
(c) 8 – 4\(\sqrt{10}\)
(d) 2\(\sqrt{10}\) – 2
Solution:
(b) 22 – 4\(\sqrt{10}\)
Hint:
(2√5 – √2)² = (2√5)² + (√2)² – 2 × 2√5 × √2
= 20 – 4\(\sqrt{10}\) + 2
= 22 – 4\(\sqrt{10}\)

Question 18.
\((0.000729)^\frac{-3}{4}\) × \((0.09)^\frac{-3}{4}\) = ……..
(a) \(\frac {10^3}{3^3}\)
(b) \(\frac {10^5}{3^5}\)
(c) \(\frac {10^2}{3^2}\)
(d) \(\frac {10^6}{3^6}\)
Solution:
(d) \(\frac {10^6}{3^6}\)
Hint:

Question 19.
If √9^x = \(\sqrt[3]{9^2}\), then x = …….
(a) \(\frac {2}{3}\)
(b) \(\frac {4}{3}\)
(c) \(\frac {1}{3}\)
(d) \(\frac {5}{3}\)
Solution:
(b) \(\frac {4}{3}\)
Hint:

Question 20.
The length and breadth of a rectangular plot are 5 × 10⁵ and 4 × 10⁴ metres respectively. Its area is ……….
(a) 9 × 10¹ m²
(b) 9 × 10⁹ m²
(c) 2 × 10¹⁰ m²
(d) 20 × 10²⁰ m²
Solution:
(c) 2 × 10¹⁰ m²
Hint:
Area of a rectangle = l × b = 5 × 10⁵ × 4 × 10⁴
= 5 × 4 × 10⁵⁺⁴
= 20 × 10⁹
= 2.0 × 10 × 10⁹
= 2 × 10¹⁰ m²

Students can download Maths Chapter 4 Geometry Additional Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 4 Geometry Additional Questions

I. Multiple Choice Questions.

Question 1.
The angle sum of a convex polygon with number of sides 7 is ………
(a) 900°
(b) 1080°
(c) 1444°
(d) 720°
Solution:
(a) 900°

Question 2.
What is the name of a regular polygon of six sides?
(a) Square
(b) Equilateral triangle
(c) Regular hexagon
(d) Regular octagon
Solution:
(c) Regular hexagon

Question 3.
One angle of a parallelogram is a right angle. The name of the quadrilateral is ………
(a) square
(b) rectangle
(c) rhombus
(d) kite
Solution:
(b) rectangle

Question 4.
If all the four sides of a parallelogram are equal and the adjacent angles are of 120° and 60°, then the name of the quadrilateral is ………
(a) rectangle
(b) square
(c) rhombus
(d) kite
Solution:
(c) rhombus

Question 5.
In a parallelogram ∠A : ∠B = 1 : 2. Then ∠A ………
(a) 30°
(b) 60°
(c) 45°
(d) 90°
Solution:
(b) 60°

Question 6.
Which of the following is a formula to find the sum of interior angles of a quadrilateral of w-sides?
(a) \(\frac{n}{2}\) × 180°
(b) (\(\frac{n+1}{2}\)) 180°
(c) (\(\frac{n-1}{2}\)) 180°
(d) (n – 2) 180°
Solution:
(d) (n – 2) 180°

Question 7.
Diagonal of which of the following quadrilaterals do not bisect it into two congruent triangles?
(a) rhombus
(b) trapezium
(c) square
(d) rectangle
Solution:
(b) trapezium

Question 8.
The point of concurrency of the medians of a triangle is known as ………
(a) circumcentre
(b) incentre
(c) orthocentre
(d) centroid
Solution:
(d) centroid

Question 9.
Orthocentre of a triangle is the point of concurrency of ………
(a) medians
(b) altitudes
(c) angle bisectors
(d) perpendicular bisectors of side
Solution:
(b) altitudes

Question 10.
ABCD is a parallelogram as shown. Find x and y.

(a) 1, 7
(b) 2, 6
(c) 3, 5
(d) 4, 4
Solution:
(c) 3, 5

Question 11.
A circle divides the plane into part ………
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(c) 3

Question 12.
The longest chord of a circle is a of the circle……….
(a) radius
(b) diameter
(c) chord
(d) secant
Solution:
(b) diameter

Question 13.
Opposite angles of a cyclic quadrilateral are ………
(a) supplementary
(b) complementary
(c) equal
(d) none of these
Solution:
(a) supplementary

Question 14.
The value of x from figure is if ‘O’ is the centre of the circle ………

(a) 20 cm
(b) 15 cm
(c) 12 cm
(d) 5 cm
Solution:
(d) 5 cm

Question 15.
If PQ = x and ‘O’ is the centre of the circle, then x= ………

(a) 7 cm
(b) 14 cm
(c) 8 cm
(d) 13 cm
Solution:
(b) 14 cm

Question 16.
In figure OM = ON = 8cm and AB = 30 cm, then CD = ………

(a) 15 cm
(b) 30 cm
(c) 40 cm
(d) 10 cm
Solution:
(b) 30 cm

Question 17.
O is the centre of a circle, ∠AOB = 100°. Then angle ∠ ACB = ………

(a) 80°
(b) 40°
(c) 50°
(d) 60°
Solution:
(c) 50°

Question 18.
In,a circle, with center O, ∠AOB = 20°, ∠BOC = 40°, arc BC = 4 Then length of arc AB will be ………

(a) 8 cm
(b) 6 cm
(c) 2 cm
(d) 1 cm
Solution:
(c) 2 cm

Question 19.
In the figure , OC = 3cm and radius of circle is 5 cm. Then AB = ………

(a) 4 cm
(b) 5 cm
(c) 6 cm
(d) 8 cm
Solution:
(d) 8 cm

Question 20.
O is the centre of the circle. The value of x in the given diagram is ………

(a) 100°
(b) 160°
(c) 200°
(d) 80°
Solution:
(d) 80°

II. Answer the following questions.

Question 1.
In the figure find x° and y°.
Solution:
∠ACD = ∠A + ∠B
(An exterior angle of a triangle is sum of its interior opposite angles)
120° = 50° + x°
x° = 120° – 50°
= 70°
In the triangle ABC

∠A + ∠B + ∠ACB = 180° (Sum of the angles of a Δ)
50° + x + ∠ACB = 180°
50° + 70° + ∠ACB = 180°
∠ACB = 180° – 120°
y = 60°
(OR)
∠ACD + ∠ACB = 180° (Angles of a linear pair)
∠ACB = 180° – 120°
= 60°
The value of x = 70° and y = 60°.

Question 2.
The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
Solution:
Let the angles of the quadrilateral be 3x, 5x, 9x and 13x.
Sum of all the angles of quadrilateral = 360°.
3x + 5x + 9x + 13x = 360°
30x = 360°
x = \(\frac{360°}{30}\)
= 12°
3x = 3 × 12 = 36°
5x = 5 × 12 = 60°
9x = 9 × 12 = 108°
13x = 13 × 12 = 156°
The required angles of quadrilateral are 36°, 60°, 108° and 156°.

Question 3.
Diagonal AC of a parallelogram ABCD bisects ∠A. Show that

(i) it bisects ∠C also
(ii) ABCD is a rhombus.
Solution:
We have a parallelogram ABCD in which diagonals AC bisect ∠A.
∠DAC = ∠BAC
(i) To prove that AC bisects ∠C
∴ ABCD is a parallelogram

∴ AB || DC and AC is a transversal
∴ ∠1 = ∠3 (Alternate interior angle) …….(1)
Also BC || AD and AC is a transversal
∴ ∠2 = ∠4 (Alternate interior angle) …….(2)
But AC bisects ∠A
∴ ∠1 = ∠2 ……… (3)
From (1), (2) and (3) we get
∠3 = ∠4
∴ AC bisects ∠C.

(ii) To prove that ABCD is a rhombus.
In ΔABC, we have ∠1 = ∠4 [∴ ∠ 1 = ∠2 = ∠4]
∴ BC = AB (side opposite to equal angles are equal) ……..(4)
Similarly AD = DC …….. (5)
But ABCD is a parallelogram AB = DC (Opposite sides of a parallelogram) ………(6)
From (4), (5) and (6) we have AB = BC = CD = DA.
Thus ABCD is a rhombus.

Question 4.
ABCD is a parallelogram and AP and CQ are perpendiculars from vertex A and C on diagonal BD.
Show that

(i) ΔAPB ≅ ΔCQD
(ii) AP = CQ.
Solution:
(i) In ΔAPB and ΔCQD we have
∠APB = ∠CQD (90° each)
AB = CD (opposite sides of parallelogram ABCD)
∠ABP = ∠CDQ (AB || CD and AD is a transversal)
Using ASA congruency we have,
ΔAPB ≅ ΔCQD

(ii) Since ΔAPB ≅ ΔCQD
∴ Their corresponding parts are equal.
∴ AP = CQ.

Question 5.
ABCD is a rectangle and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Solution:
In rectangle ABCD, P is the mid-point of AB.
Q is the mid-point of BC. R is the mid-point of CD.
S is the mid-point of DA. AC is the diagonal.
Now in ΔABC,
PQ = \(\frac{1}{2}\) AC and PQ || AC ……. (1)
Similarly in ΔACD,
SR = \(\frac{1}{2}\) AC and SR || AC ……. (2)
From (1) and (2) we get,
PQ = SR and PQ || SR
Similarly by joining BD, we have
PS = QR and PS || QR
i.e. Both pairs of opposite sides of quadrilateral PQRS are equal and parallel.
∴ PQRS is a parallelogram.

Question 6.
In the figure A, B and C are three points on a circle with centre O such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.
Solution:

O is the centre of the circle. ∠AOB = 60° and ∠BOC = 30°
Sum of all the angles of quadrilateral = 360°.
∠AOB + ∠BOC = ∠AOC
∴ ∠AOC = 30° + 60°
= 90°
∠ADC = \(\frac{1}{2}\) × 90°
= 45°

Question 7.
In the given figure A, B, C and D are four points on a circle, AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.
Solution:
In ΔCDE,

Exterior ∠BEC = Sum of interior opposite angle
130° = ∠EDC + ∠ECD
130° = ∠EDC + 20°
130° – 20° = ∠EDC
110° = ∠EDC
∴ ∠BAC = ∠BDC = 110° (Both the triangles are standing on the same base)
∠BAC = 110°

Question 8.
In the given figure KLMN is a cyclic quadrilateral. KD is the tangent at K. If ∠N is a diameter ∠NLK = 40° and ∠LNM = 50°. Find ∠MLN and ∠DKL.

Solution:
LN is a diameter
∠LMN = ∠LKN = 90° (Angle in a semi-circle)
∴ ∠MLN = 90° – 50°
= 40°
∠LNK = 90° – 40°
= 50°
∠DKL = ∠LKN = 50° (angles in the alternate segment)
∴ ∠DKL = 50°
∠MLN = 40° and ∠DKL = 50°

Question 9.
In the given figure ∠PQR = 100°, where P, Q and R are points on a circle with centre ‘O’. Find ∠OPR.

Solution:
∠PQR is the angle subtended by the chord PR in the minor segment.
reflex ∠POR = 2∠PQR
= 2 × 100°
= 200°
Now ∠POR + reflex ∠POR = 360°
∠POR + 200° = 360°
∠POR = 360° – 200°
= 160°
From the given diagram, POR is an isosceles triangle (∴ PO = OR = radii)
∴ ∠OPR = ∠ORP (angles opposite to equal sides)
In ΔOPR,
∠OPR + ∠ORP + ∠POR = 180°
∠OPR + ∠OPR + 160° = 180°
2∠OPR = 180° – 160°
2∠OPR = 20°
∠OPR = \(\frac{20°}{2}\)
= 10°
∴ ∠OPR = 10°

Question 10.
AB and CD are two parallel chords of a circle which are on either sides of the centre such that AB = 10 cm and CD = 24 cm. Find the radius if the distance between AB and CD is 17 cm.
Solution:

Given AB = 10 cm, CD = 24 cm and PQ = 17 cm.
PC = PD = \(\frac{24}{2}\)
= 12 cm
AQ = QB = \(\frac{10}{2}\)
= 5 cm
Let OP be x; OQ = (17 – x)
In the ΔOPC,
OC² = OP² + PC²
= x² + 12²
In the ΔOAQ,
OA² = AQ² + QO²
= 5² + (17 – x)²
= 25 + 289 + x² – 34x
= 314 + x² – 34x
But OA² = OC²
314 + x² – 34x = x² + 144
-34x = 144 – 314
-34x = -170
34x = 170
x = \(\frac{170}{34}\)
= \(\frac{10}{2}\)
= 5
We know,
OC² = x²+ 144
= 5² + 144
= 25 + 144
OC² = 169
But OC = \(\sqrt{169}\)
= 13
Radius of the circle = 13 cm
= x² + 144

Students can download Maths Chapter 3 Algebra Ex 3.9 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.9

Question 1.
Determine the quadratic equations, whose sum and product of roots are
(i) -9, 20
Answer:
Sum of the roots = -9 and Product of the roots = 20
The Quadratic equation is
x² – (sum of the roots) x + product of the roots = 0
x² – (-9) x + 20 = 0 ⇒ x² + 9x + 20 = 0

(ii) \(\frac { 5 }{ 3 } \), 4
Answer:
Sum of the roots = \(\frac { 5 }{ 3 } \); Product of the roots = 4
The Quadratic equation is
x² – (sum of the roots) x + product of the roots = 0
x² – (\(\frac { 5 }{ 3 } \)) x + 4 = 0 ⇒ x² – \(\frac { 5 }{ 3 } \) x + 4 = 0
3x² – 5x + 12 = 0

(iii) \(\frac { -3 }{ 2 } \), -1
Answer:
Sum of the roots = \(\frac { -3 }{ 2 } \); Product of the roots = -1
The Quadratic equation is
x² – (sum of the roots) x + product of the roots = 0
x² – (-\(\frac { 3 }{ 2 } \)) x + (-1) = 0 ⇒ x² + \(\frac { 3 }{ 2 } \) x – 1 = 0
2x² + 3x – 2 = 0

(iv) – (2 – a)², (a + 5)²
Answer:
Sum of the roots = – (2 – a)²; Product of the roots = (a + 5)²
x² – (sum of the roots) x + product of the roots = 0
x² – [-(2 – a)²] x + (a + 5)² = 0
x² + (2 – a)² x + (a + 5)² = 0

Question 2.
Find the sum and product of the roots for each of the following quadratic equations
(i) x² + 3x – 28 = 0
(ii) x² + 3x = 0
(iii) 3 + \(\frac{1}{a}=\frac{10}{a^{2}}\)
(iv) 3y² – y – 4 = 0
Solution:
(i) x² – (-3)x + (-28) = 0.
Comparing this with x² – (α + β)x + αβ = 0.
(α + β) = Sum of the roots = -3
αβ = product of the roots = -28

(ii) x² + 3x = 0 = x² – (-3)x + 0 = 0
x² – (α + β)x + αβ = 0
Sum of the roots α + β = -3
Products of the roots αβ =0

(iii) 3 + \(\frac { 1 }{ a } \) = \(\frac{10}{a^{2}}\)
Answer:
Multiply by a²
3a² + a = 10
3a² + a – 10 = 0
Sum of the roots (α + β) = \(\frac { -1 }{ 3 } \)
Product of the roots (α β) = \(\frac { -10 }{ 3 } \)

(iv) 3y² – y – 4 = 0
Answer:
Sum of the roots (α + β) = \(\frac { -(-1) }{ 3 } \) = \(\frac { 1 }{ 3 } \)
Product of the roots (α β) = \(\frac { -4 }{ 3 } \)

Students can download Maths Chapter 3 Algebra Ex 3.8 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.8

Question 1.
Find the square root of the following polynomials by division method
(i) x⁴ – 12x³ + 42x² – 36x + 9
Answer:

(ii) 31 x² – 28x³ + 4x⁴ + 42x + 9
Answer:
Rearrange the order we get
4x⁴ – 28x³ + 37x² + 42x + 9

(iii) 16x⁴ + 8x² + 1
Answer:

(iv) 121 x⁴ – 198x³ – 183x² + 216x + 144
Answer:

Question 2.
Find the square root of the expression
\(\frac{x^{2}}{y^{2}}\) – \(\frac { 10x }{ y } \) + 27 – \(\frac { 10y }{ x } \) + \(\frac{y^{2}}{x^{2}}\)
Answer:

Question 3.
Find the values of a and b if the following polynomials are perfect squares.
(i) 4x⁴ – 12x³ + 37x² + bx + a
Answer:

Since it is a perfect square
b + 42 = 0
b = – 42
a – 49 = 0
a = 49
∴ The value of a = 49 and b = – 42

(ii) ax⁴ + bx³ + 361x² + 220x + 100
Answer:
Re-arrange the order we get
100 + 220x + 361x² + bx³ + ax⁴

Since it is a perfect square
b – 264 = 0
b = 264
a – 144 = 0
a = 144
∴ The value of a = 144 and b = 264

Question 4.
Find the values of m and n if the following expressions are perfect sqaures.
(i) \(\frac{1}{x^{4}}\) – \(\frac{6}{x^{3}}\) + \(\frac{13}{x^{2}}\) + \(\frac { m }{ x } \) + n
Answer:

Since it is a perfect square
\(\frac { 1 }{ x } \) (m + 12) = 0
m + 12 = 0
m = -12
n – 4 = 0
n = 4
∴ The value of m = -12 and n = 4

(ii) x⁴ – 8x³ + mx² + nx + 16
Answer:

Since it is a perfect square
m – 16 – 8 = 0
m – 24 = 0
m = 24
n + 32 = 0
n = -32
∴ The value of m = 24 and n = -32

Students can download Maths Chapter 3 Algebra Ex 3.16 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.16

1. In the

write (i) The number of elements
(ii) The order of the matrix
(iii) Write the elements a₂₂, a₂₃, a₂₄, a₃₄, a₄₃, a₄₄.
Answer:
(i) The number of elements is 16
(ii) The order of the matrix is 4 × 4
(iii) Elements corresponds to

Question 2.
If a matrix has 18 elements, what are the possible orders it can have? What if it has 6 elements?
Answer:
The possible orders of the matrix having 18 elements are

The possible orders of the matrix having 6 elements are

Question 3.
Construct a 3 × 3 matrix whose elements are given by
(i) a_ij = |i – 2j|
Answer:
a_ij = |i – 2j|
The general 3 × 3 matrices is

a₁₁ = |1 – 2(1)| = |1 – 2| = | – 1| = 1
a₁₂ = |1 – 2(2)| = |1 – 4| = | – 3| = 3
a₁₃ = |1 – 2(3)| = |1 – 6| = | – 5| = 5
a₂₁ = |2 – 2(1)| = |2 – 2| = 0 = 0
a₂₂ = |2 – 2(2)| = |2 – 4| = | – 2| = 2
a₂₃ = |2 – 2(3)| = |2 – 6| = | – 4| = 4
a₃₁ = |3 – 2(1)| = |3 – 2| = | 1 | = 1
a₃₂ = |3 – 2(2)| = |3 – 4| = | – 1 | = 1
a₃₃ = |3 – 2(3)| = |3 – 6| = | – 3 | = 3
The required matrix

(ii) a_ij = \(\frac{(i+j)^{3}}{3}\)
Answer:
a₁₁ = \(\frac{(1+1)^{3}}{3}\) = \(\frac{2^{3}}{3}\) = \(\frac { 8 }{ 3 } \)
a₁₂ = \(\frac{(1+2)^{3}}{3}\) = \(\frac { 27 }{ 3 } \) = 9
a₁₃ = \(\frac{(1+3)^{3}}{3}\) = \(\frac { 64 }{ 3 } \) = \(\frac { 64 }{ 3 } \)
a₂₁ = \(\frac{(2+1)^{3}}{3}\) = \(\frac { 27 }{ 3 } \) = 9
a₂₂ = \(\frac{(2+2)^{3}}{3}\) = \(\frac { 64 }{ 3 } \) = \(\frac { 64 }{ 3 } \)
a₂₃ = \(\frac{(2+3)^{3}}{3}\) = \(\frac { 125 }{ 3 } \) = \(\frac { 125 }{ 3 } \)
a₃₁ = \(\frac{(3+1)^{3}}{3}\) = \(\frac { 64 }{ 3 } \) = \(\frac { 64 }{ 3 } \)
a₃₂ = \(\frac{(3+2)^{3}}{3}\) = \(\frac { 125 }{ 3 } \) = \(\frac { 125 }{ 3 } \)
a₃₃ = \(\frac{(3+3)^{3}}{3}\) = \(\frac { 216 }{ 3 } \) = 72
The required matrix

Question 4.
If  then find the tranpose of A.
Answer:

transpose of A = (A^T)

Question 5.
If then find the tranpose of – A
Answer:

Transpose of – A = (-A^T) =

Question 6.
If A = then verify (AT)T = A
Answer:

Hence it is verified

Question 7.
Find the values of x, y and z from the following equations
(i)

Answer:
Since the given matrices are equal then all the corresponding elements are equal.
y = 12, z = 3, x = 3
The value of x = 3, y = 12 and z = 3

(ii)

Answer:
x + y = 6 ……(1)
5 + z = 5
z = 5 – 5 = 0
xy = 8
y = \(\frac { 8 }{ x } \)
Substitute the value of y = \(\frac { 8 }{ x } \) in (1)
x + \(\frac { 8 }{ x } \) = 6
x² + 8 = 6x
x² – 6x + 8 = 0
(x – 4) (x – 2) = 0
∴ x – 4 = 0 or x – 2 = 0
x = 4 or x = 2
y = \(\frac { 8 }{ 4 } \) = 2 or y = \(\frac { 8 }{ 2 } \) = 4

∴ The value of x, y and z are 4, 2, 0 (or) 2, 4, 0

(iii)

Solution:
x + y + z = 9 ……….(1)
x + z = 5 ……….(2)
y + z = 7 ……….(3)

Substitute the value of y = 4 in (3)
y + z = 7
4 + z = 7
z = 7 – 4
= 3
Substitute the value of z = 3 in (2)
x + 3 = 5
x = 5 – 3
= 2
∴ The value of x = 2 , y = 4 and z = 3

Students can download Maths Chapter 4 Geometry Ex 4.1 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 4 Geometry Ex 4.1

Question 1.
In the figure, AB is parallel to CD, find x.

Solution:
(i) Through T draw TE || AB.

∴ ∠BAT + ∠ATE = 180° (AB || TE)
140° + ∠ATE = 180°
∠ATE = 180°- 140° = 40°
Similarly ∠ETC + ∠TCD = 180° (TE || CD)
∠ETC+150° = 180°
∠ETC = 180°- 150° = 30°
x = ∠ATE + ∠ETC
= 40°+ 30° = 70°
x = 70°

(ii) Draw TE || AB.

∠ABT + ∠ETB = 180° (AB || TE)
48° + ∠ETB = 180°
∠ETB = 180° – 48° = 132°
Similarly ∠CDT + ∠DTE = 180°
24° + ∠DTE = 180°
∴ ∠DTE = 180° – 24°
= 156°
∴ ∠BTE + ∠ETD = 132° + 156°
= 288°
x = 288°

(iii) In the given figure AB || CD, AD is the transversal.
∠CDA = ∠BAD
= 53° (alternate angles are equal)
In ΔECD, ∠D = ∠A = 53° (Alternate angles are equal)
∠E + ∠C + ∠D = 180° (sum of the angles of a triangle)
x° + 38° + 53° = 180°
x° = 180°- 91°
= 89°
x = 89°

Question 2.
The angles of a triangle are in the ratio 1 : 2 : 3, find the measure of each angle of the triangle.
Solution:
The ratio of the angles of a triangle = 1 : 2 : 3.
Let the angles of a triangle be x, 2x and 3x.
x + 2x + 3x = 180° (Total angle of a triangle is 180°)
6x = 180°
x = \(\frac{180°}{6}\)
= 30°
x = 30°; 2x = 2 × 30° = 60°; 3x = 3 × 30° = 90°
Measures of the angles of a triangle = 30°, 60° and 90°.

Question 3.
Consider the given pairs of triangles and say whether each pair is that of congruent triangles. If the triangles are congruent, say ‘how’; if they are not congruent say ‘why’ and also say if a small modification would make them congruent:

(i) In ΔPQR and ΔABC
PQ = AB (Given)
RQ = BC (Given)
ΔABC is not congruent to ΔPQR.
If PR = AC then ΔABC ≅ ΔPQR

(ii) In ΔABD and ΔCDB
AB = CD (Given)
AD = BC (Given)
BD is common
By SSS congruency
ΔABD ≅ ΔCDB

(iii) In ΔPXY and ΔPXZ
PX is common.
XY = XZ (Given)
PY = PZ (Given)
By SSS congruency
ΔPXY ≅ ΔPXZ

(iv) In the given figure BD bisect AC
In ΔAOB and ΔOCD
OA = OC (Given)
∠AOB = ∠DOC (vertically opposite angles)
∠B = ∠D (Given)
By ASA congruency ΔAOB ≅ ΔOCD

(v) In the given figure AC and BD bisect each other at O.
∴ OA = OC (Given); OB = OD (Given)
∠AOB = ∠COD (vertically opposite angles)
By SAS congruency
ΔAOB ≅ ΔOCD

(vi) In the given figure
AB = AC (Given)
BM = MC (AM is the median of the ΔABC)
AM is common (By SSS congruency)
∴ ΔABM ≅ ΔACM

Question 4.
ΔABC and ΔDEF are two triangles in which AB = DF, ∠ACB = 70°, ∠ABC = 60°; ∠DEF = 70° and ∠EDF = 60°. Prove that the triangles are congruent.
Solution:

In ΔABC ∠B = 60° and ∠C = 70°
∴ ∠A = 180° – (60° + 70°)
= 180° – 130°
= 50°
In ΔDEF ∠E = 70° and ∠D = 60°
∠F = 180° – (70° + 60°)
= 180° – 130°
= 50°
∠A = ∠F = 50°
∠B = ∠D = 60°
∠C = ∠E = 70°
By AAA congruency
ΔABC ≅ ΔFDE
(or)
∠B = ∠D = 60°
∠C = ∠E = 70°
AB = FE
By ASA congruency
ΔABC ≅ ΔFDE

Question 5.
Find all the three angles of the ΔABC.

Solution:
∠A + ∠B = ∠ACD (An exterior angle of a triangle is sum of its interior opposite angles)
x + 35 + 2x – 5 = 4x – 15
3x + 30 = 4x – 15
30 + 15 = 4x – 3x
45° = x
∠A = x + 35°
= 45° + 35°
= 80°
∠B = 2x – 5
= 2(45°) – 5°
= 90° – 5°
= 85°
∠ACD = 4x – 15
= 4 (45°) – 15°
= 180° – 15°
= 165°
∠ACB = 180° – ∠ACD
= 180° – 165°
= 15°
∠A = 80°, ∠B = 85° and ∠C = 15°.

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.3 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.3

Question 1.
Find the mid-points of the line segment joining the points.
(i) (-2, 3) and (-6, -5)
(ii) (8, -2) and (-8, 0)
(iii) (a, b) and (a + 2b, 2a – b)
(iv) (\(\frac{1}{2},\frac{-3}{7}\)) and (\(\frac{3}{2},\frac{-11}{7}\))
Solution:

Question 2.
The centre of a circle is (-4, 2). If one end of the diameter of the circle is (-3, 7) then find the other end.
Solution:
Let the other end of the diameter B be (a, b)

∴ \(\frac{-3+a}{2}\) = -4
-3 + a = -8
a = -8 + 3
a = -5
\(\frac{7+b}{2}\)
7 + b = 4
b = 4 – 7 ⇒ b = -3
The other end of the diameter is (-5, -3).

Question 3.
If the mid-point (x, y) of the line joining (3, 4) and (p, 7) lies on 2x + 2y + 1 = 0, then what will be the value of p?
Solution:

3 + p + 11 + 1 = 0 ⇒ p + 15 = 0 ⇒ p = -15
The value of p is -15.

Question 4.
The mid-point of the sides of a triangle are (2, 4), (-2, 3) and (5, 2). Find the coordinates of the vertices of the triangle.
Solution:
Let the vertices of the ΔABC be A (x₁ y₁), B (x₂, y₂) and C (x₃, y₃)

\(\frac{x_{2}+x_{3}}{2}\) = -2 ⇒ x₂ + x₃ = -4 → (3)
\(\frac{y_{2}+y_{3}}{2}\) = 3 ⇒ y₂ + y₃ = 6 → (4)

\(\frac{x_{1}+x_{3}}{2}\) = 5 ⇒ x₁ + x₃ = 10 → (5)
\(\frac{y_{1}+y_{3}}{2}\) = 2 ⇒ y₁ + y₃ = 4 → (6)
By adding (1) + (3) + (5) we get
2x₁ + 2x₂ + 2x₃ = 4 – 4 + 10
2(x₁ + x₂ + x₃) = 10 ⇒ x₁ + x₂ + x₃ = 5
From (1) x₁ + x₂ = 4 ⇒ 4 + x₃ = 5
x₃ = 5 – 4 = 1
∴ The vertices of the ΔABC are
A (9, 3)
B (-5, 5), C (1, 1)
From (3) x₂ + x₃ = -4 ⇒ x₁ + (-4) = 5
x₁ = 5 + 4 = 9
From (5) ⇒ x₁ + x₃ = 8
x₂ + 10 = 5
x₂ = 5 – 10 = -5
∴ x₁ = 9, x₂ = -5, x₃ = 1
By adding (2) + (4) + (6) we get
2y₃ + 2y₂ + 2y₃ = 8 + 6 + 4
2(y₁ +y₂ + y₃) = 18 ⇒ y₁ + y₂ + y₃ = 9
From (2) ⇒ y₁ + y₂ = 8
8 + y₃ = 9 ⇒ y₃ = 9 – 8 = 1
From (4)
y₂ + y₃ = 6 ⇒ y₁ + 6 = 9
y₁ = 9 – 6 = 3
From (6)
y₁ + y₃ = 4 ⇒ y₂ + 4 = 9
y₂ = 9 – 4 = 5
∴ y₁ = 3, y₂ = 5, y₃ = 1

Question 5.
O (0, 0) is the centre of a circle whose one chord is AB, where the points A and B (8, 6) and (10, 0) respectively. OD is the perpendicular from the centre to the chord AB. Find the coordinates of the mid-point of OD.
Solution:
Note: Since OD is perpendicular to AB, OD bisect the chord
D is the mid-point of AB

Question 6.
The points A(-5, 4) , B(-1, -2) and C(5, 2) are the vertices of an isosceles right-angled triangle where the right angle is at B. Find the coordinates of D so that ABCD is a square.
Solution:
Since ABCD is a square
Mid-point of AC = mid-point of BD
Let the point D be (a, b)

But mid-point of BD = Mid-point of AC
(\(\frac{-1+a}{2}, \frac{-2+b}{2}\)) = (0, 3)
\(\frac{-1+a}{2}\) = 0
-1 + a = 0
a = 1
(\(\frac{-2+b}{2}\))
-2 + b = 6
b = 6 + 2 = 8
∴ The vertices D is (1, 8).

Question 7.
The points A (-3, 6), B (0, 7) and C (1, 9) are the mid points of the sides DE, EF and FD of a triangle DEF. Show that the quadrilateral ABCD is a parallellogram.
Solution:
Let D be (x₁ y₁), E (x₂, y₂) and F (x₃, y₃)


x₁ + x₂ = -6 → (1)
y₁ + y₂ = 12 → (2)

x₁ + x₃ = 2 → (5)
y₁ + y₃ = 18 → (6)
Add (1) + (3) + (5)
2x₁ + 2x₂ + 2x₃ = -6 + 0 + 2
2 (x₁ + x₂ + x₃) = -4
x₁ + x₂ + x₃ = -2
From (1) x₁ + x₂ = -6
x3 = -2 + 6 = 4
From (3) x₂ + x₃ = 0
x₁ = -2 + 0
x₁ = -2
From (5) x₁ + x₃ = 2
x₂ + 2 = -2
x₂ = -2 -2 = -4
∴ x₁ = -2, x₂ = -4, x₃= 4
Add (2) + (4) + 6
2y₁ + 2y₂ + 2y₃ = 12 + 14 + 18
2(y₁ + y₂ + y₃) = 44
y₁ + y₂+ y₃ = 44/2 = 22
From (2) y₁ + y₂ = 12
12 + y₃ = 22 ⇒ y₃ = 22 – 12 = 11
From (4) y₂ + y₃ = 14
y₁ + 14 = 22 ⇒ y₁ = 22 – 14
y₁ = 8
From (6)
y₁ +y₃ = 18
y2 + 18 = 22 ⇒ = 22 – 18
= 4
∴ y₁ = 8, y₂ = 4, y₃ = 11
The vertices D is (-2, 8)

Mid-point of diagonal AC = Mid-point of diagonal BD
The quadrilateral ABCD is a parallelogram

Question 8.
A(-3, 2) , B(3, 2) and C(-3, -2) are the vertices of the right triangle, right angled at A. Show that the mid point of the hypotenuse is equidistant from the vertices.
Solution:

AD = CD = BD = \(\sqrt{13}\)
Mid-point of BC is equidistance from the centre.