Bhagya

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.3

Question 1.
Find the value of the following:

(iii) tan 15° tan 30° tan 45° tan 60° tan 75°

Solution:
(i) cos 45° = \(\frac{1}{√2}\)

= 1² + 1² – 2(\(\frac{1}{√2}\))²
= 1 + 1 – 2(\(\frac{1}{2}\))
= 2 – 1
= 1

(ii) cos 60° = \(\frac{1}{2}\)

= 1 + 1 + 1 – 8(\(\frac{1}{2}\))²
= 3 – 8 × \(\frac{1}{4}\)
= 3 – 2
= 1

(iii) tan 30° = \(\frac{1}{√3}\), tan 45° = 1, tan 60° = √3
tan 15° . tan 30°. tan 45° . tan 60°. tan 75° = tan 15° . \(\frac{1}{√3}\) . 1 . √3 tan 75°
= tan 15° × tan 75° × \(\frac{1}{√3}\) × 1 × √3
= tan(90° – 75°) × \(\frac{1}{cot 75°}\) × 1 [tan 90° – θ = cot θ]
= cot 75° × \(\frac{1}{cot 75°}\) × 1
= 1

= 1 + 1
= 2

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.2

Question 1.
Verify the following equalities:
(i) sin² 60° + cos² 60° = 1
Solution:
sin 60° = \(\frac{√3}{2}\); cos 60° = \(\frac{1}{2}\)
L.H.S = sin² 60° + cos² 60°

L.H.S = R.H.S
Hence it is proved.

(ii) 1 + tan² 30° = sec² 30°
Solution:
tan 30° = \(\frac{1}{√3}\); sec 30° = \(\frac{2}{√3}\)
L.H.S = 1 + tan² 30°

∴ L.H.S = R.H.S
Hence it is proved.

(iii) cos 90° = 1 – 2sin² 45° = 2cos² 45° – 1
Solution:
cos 90° = 0, sin 45° = \(\frac{1}{√2}\), cos 45° = \(\frac{1}{√2}\)
cos 90° = 0 ……. (1)
1 – 2 sin² 45° = 1 – 2 (\(\frac{1}{√2}\))²
= 1 – 2 × \(\frac{1}{2}\)
= 1 – 1 = 0 → (2)
2 cos² 45° – 1 = 2(\(\frac{1}{√2}\))² – 1
= \(\frac{2}{2}\) – 1
= \(\frac{2 – 2}{2}\) = 0 → (3)
From (1), (2) and (3) we get
cos 90° = 1 – 2 sin² 45° = 2 cos² 45° – 1
Hence it is proved.

(iv) sin 30° cos 60° + cos 30° sin 60° = sin 90°
Solution:
sin 30° = \(\frac{1}{2}\); cos 60° = \(\frac{1}{2}\); cos 30° = \(\frac{√3}{2}\); sin 60° = \(\frac{√3}{2}\); sin 90° = 1
L.H.S = sin 30° cos 60° + cos 30° sin 60°

= 1
R.H.S = sin 90° = 1
L.H.S = R.H.S
Hence it is proved.

Question 2.
Find the value of the following:
(i) \(\frac{tan 45°}{cosec 30°}\) + \(\frac{sec 60°}{cot 45°}\) – \(\frac{5 sin 90°}{2 cos 0°}\)
(ii) (sin 90° + cos 60° + cos 45°) × (sin 30° + cos 0° – cos 45°)
(iii) sin²30° – 2 cos³ 60° + 3 tan⁴ 45°
Solution:
(i) tan 45° = 1, cosec 30° = 2; sec 60° = 2; cot 45° = 1; tan 45°, sin 90° = 1; cos 0° = 1

= 0

(ii) sin 90° = 1; cos 60° = \(\frac{1}{2}\); cos 45° = \(\frac{1}{√2}\); sin 30° = \(\frac{1}{2}\); cos 0° = 1
(sin 90° + cos 60° + cos 45°) × (sin 30° + cos 0° – cos 45°)

= \(\frac{7}{4}\)

(iii) sin 30° = \(\frac{1}{2}\); cos 60° = \(\frac{1}{2}\); tan 45° = 1
sin² 30° – 2 cos³ 60° + 3 tan⁴ 45°

= 3

Question 3.
Verify cos 3 A = 4 cos³ A – 3 cos A, when A = 30°
Solution:
L.H.S = cos 3 A
= cos 3 (30°)
= cos 90°
= 0
R.H.S = 4 cos³ A – 3 cos A
= 4 cos³ 30° – 3 cos 30°

= 0
∴ L.H.S = R.H.S
Hence it is proved.

Question 4.
Find the value of 8 sin2x, cos 4x, sin 6x, when x = 15°.
Solution:
8 sin 2x cos 4x sin 6x = 8 sin 2 (15°) × cos 4 (15°) × sin (6 × 15°)
= 8 sin 30° × cos 60° × sin 90°
= 8 × \(\frac{1}{2}\) × \(\frac{1}{2}\) × 1
= 2

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.1

Question 1.
From the given figure, find all the trigonometric ratios of angle B.

Solution:

Question 2.
From the given figure, find the values of

(i) sin B
(ii) sec B
(iii) cot B
(v) tan C
(vi) cosec C
Solution:
In the right ΔABD,
AD² = AB² – BD²
= 13² – 5²
= 169 – 25
= 144
AD = \(\sqrt{144}\)
= 12
In the right ΔADC,
AC² = AD² + DC²
= 12² + 16²
= 144 + 256
= 400
AC = \(\sqrt{400}\)
= 20

Question 3.
If 2 cos θ = √3, then find all the trigonometric ratios of angle θ.
Solution:

2 cos θ = √3 ⇒ cos θ = \(\frac{√3}{2}\)
AB² = AC² – BC²
= 2² – (√3)² ⇒ = 4 – 3 = 1
AB = √1 = 1

Question 4.
If cos A =\(\frac{3}{5}\), then find the value of \(\frac{sin A-cos A}{2 tan A}\)
Solution:

cos A = \(\frac{3}{5}\)
ΔABC
BC² = AC² – AB²
= 5² – 3²
= 25 – 9
= 16
BC = \(\sqrt{16}\) = 4
sin A = \(\frac{4}{5}\); tan A = \(\frac{4}{3}\)

∴ The value of \(\frac{sin A-cos A}{2 tan A}\) = \(\frac{3}{40}\)

Question 5.
If cos A = \(\frac{2x}{1+x_{2}}\) then find the values of sin A and tan A in terms of x.
Solution:
cos A = \(\frac{2x}{1+x_{2}}\)
In the triangle ABC

BC² = AC² – AB²
= (1 + x²)² – (2x)²
= 1 + x⁴ + 2x² – 4x²
= x⁴ – 2x² + 1
= (x² – 1)² (or) (1 – x²)² (using (a – b)²)
BC = \(\sqrt{(x^{2}-1)^{2}}\) (or) \(\sqrt{(1-x^{2})^{2}}\)
BC = x² – 1
The value of sin A = \(\frac{BC}{AC}\) = \(\frac{x²-1}{x²+1}\)
tan A = \(\frac{BC}{AB}\) = \(\frac{x²-1}{2x}\)
and
BC = 1 – x²
The value of sin A = \(\frac{1-x²}{1+x²}\)
tan A = \(\frac{1-x²}{2x}\)

Question 6.
If sin θ = \(\frac{a}{\sqrt{a²+b²}}\) then show that b sin θ = a cos θ.
Solution:

sin θ = \(\frac{a}{\sqrt{a²+b²}}\)
In the triangle ΔABC
BC² = AC² – AB²

L. H. S = R. H. S

Question 7.
If 3 cot A = 2, then find the value of \(\frac{4sin A-3cos A}{2 sin A+3 Cos A}\)
Solution:

3 cot A = 2 ⇒ cot A = \(\frac{2}{3}\)
AC² = AB² + BC²
= 3² + 2²
= 9 + 4
AC = \(\sqrt{13}\)
cos A = \(\frac{AB}{AC}\) = \(\frac{3}{\sqrt{13}}\)
sin A = \(\frac{BC}{AC}\) = \(\frac{2}{\sqrt{13}}\)
The value of \(\frac{4sin A-3cos A}{2 sin A+3 Cos A}\)

The value is (\(\frac{-1}{13}\))

Question 8.
If cos θ : sin θ = 1 : 2, then find the value of \(\frac{8cos θ-2cos θ}{4 cos θ+2 sin θ}\)
Solution:
cos θ : sin θ = 1 : 2

Aliter:
cos θ : sin θ = 1 : 2
2 cos θ = sin θ ⇒ 2 = \(\frac{sin θ}{cos θ}\) ⇒ 2 = tan θ

∴ The value is \(\frac{1}{2}\)

Question 9.
From the given figure, prove that θ + ∅ = 90°. Also prove that there are two other right angled triangles. Find sin α, cos β and tan ∅.

Solution:
In the ΔABC,
AB = 9 + 16 = 25
AC = 15; BC = 20
AB² = 25²
= 625 ……. (1)
AC² + BC² = 15² + 20²
= 225 + 400
= 625 …….. (2)
From (1) and (2) we get
AB² = AC² + BC²
ABC is a right angle triangle at C (Pythagoras theorem)
∴ ∠C = 90°
θ + ∅ = 90°
Also ADC is a right angle triangle ∠ADC = 90° (Given)
BDC is also a right angle triangle ∠BDC = 90° (since ADB is a straight line sum of the two angle is 180°)
From the given diagram

Question 10.
A boy standing at a point O finds his kite flying at a point P with distance OP = 25 m. It is at a height of 5 m from the ground. When the thread is extended by 10 m from P, it reaches a point Q. What will be the height QN of the kite from the ground? (use trigonometric ratios).

Solution:
Let the angle O be “θ”
In ΔONQ

In ΔOMP
sin θ = \(\frac{PM}{OP}\) ⇒ sin θ = \(\frac{5}{25}\)
sin θ = \(\frac{1}{5}\) ……… (2)
From (1) and (2) we get
\(\frac{h}{35}\) = \(\frac{1}{5}\)
5 h = 35 ⇒ h = \(\frac{35}{5}\) = 7
The height of the kite from the ground is 7m.

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Additional Questions

I. Choose the correct answer.

Question 1.
The HCF of x² – y²; x³ – y³, …………. xn – yn where n ∈ N is
(1) x – y
(2) x + y
(3) xn – yn
(4) do not intersect
Answer:
(1) x – y

Question 2.
Which of the following is correct.
(i) Every polynomial has finite number of multiples
(ii) LCM of two polynomials of degree “2” may be a constant
(iii) HCF of 2 polynomials may be a constant
(iv) Degree of HCF of two polynomials is always less than degree of L.C.M.
(1) (i) and (iii)
(2) (iii) and (iv)
(3) (iii) only
(4) (iv) only
Answer:
(3) (iii) only

Question 3.
The HCF of x² – 2xy + y² and x⁴ – y⁴ is …………….
(1) 1
(2) x + y
(3) x – y
(4) x² – y²
Answer:
(3) x – y

Question 4.
The L.C.M. of a^k a^k+3, a^k+5 where K ∈ N is …………
(1) a^k+5
(2) a^k
(3) a^k+6
(4) a^k+9
Answer:
(1) a^k+5

Question 5.
The LCM of (x + 1)² (x – 3) and
(x² – 9) (x + 1) is
(1) (x + 1)³ (x² – 9)
(2) (x + 1)² x² – 9)
(3) (x + 1)² (x – 3)
(4) (x – 9) (x + 1)
Answer:
(2) (x + 1)²(x² – 9)

Question 6.
If \(\frac{a^{3}}{a-b}\) is added with \(\frac{b^{3}}{b-a}\) then the new expressions is …………
(1) a² – ab + b²
(2) a² + ab + b²
(3) a³ + b³
(4) a³ – b³
Answer:
(2) a² + ab + b²

Question 7.
The solution set of x + \(\frac { 1 }{ x } \) = \(\frac { 5 }{ 2 } \) is ………….
(1) 2,\(\frac { 1 }{ 2 } \)
(2) 2,-\(\frac { 1 }{ 2 } \)
(3) -2, – \(\frac { 1 }{ 2 } \)
(4) -2, \(\frac { 7 }{ 2 } \)
Answer:
(1) 2,\(\frac { 1 }{ 2 } \)

Question 8.
On dividing \(\frac{x^{2}-25}{x+3}\) by \(\frac{x+5}{x^{2}-9}\) is equal to ……………….
(1) (x – 5)(x + 3)
(2) (x + 5) (x – 3)
(3) (x – 5)(x – 3)
(4) (x + 5)(x + 3)
Answer:
(3) (x – 5)(x – 3)

Question 9.
The square root of (x + 11)² – 44x is ………….
(1)|(x – 11)²
(2) |x + 11|
(3) |11 – x²|
(4) |x – 11|
Answer:
(4) |x – 11|

Question 10.
If α, β are the zeros of the polynomial p(x) = 4x² + 3x + 7 then \(\frac{1}{\alpha}\) + \(\frac{1}{\beta}\) is equal to …………
(1) \(\frac { 7 }{ 3 } \)
(2) – \(\frac { 7 }{ 3 } \)
(3) \(\frac { 3 }{ 7 } \)
(4) – \(\frac { 3 }{ 7 } \)
Answer:
(4) – \(\frac { 3 }{ 7 } \)

Question 11.
The value of is  ……….
(1) -5
(2) 5
(3) 4
(4) -3
Answer:
(2) 5

Question 12.
If α and β are the roots of the equation ax² + bx + c = 0 then (α + β)² is ……………..
(1) \(\frac{-b^{2}}{a^{2}}\)
(2) \(\frac{-c^{2}}{a^{2}}\)
(3) \(\frac{-b^{2}}{a^{2}}\)
(4) \(\frac { bc }{ a } \)
Answer:
(3) \(\frac{-b^{2}}{a^{2}}\)

Question 13.
The roots of the equation x² – 8x + 12 = 0 are
(1) real and equal
(2) real and rational
(3) real and irrational
(4) unreal
Answer:
(2) real and rational

Question 14.
If one root of the equation is the reciprocal of the other root in ax² + bx + c = 0 then …………
(1) a = c
(2) a = b
(3) b = c
(4) c = 0
Answer:
(1) a = c

Question 15.
If α and β are the roots of the equation x² + 2x + 8 = 0 then the value of \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\) is ………………
(1) \(\frac { 1 }{ 2 } \)
(2) 6
(3) \(\frac { 3 }{ 2 } \)
(4) –\(\frac { 3 }{ 2 } \)
Answer:
(4) –\(\frac { 3 }{ 2 } \)

Question 16.

are………
(1) 4, 6, 6
(2) 6, 6, 4
(3) 6, 4, 6
(4) 4, 4, 6
Answer:
(3) 6, 4, 6

Question 17.
If [-1 -2 4] then the value of “a” is ………….
(1) 2
(2) -4
(3) 4
(4) -2
Answer:
(4) -2

Question 18.
The matrix A given by (a_ij)_2×2 if a_ij = i – j is …………

Answer:

Question 19.
If A is of order 4 × 3 and B is of order 3 × 4 then the order of BA is ………………….
(1) 3 × 4
(2) 4 × 4
(3) 3 × 3
(4) 4 × 1
Answer:
(3) 3 × 3

Question 20.
If then “x” is ……………..
(1) 1
(2) 2
(3) 3
(4) 4
Answer:
(4) 4

II. Answer the following.

Question 1.
Solve x + y = 7; y + z = 4; z + x = 1
Answer:
x + y = 7 ……(1)
y + z = 4 ………(2)
z + x = 1 …………(3)
Adding (1); (2) and (3)
2x + 2y + 2z = 12
x + y + z = 6 ….(4)
From (1) ⇒ x + y = 7
7 + z = 6
z = 6 – 7 = -1
From (2) ⇒ x + 4 = 6
x = 6 – 4 = 2
From (3) ⇒ y + 1 = 6
y = 6 – 1 = 5
The value of x = 2, y = 5 and z = -1

Question 2.
Find the HCF of 25x⁴y⁷; 35x³y⁸; 45x³y³
Answer:
25x⁴y⁷ = 5 × 5 × x⁴ × y⁷
35x³y⁸ = 5 × 7 × x³ × y⁸
45 x³y³ = 3 × 3 × 5 × x³ × y³
H.C.F. = 5x³y³

Question 3.
Find the values of k for which the following equation has equal roots.
(k – 12)x² + 2(k – 12)x + 2 = 0
Solution:
\(\frac{(k-12)}{a} x^{2}+\frac{2(k-12)}{b} x+\frac{2}{c}=0\)
Δ = b² – 4ac = (2(k – 12))² – 4(6 – 12)(2)
= 4(k – 12)[(k – 12) – 2]
= 4(k – 12)(k – 14)
The given equation will have equal roots, if A = 0
⇒ 4(k – 12)(k – 14) = 0
k – 12 = 0 or k – 14 = 0
k = 12, 14

Question 4.
Find the LCM of x³ + y³; x³ – y³; x⁴ + x²y² + y⁴
Answer:
x³ + y³ = (x + y) (x² – xy + y²)
x³ – y³ = (x – y)(x² + xy + y²)
x⁴ + x²y² + y⁴ = (x² + y²)² – (xy)²
= (x² + y² + xy)
L.C.M. = (x + y)(x – y) (x² + xy + y²)
(x² – xy + y²)
= [(x + y) (x² – xy +y²)]
[(x – y) (x² + xy + y²)]
= (x³ + y³) (x³ – y³)
L.C.M. = x⁶ – y⁶

Question 5.
The sum of two numbers is 15. If the sum of their reciprocals is \(\frac{3}{10}\), find the numbers.
Solution:
Let the numbers be α, β
Sum of the roots = α + β = 15 ………….. (1)
sum of their reciprocals = \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{10}\) ……….. (2)
\(\frac{\beta+\alpha}{\alpha \beta}=\frac{3}{10}\)
10(α + β) = 3αβ …………. (3)
3αβ = 10 × 15 = 150
Products of the roots = αβ = 50 ………….. (4)
∴ From (1) & (4), we have
x² – 15x + 50 = 0
(x – 10)(x – 5) = 0 ⇒ x = 10, 5
∴ he numbers are 10, 5.

Question 6.
For What value of k, the G.C.D. of [x² + x – (2k + 2)] and 2x² + kx – 12 is (x + 4)?
Answer:
p(x) = x² + x – (2k + 2)
g(x) = 2x² + kx – 12
G.C.D. = x + 4
when x + 4 is the G.C.D.
p(-4) = 0 or g(-4) = 0
[Hint: Take any one of the polynomial]
g(x) = 2x² + kx – 12 = 0
2(-4)² + k (-4) – 12 = 0
2(16) – 4x – 12 = 0
32 – 4k – 12 = 0
20 = 4k
k = \(\frac { 20 }{ 4 } \) = 5
The value of k = 5

Question 7.
Simplify \(\frac{x^{2}+x-6}{x^{2}+4 x+3}\)
Answer:
x² + x – 6 = (x + 3) (x – 2)
x² + 4x + 3 = (x + 3) (x + 1)

Question 8.
Multiply

Answer:

Question 9.
if P = \(\frac{x^{3}-36}{x^{2}-49}\) and Q = \(\frac { x+6 }{ x+7 } \) find the value of \(\frac { P }{ Q } \).
Answer:

Question 10.
Simplify
\(\frac { x }{ x+y } \) – \(\frac { y }{ x-y } \)
Answer:

Question 11.
Find the square root of (x + 11)² – 44x
Answer:

Question 12.
Find the square root of x⁴ + \(\frac{1}{x^{4}}\) + 2
Answer:

Question 13.
Solve the equation 2x – 1 – \(\frac { 2 }{ x-2 } \) = 3
Answer:
2x – 1 – \(\frac { 2 }{ x-2 } \) = 3

(x – 3) (x – 1) = 0
x – 3 = 0 or x – 1 = 0
x = 3 or x = 1
The solution set is (1,3)

Question 14.
Find the roots of \(\sqrt { 2 }\) x² + 7x + 5\(\sqrt { 2 }\) = 0
Answer:
\(\sqrt { 2 }\) x² + 7x + 5 \(\sqrt { 2 }\) = 0
\(\sqrt { 2 }\) x² + 2x + 5x + 5 \(\sqrt { 2 }\) = 0

\(\sqrt { 2 }\) x (x + \(\sqrt { 2 }\)) + 5 (x + \(\sqrt { 2 }\)) = 0
(x + \(\sqrt { 2 }\)) (\(\sqrt { 2 }\) x + 5) = 0
(x + \(\sqrt { 2 }\) ) = 0 or \(\sqrt { 2 }\) x + 5 = 0
x = – \(\sqrt { 2 }\) or \(\sqrt { 2 }\) x + 5 = 0
x = – \(\sqrt { 2 }\) or \(\sqrt { 2 }\) x = -5
x = \(\frac{-5}{\sqrt{2}}\)
The roots are and – \(\sqrt { 2 }\) and \(\frac{-5}{\sqrt{2}}\)

Question 15.
Solve \(\sqrt { x+5 }\) = 2x + 3 using formula method.
Answer:
\(\sqrt { x+5 }\) = 2x + 3
(\(\sqrt { x+5 }\))² = (2x + 3)²
x + 5 = 4x² + 9 + 12x
0 = 4x² + 12x – x + 9 – 5
0 = 4x² + 11x + 4
Here a = 4, b = 11, c = 4

Question 16.
The sum of a number and its reciprocal is \(\frac { 37 }{ 6 } \). Find the number.
Answer:
Let the require number be “x”
Its reciprocal is \(\frac { 1 }{ x } \)
By the given data

The required number is \(\frac { 1 }{ 6 } \) or 6

Question 17.
Determine the nature of the roots of the equation 2x² + x – 1 = 0
Answer:
2x² + x – 1 = 0
Here a = 2,b = 1,c = -1
∆ = b² – 4 ac
= 1² – 4(2) (-1)
= 1 + 8
= 9
Since b² – 4ac > 0 the roots are real and unequal

Question 18.
Find the value of k for which the given equation 9x² + 3kx + 4 = 0 has real and equal roots.
Answer:
9x² + 3 kx + 4 = 0
a = 9, b = 5k, c = 4
since the equation has real and equal roots
b² – 4ac = 0
(3k)² – 4(9) (4) = 0
9k² – 144 = 0
9k² = 144
k² = \(\frac { 144 }{ 9 } \) = 16
k = \(\sqrt { 16 }\)
k = ± 4

Question 19.
If one root of the equation
3x² – 10x + 3 = 0 is \(\frac { 1 }{ 3 } \) find the other root
Answer:
α and β are the roots of the equation 3x² – 10x + 3 = 0
Sum of the roots (α + β) = \(\frac { 10 }{ 3 } \)
Product of the roots (αβ) = \(\frac { 3 }{ 3 } \) = 1
one of the roots is \(\frac { 1 }{ 3 } \) (say α = \(\frac { 1 }{ 3 } \))
αβ = 1
\(\frac { 1 }{ 3 } \) × β = 1
β = 3
The other roots is 3

Question 20.
Form the quadratic equation whose roots are 3 + \(\sqrt { 7 }\); 3 – \(\sqrt { 7 }\)
Answer:
Sum of the roots = 3 + \(\sqrt { 7 }\) + 3 – \(\sqrt { 7 }\)
= 6
Product of the roots = (3 + \(\sqrt { 7 }\)) (3 – \(\sqrt { 7 }\) )
= 32 – (\(\sqrt { 7 }\))2
= 9 – 7
= 2
The required equation is
x² – (sum of the roots) x + product of the roots = 0
x² – (6)x + 2 = 0
x – 6x + 2 = 0

Question 21.
If α and β are the roots of the equation 3x² – 5x + 2 = 0, then find the value of α – β.
Answer:
α and β are the roots of the equation
3x² – 5x + 2 = 0
α + β = \(\frac { 5 }{ 3 } \), αβ = \(\frac { 2 }{ 3 } \)

Question 22.
Determine the matrix A = (a_ij)_3×2 if a_ij = 3i – 2j
Answer:

a_ij = 3_i – 2_j
a₁₁ = 3(1) – 2(1) = 3 – 2 = 1
a₁₂ = 3(1) – 2(1) = 3 – 4 = 1
a₂₁ = 3(2) – 2(1) = 6 – 2 = 4
a₂₂ = 3(2) – 2(2) = 6 – 4 = 2
a₃₁ = 3(3) – 2(1) = 9 – 2 = 7
a₃₂ = 3(3) – 2(2) = 9 – 4 = 5

Question 23.

Answer:

Question 24.
Find if

Answer:

Question 25.

find BA
Answer:

Question 26.
Find the unknowns a, b, c, d, x, y in the given matrix equation.

Answer:

Equating the corresponding elements of the two matrices we get
d + 1 = 2
d = 2 – 1 = 1
10 + a = 2a + 1
10 – 1 = 2a – a
9 = a
36 – 2 = b – 5
3b – b = -5 + 2
2b = -3 ⇒ b = \(\frac { -3 }{ 2 } \)
a – 4 = 4c ⇒ a – 4c = 4
9 – 4c = 4 ⇒ 4c = 4 – 9
-4c = -5 ⇒ c = \(\frac { 5 }{ 4 } \)
The value of a = 9, b = \(\frac { -3 }{ 2 } \), c = \(\frac { 5 }{ 4 } \) and d = 1

Question 27.
Prove that

multiplication is inverse to each other.
Answer:

AB = BA = I
Multiplication of matrices are iverse to each other.

III. Answer the following questions.

Question 1.
Solve x – \(\frac { y }{ 5 } \) = 6; y – \(\frac { z }{ 7 } \) = 8; z – \(\frac { x }{ 2 } \) = 10
Answer:
x – \(\frac { y }{ 5 } \) = 6
multiply by 5
5x – y = 30 …….(1)
y – \(\frac { z }{ 7 } \) = 8

Substitute the value of x = 8 in (1)
5(8) – y = 30
– y = 30 – 40 = -10
∴ y = 10
Substitute the value of x = 8 in (3)
2z – 8 = 20
2z = 20 + 8
z = \(\frac { 28 }{ 2 } \) = 14
The value of x = 8, y = 10 and z = 14

Question 2.
Solve for x,y and z using the given 3 equations
\(\frac { 2 }{ y } \) – \(\frac { 4 }{ z } \) + \(\frac { 3 }{ x } \) = 3; \(\frac { 5 }{ x } \) – \(\frac { 4 }{ y } \) – \(\frac { 8 }{ z } \) = 8 ; \(\frac { 6 }{ y } \) + \(\frac { 6 }{ z } \) +\(\frac { 1 }{ x } \) = 2
Answer:
Let \(\frac { 1 }{ x } \) = a, \(\frac { 1 }{ y } \) = b, \(\frac { 1 }{ z } \) = c
3a + 2b – 4c = 3 ………(1)
5a – 4b – 8c = 8 ………(2)
a + 6b + 6c = 2 ………(3)
(1) × 2 ⇒ 6a + 4b – 8c = 6 …..(1)

Question 3.
100 pencils are to be kept inside three types of boxes A, B and C. If 5 boxes of type A, 3 boxes of type B, 2 boxes of type C are used 6 pencils are left out. If 3 boxes of type A, 5 boxes of type B, 2 boxes of type C are used 2 pencils are left out. If 2 boxes of type A, 4 boxes of type B and 4 boxes of type C are used, there is a space for 4 pencils. Find the number of pencils that each box can hold.
Answer:
Let the number of pencil in the box A be “x”
Let the number of pencil in the box B be “y”
Let the number of pencil in the box C be “z”
By the given first condition
5x + 3y + 2z = 94 ….(1)
By the given second condition
3x + 5y + 2z = 98 ….(2)
By the given third condition
2x + 4y + 4z = 104 ….(3)
subtract (1) and (3)

substitute x = 8 and y = 10 in (1)
5(8) + 3(10) + 2z = 94
40 + 30 + 2z = 94
2z = 94 – 70
2z = 24
z = \(\frac { 24 }{ 2 } \) = 12
Number of pencil in box A = 8
Number of pencil in box B = 10
Number of pencil in box C = 12

Question 4.
What 2 masons earn in a day is earned by 3 male workers in a day. The daily wages of 15 female workers is ₹30 more than the total daily wages of 5 masons and 3 male workers. If one mason, one male worker and 2 female workers are engaged for a day, the builder has to pay ?160 as wages. Find the daily wages of a mason, a male worker and a female worker.
Answer:
Let the daily wage of a mason be ₹ x
Let the daily wage of a male worker be ₹ y
Let the daily wage of a female worker be ₹ z
By the given first condition
2x = 3y
2x – 3y = 0 …..(1)
By the given second condition
15z = 5x + 3y + 30
-5x – 3y + 15z = 30
5x + 3y – 15z = -30 ………(2)
By the given third condition

substitute the value of x = 60 in ….(1)
2(60) – 3y = 0
120 = 3y
y = \(\frac { 120 }{ 3 } \) = 40
substitute the value of x = 60 and y = 40 in (3)
60 + 40 + 2z = 160
2z = 160 – 100
2z = 60
z = \(\frac { 60 }{ 2 } \) = 30
Daily wages of a manson = ₹60
Daily wages of a male worker = ₹40
Daily wages of a female worker = ₹30

Question 5.
Find the G.C.D. of x³ – 10x² + 31x – 30 and 2x³ – 8x² + 2x + 12
Answer:
p(x) = x³ – 10x² + 31x – 30
g(x) = 2x³ – 8x² + 2x + 12
= 2(x³ – 4x² + x + 6)

G.C.D. = x² – 5x + 6

Question 6.
The G.C.D of x⁴ + 3x³ + 5x² + 26x + 56 and x⁴ + 2x³ – 4x² – x + 28 is x² + 5x + 7. Find their L.C.M.
Answer:
p(x) = x⁴ + 3x³ + 5x² + 26x + 56
g(x) = x⁴ + 2x³ – 4x² – x + 28
G.C.D. = x² + 5x + 7

Question 7.
Find the values of “a” and “b” given that p(x) = (x² + 3x + 2) (x² – 4x + a); g(x) = (x² – 6x + 9) × (x² + 4x + b) and their G.C.D is (x + 2) (x – 3)
Answer:
p(x) = (x² + 3x + 2) (x² – 4x + a)
= (x + 1) (x + 2) (x² – 4x + a)
G.C.D is given as (x + 2) (x – 3)
x – 3 is a factor of x² – 4x + a

p(3) = 0
9 – 4(3) + a = 0
9 – 12 + a = 0
– 3 + a =0
a = 3

g(x) = (x² – 6x + 9) (x² + 4x + 6)
= (x – 3) (x – 3) (x² + 4x + b)
But G.C.D. is (x + 2) (x – 3)
∴ x + 2 is a factor of x² + 4x + 6
g(-2) = 0
4 + 4(-2) + b = 0
4 – 8 + 6 = 0
-4 + b = 0
b = 4
The value of a = 3 and b = 4

Question 8.
Find the other polynomial g(x), given that LCM, HCF and p(x) as (x – 1) (x – 2) (x² – 3x + 3); x – 1 and x³ – 4x² + 6x – 3 respectively.
Answer:
LC.M. = (x – 1) (x – 2) (x² – 3x + 3)
HCF = (x – 1)
p(x) = x³ – 4x² + 6x – 3

p(x) = (x – 1) (x² – 3x + 3)
p(x) × g(x) = LCM × HCF
(x – 1) (x² – 3x + 3) × g(x)
= (x – 1) (x – 2) (x² – 3x + 3) (x – 1)

The other polynominal g(x) x² – 3x + 2

Question 9.
Divide

Answer:
2x² + x – 3 = 2x² + 3x – 2x – 3
= x(2x + 3) – 1 (2x + 3)
= (2x + 3) (x – 1)

2x² + 5x + 3 = 2x² + 3x + 2x + 3
= x(2x + 3) + 1 (2x + 3)
= (2x + 3) (x + 1)
x² -1 = (x + 1) (x – 1)

Question 10.
Simplify

Answer:
(x² – x – 6) = (x – 3) (x + 2)
2x² + 5x – 3 = 2×2 + 6x – x – 3

= 2x (x + 3) -1 (x + 3)
= (x + 3) (2x – 1)
x² + 5x + 6 = (x + 2) (x + 3)

Question 11.
Find the square root of (6x² + 5x – 6) (6x² – x – 2) (4x² + 8x + 3)
Answer:
6x² + 5x – 6 = 6x² + 9x – 4x – 6
= 3x(2x + 3) -2 (2x + 3)
= (2x + 3) (3x – 2)

6x² – x – 2 = 6x² – 4x + 3x – 2
= 2x (3x – 2) + 1 (3x – 2)
= (3x – 2) (2x + 1)

4x² + 8x + 3 = 4x² + 6x + 2x + 3
= 2x(2x + 3) + 1 (2x + 3)
= (2x + 3) (2x + 1)

Question 12.
Find the square root of the polynomial
\(\frac{4 x^{2}}{y^{2}}\) + \(\frac { 8x }{ y } \) + 16 + 12 \(\frac { y }{ x } \) + \(\frac{9 y^{2}}{x^{2}}\)
Answer:

Question 13.
If m – nx + 28x² + 12x³ + 9x⁴ is a perfect square, then find the values of m and n.
Answer:
Arrange the polynomial in descending power of x.
9x⁴ + 12x³ + 28x² – nx + m
Now,

Since the given polynomial is a perfect square,
-nx – 16x = 0
-x (n + 16) = 0
n + 16 = 0 ⇒ n = -16
m – 16 = 0 ⇒ m = 16
The value m = 16 and n = -16

Question 14.
If b + \(\frac { a }{ x } \) + \(\frac{13}{x^{2}}\) – \(\frac{6}{x^{3}}\) + \(\frac{1}{x^{4}}\) is a perfect square, find the values of “a” and “b”
Answer:
Arrange the values of “a” and “b”

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

Question 15.
Solve
\(\frac { 1 }{ x + 1 } \) + \(\frac { 4 }{ 3x+6 } \) = \(\frac { 2 }{ 3 } \)
Answer:

6x² – 12x + 9x – 18 = 0
6x(x – 2) + 9(x – 2) = 0

(x – 2) (6x + 9) = 0
x – 2 = 0 or 6x + 9 = 0
x = 2 or 6x + 9 = 0
x = 2 or 6x = -9
x = – \(\frac { 9 }{ 6 } \) = \(\frac { -3 }{ 2 } \)
The solution is \(\frac { -3 }{ 2 } \) or 2

Question 16.
A two-digit number is such that the product of the digits is 14. When 45 is added to the number, the digits interchange their places. Find the number (solve by completing square method)
Answer:
Let the ten’s digit be “x”
∴ The unit digit = \(\frac { 14 }{ x } \)
∴ The number is 10x + \(\frac { 14 }{ x } \)
If the digits are interchanged the number is \(\frac { 140 }{ x } \) + x
By the given condition
10x + \(\frac { 14 }{ x } \) + 45 = \(\frac { 140 }{ x } \) + x
multiply by x
10x² + 14 + 45x = 140 + x²
9x² + 45x + 14 – 140 = 0
9x² + 45x – 126 = 0
Divided by 9
x² + 5x – 14 = 0
x² + 5x = 14

Since the digit of the number can not be negative
∴ x = 2
The number = 10x + \(\frac { 14 }{ x } \)
= 20 + \(\frac { 14 }{ 2 } \)
= 20 + 7
= 27
The number is 27

Question 17.
A rectangular garden 10 m by 16 m is to be surrounded by a concreate walk of uniform width. Given that the area of walk is 120 sqm assuming the width of walk be ‘V form the equation then solve it by formula method.
Answer:
Area of the garden = 16 × 10
= 160 sq.m

Area of the garden with walking area
= (1.6 + 2x) (10 + 2x)
= 160 + 32x + 20x + 4x²
= 4x² + 52x + 160
Area of the concrete walk = Area of the garden with walk – Area of garden
= 4x² + 52x + 160 – 160
120 = 4x² + 52x
4x² + 52x – 120 = 0
(÷ by 4) ⇒ x² + 13x – 30 = 0
Here a = 1, b = 13, c = -30
(comparing with ax² + bx + c = 0)

Since the width cannot be negative. Width of the garden = 2 m

Question 18.
If α and β are the roots of the equation 3x² – 5x + 2 = 0 find the value of
(i) \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\)
(ii) α – β
(iii) \(\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha}\)
Answer:
Comparing with ax² + bx + c = 0
a = 3, b = -5, c = 2
α and β are the roots of the equation
3x² – 5x + 2 = 0



(iii) \(\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha}=\frac{\alpha^{3}+\beta^{3}}{\alpha \beta}\)

Question 19.
If α and β are the roots of the equation 3x² – 6x + 1 = 0 from the equation whose roots are
(i) α² β;β²α
(ii) 2α + β; 2β + a
Answer:
α and β are the roots of 3x² – 6x + 1 = 0
α + β = \(\frac { 6 }{ 3 } \) = 2
αβ = \(\frac { 1 }{ 3 } \)
(i) Given the roots are α²β and β²α
Sum of the roots = α²β + β²α
= αβ (α + β)
= \(\frac { 1 }{ 3 } \)(2)
= \(\frac { 2 }{ 3 } \)
Product of the roots = (α²β) x (β²α)
= α²β²
= (αβ)³
= (\(\frac { 1 }{ 3 } \))³
= \(\frac { 1 }{ 27 } \)
The quadratic equation is
x² – (sum of the roots) x + product of the roots = 0
x² – (\(\frac { 2 }{ 3 } \)) x + \(\frac { 1 }{ 27 } \) = 0
multiply by 27
27x² – 18x + 1 = 0

(ii) Given the roots are 2α + β; 2 β + α
Sum of the roots = 2α + β + 2 β + α
= 2(α + β) + (α + β)
= 2(2) + 2
= 6

The quadratic polynomial is
x² – (sum of the roots) x + product of the roots = 0
x² – 6x + \(\frac { 25 }{ 3 } \) = 0
3x² – 18x + 25 = 0

Question 20.
Find X and Y if

Answer:

Question 21.
Solve for x,y

Answer:

Question 22.

Show that A² – 7A + 101₃ = 0
Answer:

Question 23.
Verify that (AB)^T = B^T A^T if

Answer:

From (1) and (2) we get
(AB)^T = B^TA^T

Question 24.
Draw the graph of y = x² and hence solve x² – 4x – 5 = 0.
Answer:
Given equations are y = x² and x² – 4x – 5 = 0
(i) Assume the values of x from – 4 to 5.

(ii) Plot the points (- 4,16), (- 3, 9), (- 2,4), (-1, 1), (0,0), (1, 1), (2,4), (3, 9), (4,16), (5,25).
(iii) Join the points by a smooth curve.
(iv) Solve the given equations


(v) The points of intersection of the line and the parabola are (-1, 1) and (5, 25).
The x-coordinates of the points are -1 and 5.
Thus solution set is {- 1, 5}.

Question 25.
Draw the graph of y = 2x² + x – 6 and hence solve 2x² + x – 10 = 0.
Answer:
Given equations are y = x² and x² – 4x – 5 = 0
(i) Assume the values of x from – 4 to 5.

(ii) Plot the points (- 4, 22), (- 3, 9), (- 2, 0), (-1, -5), (0, -6), (1, -3), (2, 4), (3, 15), (4, 30).
(iii) Join the points by a smooth curve.
(iv) Solve the given equations: Subtract 2x² + x – 10 = 0

y = 4 is a straight line parallel to X-axis
(v) The straight line and parabola intersect at point (-2.5, 4) and (2, 4).
The x-coordinates of the points are -2.5 and 2.
The solution set is {- 2.5, 2}.

Students can download Maths Chapter 3 Algebra Unit Exercise 3 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 Unit Exercise 3

Question 1.
Solve
\(\frac { 1 }{ 3 } \) (x + y – 5) = y – z = 2x – 11 = 9 – (x + 2z)
Answer:
\(\frac { 1 }{ 3 } \) (x + y – 5) = y – z
x + y – 5 = 3y – 3z
x + y – 3y + 3z = 5
x – 2y + 3z = 5 ….(1)
y – z – 2x – 11
-2 x + y – z = -11
2x – y + z = 11 …..(2)
2x – 11 = 9-(x + 2 z)
2x – 11 = 9 – x – 2z
2x + x + 2z = 9 + 11
3x + 2z = 20 ….(3)

3x – z = 17 …. (5)


Substitute the value of z = 1 in (3)
3x + 2(1) = 20
3x = 20 – 2
3x = 18
x = \(\frac { 18 }{ 3 } \) = 6
substitute the value of x = 6, z = 1 in (2)
2(6) – y + 1 = 11
12 – y + 1 = 11
13 – y = 11
-y = 11 – 13
-y = -2
y = 2
∴ The value of x = 6, y = 2 and z = 1

Question 2.
One hundred and fifty students are admitted to a school. They are distrbuted over three sections A, B and C. If 6 students are shifted from section A to section C, the sections will have equal number of students. If 4 times of students of section C exceeds the number of students of section A by the number of students in section B, find the number of students in the three sections.
Answer:
Let the number of students in section A be “x”
Let the number of students in section B be “y”
Let the number of students in section C be “z”
By the given first condition
x + y + z = 150 ……(1)
again by the second condition
x – 6 = z + 6
x – z = 6 + 6
x – z = 12 ….(2)
again by the third condition
x + y = 4z
x + y – 4z = 0
x + y – 4z = 0 ….(3)
Subtracting (1) and (3)

Substitute the value of z = 30 in (2)
x – 30 = 12
x = 12 + 30
= 42
Substitute the value of x = 42 and z = 30 in (1)
42 + y + 30 = 150
y + 72 = 150
y = 150 – 72
= 78
Number of students in section A, B and C are = 42, 78 and 30.

Question 3.
In a three-digit number, when the tens and the hundreds digit are interchanged the new number is 54 more than three times the original number. If 198 is added to the number, the digits are reversed. The tens digit exceeds the hundreds digit by twice as that of the tens digit exceeds the unit digit. Find the original number.
Answer:
Let the hundred digit be x
the tens digit be y and the unit digit be z
∴ The number is 100x + 10y + z
By the given first condition
100y + 10x + z = 54 + 3 (100x + 10y + z)
100y + 10x + z = 54 + 300x + 30y + 3z
-290x + 70y – 2z = 54 (÷ -2)
145x-35y + z = -27 ….(1)
Again by the second condition
198 + 100x + 10y + z = 100z + 10y + x
99x – 99z = -198 (÷ 99)
x – z = -2 ….(2)
Again by the third condition
y – x = 2(y – z)
y – x = 2y – 2z
– x – y + 2z = 0
x + y – 2z = 0 ….(3)

substitute the value of x = 1 in …….(2)
1 – z = -2
3 = z
∴ z = 3
substitute the value of x = 1 and z = 3 in …….(3)
1 – y – 6 = 0
y – 5 = 0
y = 5
∴ The number is 153

Question 4.
Find the least common multiple of xy(k² +1) + k(x² + y²) and xy(k² – 1) + k(x² – y²).
Solution:
xy (k² + 1) + k (x² + y²) …………… (1)
xy(k² – 1) + k(x² – y²) …………… (2)
(1) ⇒ xyk² + xy + kx² + ky²
(2) ⇒ xyk² – xy + kx² – ky²
(1) ⇒ yk (xk + y) + x (xk + y)
= (xk + y) (x + yk)
(2) ⇒ yk (xk – y) + x (xk – y)
= (x + yk) (xk – y)
∴ L.C.M. : (x + yk) (xk + y) (xk – y)
= (x + yk) (x²k² – y²)

Question 5.
Find the GCD of the following by division algorithm
2x⁴ + 13x³ + 27x² + 23x + 7,
x³ + 3x² + 3x + 1, x² + 2x + 1
Answer:
p(x) = 2x⁴ + 13x³ + 27x² + 23x + 7
g(x) = x³ + 3x² + 3x + 1
r(x) = x² + 2x + 1
(i) Find the G.C.D. of p(x) and g(x)

(ii) Find the G.C.D. of r(x) and the G.C.D. of p(x) and g(x)

∴ G.C.D.= x² + 2x + 1
∴ G.C.D. of the three
polynomials = x² + 2x + 1

Question 6.
Reduce the given Rational expressions to its lowest form
(i) \(\frac{x^{3 a}-8}{x^{2 a}+2 x^{a}+4}\)
Answer:
x^3a – 8 = (x^a)³ – 2³
(using the formula a³ – b³ = (a – b)(a² + ab + b²)
= (x^a – 2)[(x^a)² + x^a × 2 + 2²]
= (x^a – 2) (x^2a + 2x^a + 4)

(ii) \(\frac{10 x^{3}-25 x^{2}+4 x-10}{-4-10 x^{2}}\)
Answer:
10x³ – 25x² + 4x – 10 = 5x²(2x – 5) + 2 (2x – 5)
= (2x – 5) (5x² + 2)
– 4 – 10x² = -2 (2 + 5x²)
= -2(5x² + 2)

Question 7.
Simplify

Answer:

Question 8.
Arul, Ravi and Ram working together can clean a store in 6 hours. Working alone, Ravi takes twice as long to clean the store as Arul does. Ram needs three times as long as Arul does. How long would it take each if they are working alone?
Answer:
Let the time taken by Arul be “x” hours
Let the time taken by Ravi be “y” hours
Let the time taken by Ram be “z” hours
By the given first condition
\(\frac { 1 }{ x } \) + \(\frac { 1 }{ y } \) + \(\frac { 1 }{ z } \) = \(\frac { 1 }{ 6 } \)
Again by the given second condition
\(\frac { 1 }{ x } \) = 2 × \(\frac { 1 }{ y } \)
\(\frac { 1 }{ x } \) – \(\frac { 2 }{ y } \) = 0
By the given third condition
3 × \(\frac { 1 }{ z } \) = \(\frac { 1 }{ x } \)
– \(\frac { 1 }{ x } \) + \(\frac { 3 }{ z } \) = 0
Let \(\frac { 1 }{ x } \) = a, \(\frac { 1 }{ y } \) = b, \(\frac { 1 }{ z } \) = c
a + b + c = \(\frac { 1 }{ 6 } \)
6a + 6b + 6c = 1 …….(1)
a – 2b = 0 ……….(2)
-a + 3c = 0 …………(3)

Arul take 11 hours, Ravi take 22 hours and Ram takes 33 hours.

Question 9.
Find the square root of 289x⁴ – 612x³ + 970x² – 684x + 361.
Answer:

Question 10.
Solve \(\sqrt { y+1 }\) + \(\sqrt { 2y-5 }\) = 3
Answer:
\(\sqrt { y+1 }\) + \(\sqrt { 2y-5 }\) = 3
(squaring on bothsides)

8y² – 9y² – 12y + 78y – 20 – 169 = 0
-y² – 66y – 189 = 0
y² – 66y + 189 = 0
(y – 3) (y – 63) = 0

y – 3 or y = 63
The value of y is 3 and 63

Question 11.
A boat takes 1.6 hours longer to go 36 kins up a river than down the river. If the speed of the water current is 4 km per hr, what is the speed of the boat in still water?
Answer:
Let the speed of the boat in still water be “x”
Time taken to go for up of a river = \(\frac { 36 }{ x+4 } \)
By the given condition
\(\frac { 36 }{ x-4 } \) – \(\frac { 36 }{ x+4 } \) = 1.6

The speed of the boat in still water = 14 km/hr

Question 12.
Is it possible to design a rectangular park of perimeter 320 m and area 4800 m2? If so find its length and breadth.
Answer:
Let the length of the rectangular park be “l”
and the breadth of the rectangular park be “b”
Perimeter of the park = 320 m
2 (l + b) = 320
l + b = 160
l = 160 – b ……….(1)
Area of the park = 4800 m²
l × b = 4800 ….(2)
substitute the value of l = 160 – b in (2)
(160 – b)b = 4800
160b – b² = 4800
b² – 160b + 4800 = 0
(b – 120) (b – 40) = 0

b = -120 = 0 or b – 40 = 0
b = 120 or b = 40
If breadth is 120 length is 40
If breadth is 40 length is 120
Length of the park = 120 m
Breadth of the park = 40 m

Question 13.
At t minutes past 2 pm, the time needed to 3 pm is 3 minutes less than \(\frac{t^{2}}{4}\) Find t.
Answer:
Time needed by the minutes hand to show
3 pm = (60 – 1) minutes
By the given condition


∴ The value of t = 14 minutes

Question 14.
The number of seats in a row is equal to the total number of rows in a hall. The total number of seats in the hall will increase by 375 if the number of rows is doubled and the number of seats in each row is reduced by 5. Find the number of rows in the hall at the beginning.
Answer:
Let the number of rows in the hall be “x”
∴ Total number of rows = x
Total number of seats in the hall is “x2”
By the given condition
x² + 375 = 2x (x – 5)
x² + 375 = 2x²– 10x
x² – 2x² + 10x + 375 = 0
– x² + 10x + 375 = 0
– x² – 10x – 375 = 0
(x – 25) (x + 15)
x – 25 = 0 or x + 15 = 0
x = 25 or x = – 15

Number of rows in the hall = 25

Question 15.
If α and β are the roots of the polynomial f(x) – x² – 2x + 3, find the polynomial whose roots are
(i) α + 2, β + 2
Answer:
α and β are the roots of the polynomial
x² – 2x + 3 = 0
α + β = 2; αβ = 3
(i) Sum of the roots = α + 2 + β + 2
= α + β + 4
= 2 + 4
= 6
Product of the roots = (α + 2) (β + 2)
= αβ + 2α + 2β + 4
= αβ + 2(α + β) + 4
= 3 + 4 + 4
= 11
The quadratic polynomial
x² – (sum of the roots) x + product of the roots = 0
x² – (6) x + 11 = 0
x² – 6x + 11 = 0

(ii) \(\frac{\alpha-1}{\alpha+1}, \frac{\beta-1}{\beta+1}\)
Answer:
Sum of the roots


The quadratic polynomial is
x² – (sum of the roots) + products of the roots = 0
x² – (\(\frac { 2 }{ 3 } \)) x + \(\frac { 1 }{ 3 } \) = 0
3x² – 2x + 1 = 0

Question 16.
If -4 is a root of the equation x² + px – 4 = 0 and if the equation x² + px + q = 0 has equal roots, find the values of p and q.
Solution:
f(x) = x² + px – 4 = 0
If -4 is a root, then
f(-4) = (-4)² + P(-4) – 4 = 16 – 4p – 4 = 0
12 – 4p = 0
-4p = -12
p = 3
x² + 3x + q =0 has equal roots,
∆ = b² – 4ac = 0
3² – 4 × 1 × q = 0
9 – 4q = 0
-4 q = -9
q = \(\frac{9}{4}\)
p = 3, q = \(\frac{9}{4}\)

Question 17.
Two farmers Senthil and Ravi cultivates three varieties of grains namely rice, wheat and ragi. If the sale (in ₹) of three varieties of grains by both the farmers in the month of April is given by the matrix.

and the May month sale (in ₹) is exactly twice as that of the April month sale for each variety.
(i) What is the average sales of the months April and May.
Answer:
(i) Let A represent the sale on April

Let B represent the sale on May

Average sale of the month April and May

(ii) If the sales continue to increase in the same way in the successive months, what will be sales in the month of August?
Answer:
If it increasing in the successive months of
May sale is 2 (April sale)
June sale is 4 (April sale)
July sale is 8 (April sale)
August sale is 16 (April sale)
Sales in the month of August

Question 18.
If cos = I₂, find x.
Answer:

Question 19.
Given

and if BA = C², find p and q
Answer:

∴The value of p = 8 and q = 4

Question 20.

find the matrix D, such that CD – AB = 0
Answer:
Given

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.5

Question 1.
If sin 30° = x and cos 60° = y, then x² + y² is …….
(a) \(\frac{1}{2}\)
(b) 0
(c) sin 90°
(d) cos 90°
Solution:
(a) \(\frac{1}{2}\)
Hint:
sin 30° = x = \(\frac{1}{2}\)
cos 60° = y = \(\frac{1}{2}\)
x² + y²

Question 2.
If tan θ = cot 37°, then the value of θ is ………
(a) 37°
(b) 53°
(c) 90°
(d) 1°
Solution:
(b) 53°
Hint:
tan θ = cot 37°
= cot (90° – 53°)
= tan 53°
The value of θ is 53°

Question 3.
The value of tan 72°. tan 18° is ………
(a) 0
(b) 1
(c) 18°
(d) 72°
Solution:
(b) 1
Hint:
tan 72° . tan 18° = tan 72° . tan (90° – 72°)
= tan 72° . cot 72°
= tan 72° × \(\frac{1}{tan 72°}\)
= 1

Question 4.
The value of \(\frac{2 tan 30°}{1 – tan^{2} 30°}\) is equal to ………
(a) cos 60°
(b) sin 60°
(c) tan 60°
(d) sin 30°
Solution:
(c) tan 60°
Hint:

= √3 = tan 60°

Question 5.
If 2 sin 2θ = √3 then the value of θ is ………
(a) 90°
(b) 30°
(c) 45°
(d) 60°
Solution:
(b) 30°
Hint:
2 sin 2θ = √3 ⇒ sin 2θ = \(\frac{√3}{2}\)
sin 2θ = sin 60° ⇒ 2θ = 60°
θ = 30°

Question 6.
The value of 3 sin 70° sec 20° + 2 sin 39° sec 51° is ………
(a) 2
(b) 3
(c) 5
(d) 6
Solution:
(c) 5
Hint:
3 sin 70° sec 20° + 2 sin 39° sec 51°
= 3. sin 70° . sec (90° – 70°) + 2 sin 39° . sec (90° – 39°)
= 3. sin 70° . cosec 70° + 2 sin 39° . cosec 39°
= 3. sin70° × \(\frac{1}{sin 70°}\) + 2 sin 39° × \(\frac{1}{sin 39°}\)
= 3 + 2
= 5

Question 7.
The value of \(\frac{1-tan^{2}45°}{1+tan^{2}45°}\) is ……..
(a) 2
(b) 1
(c) 0
(d) \(\frac{1}{2}\)
Solution:
(c) 0
Hint:
\(\frac{1-tan^{2}45°}{1+tan^{2}45°}\) = \(\frac{1-1}{1+1}\)
= \(\frac{0}{2}\) = 0

Question 8.
The value of cosec (70° + θ) – sec (20° + θ) + tan (65° + θ) – cot (25° + θ) is ……..
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(a) 0
Hint:
cosec (70° + θ) – sec (20° – θ) + tan (65° + θ) – cot (25° – θ)
= sec [90° – (70° + θ)] – sec (20° – θ) + tan (65° + θ) – tan [90° – (25° – θ)]
= sec (20° – θ) – sec (20° – θ) + tan (65° + θ) – tan (65° + θ)
= 0

Question 9.
The value of tan 1° . tan 2° . tan 3°…. tan 89° is ………
(a) 0
(b) 1
(c) 2
(d) \(\frac{√3}{2}\)
Solution:
(b) 1
Hint:
tan 1° . tan 2° . tan 3° …….. tan 89°
= tan (90° – 89°). tan (90° – 88°) .tan (90° – 87°) …….. tan 45° . tan (89°)
= cot 89° . cot 88°. cot 87°. ……. tan 45° …….. tan 87°. tan 88°. tan 89°
= 1

Question 10.
Given that sin α = \(\frac{1}{2}\) and cos β = \(\frac{1}{2}\), then the value of α + β is ………
(a) 0°
(b) 90°
(c) 30°
(d) 60°
Solution:
Hint:
sin α = \(\frac{1}{2}\)
sin 30° = \(\frac{1}{2}\) ∴ α = 30°
cos β = \(\frac{1}{2}\) ∴ β = 60°
∴ α + β = 30° + 60°
= 90°

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.6

Question 1.
If the y-coordinate of a point is zero, then the point always lies ……..
(a) in the I quadrant
(b) in the II quadrant
(c) on x-axis
(d) on y-axis
Solution:
(c) on x-axis

Question 2.
The points (-5, 2) and (2, -5) lie in the ……….
(a) same quadrant
(b) II and III quadrant respectively
(c) II and IV quadrant respectively
(d) IV and II quadrant respectively
Solution:
(c) II and IV quadrant respectively

Question 3.
On plotting the points O (0, 0), A (3, -4), B (3, 4) and C (0, 4) and joining OA, AB, BC and CO, which of the following figure is obtained?
(a) Square
(b) Rectangle
(c) Trapezium
(d) Rhombus
Solution:
(c) Trapezium

Question 4.
If P (-1, 1), Q ( 3, -4), R (1, -1), S (-2, -3) and T (- 4, 4) are plotted on a graph paper, then the points in the fourth quadrant are ………
(a) P and T
(b) Q and R
(c) only S
(d) P and Q
Solution:
(b) Q and R

Question 5.
The point whose ordinate is 4 and which lies on the v-axis is ……….
(a) (4, 0)
(b) (0, 4)
(c) (1, 4)
(d) (4, 2)
Solution:
(b) (0, 4)

Question 6.
The distance between the two points (2, 3) and (1, 4) is ……..
(a) 2
(b) \(\sqrt{56}\)
(c) \(\sqrt{10}\)
(d) √2
Solution:
(d) √2
Hint:
\(\sqrt{(1-2)^{2}+(4+3)^{2}}\)
= \(\sqrt{(-1)^{2}+1^{2}}\)
= \(\sqrt{1+1}\)
= √2

Question 7.
If the points A (2, 0), B (-6, 0), C (3, a – 3) lie on the x-axis then the value of a is ……..
(a) 0
(b) 2
(c) 3
(d) -6
Solution:
(c) 3
Hint:
a – 3 = 0 ⇒ a = 3

Question 8.
If (x + 2, 4) = (5, y – 2), then the coordinates (x, y) are ………
(a) (7, 12)
(b) (6, 3)
(c) (3, 6)
(d) (2, 1)
Solution:
(c) (3, 6)
Hint:
x + 2 = 5
∴ x = 5 – 2 = 3
and
4 = y – 2
4 + 2 = y
∴ y = 6

Question 9.
If Q₁, Q₂, Q₃, Q₄ are the quadrants in a Cartesian plane then Q₂ ∩ Q₃ is ……..
(a) Q₁ U Q₂
(b) Q₂ U Q₃
(C) Null set
(d) Negative x-axis
Solution:
(c) Null set

Question 10.
The distance between the point (5, -1 ) and the origin is ………
(a) \(\sqrt{24}\)
(b) \(\sqrt{37}\)
(c) \(\sqrt{26}\)
(d) \(\sqrt{17}\)
Solution:
(c) \(\sqrt{26}\)
Hint:

Question 11.
The coordinates of the point C dividing the line segment joining the points P (2, 4) and Q (5, 7) internally in the ratio 2 : 1 is ……..
(a) (\(\frac{7}{2}, \frac{11}{2}\))
(b) (3, 5)
(c) (4, 4)
(d) (4, 6)
Solution:
(d) (4, 6)
Hint:
A line divides internally in the ratio m : n
m = 2, n = 1
x₁ = 2, x₂ = 5
y₁ = 4, y₂ = 7

= (4, 6)

Question 12.
If p (\(\frac{a}{3}, \frac{b}{2}\)) is the mid-point of the line segment joining A (-4, 3) and B (-2, 4) then (a, b) is ………
(a) (-9, 7)
(b) (-3, \(\frac{7}{2}\))
(c) (9, -7)
(d) (3, –\(\frac{7}{2}\))
Solution:
(a) (-9, 7)
Hint:
Mid point of a line

\(\frac{a}{3}\) = -3 ⇒ a = -9
\(\frac{b}{2}\) = \(\frac{7}{2}\) ⇒ b = 7
(a, b) is (-9, 7)

Question 13.
In what ratio does the point Q (1, 6) divide the line segment joining the points P (2, 7) and R(-2, 3) ………
(a) 1 : 2
(6) 2 : 1
(c) 1 : 3
(d) 3 : 1
Solution:
(c) 1 : 3
Hint:
A line divides internally in the ratio m : n the point P =
(\(\frac{mx_{2}+nx_{1}}{m+n}\), \(\frac{my_{2}+ny_{1}}{m+n}\))

(1, 6) = (\(\frac{-2m+2n}{m+n}\), \(\frac{3m+7n}{m+n}\))
\(\frac{-2m+2n}{m+n}\) = 1
-2m + 2n = m + nx
-3m = n – 2n
3m = n
\(\frac{m}{n}\) = \(\frac{1}{3}\)
∴ m : n = 1 : 3
and
\(\frac{3m+7n}{m+n}\) = 6
3m + 7n = 6m + 6n
6m – 3n = 7n – 6n
3m = n
\(\frac{m}{n}\) = \(\frac{1}{3}\)

Question 14.
If the coordinates of one end of a diameter of a circle is (3, 4) and the coordinates of its centre is (-3, 2), then the coordinate of the other end of the diameter is ……..
(a) (0, -3)
(b) (0, 9)
(c) (3, 0)
(d) (-9, 0)
Solution:
(d) (-9, 0)
Hint:
Let the other end of the diameter be (a, b)

Mid point of a line =
(\(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\))
(-3, 2) = \(\frac{3+a}{2}, \frac{4+b}{2}\)
\(\frac{3+a}{2}\) = -3
3 + a = -6
a = -6 – 3 = -9
and
\(\frac{4+b}{2}\) = 2
4 + b = 4
b = 4 – 4 = 0
The other end is (-9, 0)

Question 15.
The ratio in which the x-axis divides the line segment joining the points A (a₁, b₁) and B (a₂, b₂) is ……..
(a) b₁ : b₂
(b) -b₁ : b₂
(c) a₁ : a₂
(d) -a₁ : a₂
Solution:
(b) -b₁ : b₂
Hint:
A line divides internally in the ratio m : n the point P is,
(\(\frac{mx_{2}+nx_{1}}{m+n}, \frac{my_{2}+ny_{1}}{m+n}\))

The point P is (a, 0) = (\(\frac{ma_{2}+na_{1}}{m+n}, \frac{mb_{2}+nb_{1}}{m+n}\))
∴ \(\frac{mb_{2}+nb_{1}}{m+n}\) = 0
mb₂ + nb₁ = 0 ⇒ mb₂ = -nb₁
\(\frac{m}{n}\) = \(\frac{b_{1}}{b_{2}}\)
∴ m : n = -b₁ : b₂

Question 16.
The ratio in which the x-axis divides the line segment joining the points (6, 4) and (1, -7) is ………
(a) 2 : 3
(b) 3 : 4
(c) 4 : 7
(d) 4 : 3
Solution:
(c) 4 : 7
Hint:
A line divides internally in the ratio m : n the point P

-7m + 4n = 0
4n = 7m
\(\frac{m}{n}\) = \(\frac{4}{7}\)
The ratio is 4 : 7

Question 17.
If the coordinates of the mid-points of the sides AB, BC and CA of a triangle are (3, 4), (1, 1) and (2, -3) respectively, then the vertices A and B of the triangle are ……….
(a) (3, 2), (2, 4)
(b) (4, 0), (2, 8)
(c) (3, 4), (2, 0)
(d) (4, 3), (2, 4)
Solution:
(b) (4, 0), (2, 8)
Hint:

x₁ + x₂ = 6 → (1)
y₁ + y₂ = 8 → (2)
x₂ + x₃ = 2 → (3)
y₂ + y₃ = 2 → (4)
x₁ + x₃ = 4 → (5)
y₁ + y₃ = -6 → (6)
Adding (1) + (3) + (5) we get,
2(x₁ + x₂ + x₃) = 3
x₁ + x₂ + x₃ = 6
x₁ + x₃ = 4
∴ x₂ = 6 – 4 = 2
x₂ + x₃ = 2
x₁ = 6 – 2 = 4
Adding (2) + (4) + (6) we get,
2 (y₁ + y₂ + y₃) = 4
y₁ + y₂ + y₃ = 2
y₂ + y₃ = 2
∴ y₁ = 2 – 2 = 0
y₁ + y₃ = -6
y₂ = 2 + 6 = 8
∴ The vertices A is (4, 0) and B is (2, 8).

Question 18.
The mid-point of the line joining (-a, 2b) and (-3a, -4b) is ……..
(a) (2a, 3b)
(b) (-2a, -b)
(c) (2a, b)
(d) (-2a, -3b)
Solution:
(b) (-2a, -b)
Mid points of line

= (-2a, -b)

Question 19.
In what ratio does the y-axis divides the line joining the points (-5, 1) and (2, 3) internally ………
(a) 1 : 3
(b) 2 : 5
(c) 3 : 1
(d) 5 : 2
Solution:
(d) 5 : 2
Hint:
When it cut the y-axis the point P is (0, a)
A line divides internally in the ratio m : n the point

2m – 5n = 0 ⇒ 2m = 5n
\(\frac{m}{n}\) = \(\frac{5}{2}\) ⇒ m : n = 5 : 2

Question 20.
If (1, -2), (3, 6), (x, 10) and (3, 2) are the vertices of the parallelogram taken in order, then the value of x is ………
(a) 6
(b) 5
(c) 4
(d) 3
Solution:
(b) 5
Hint:
Since ABCD is a parallelogram

Mid-point of AC = Mid-point of BD

\(\frac{1+x}{2}\) = \(\frac{6}{2}\) ⇒ 1 + x = 6 ⇒ x = 6 – 1 = 5
The value of x = 5

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.4

Question 1.
Find the value of the following:
(i) sin 49°
(ii) cos 74° 39′
(iii) tan 54° 26′
(iv) sin 21° 21′
(v) cos 33° 53′
(vi) tan 70° 17′
Solution:
(i) sin 49° = 0.7547

(ii) cos 74° 39′ = cos 74° 39′ + 3′
From the table we get,
cos 74° 36′ = 0.2656
Mean difference of 3 = 8
cos 74° 39′ = 0.2656 – 8
= 0.2648

(iii) tan 54° 26′ = tan 54° 24′
tan 54° 24′ = 1.3968
Mean difference of 2 = 17
tan 54° 26’ = 1.3968 + 17
= 1.3985

(iv) sin 21° 21′ = sin 21° 18’+ 3’
sin 21° 18’ = 0.3633
Mean difference of 3 = 8(+)
sin 21° 21′ = 0.3633 + 8
= 0.3641

(v) cos 33° 53′ = cos 33° 48’ + 5′
cos 33° 48′ = 0.8310
Mean difference of 5 = 8
cos 33° 53′ = 0.8310 – 8
= 0.8302

(vi) tan 70° 17′ = tan 70° 12’+ 5’
tan 70° 12′ = 2. 7776
Mean difference of 5 = 131
tan 70° 17′ = 2. 7776 + 131
= 2.7907

Question 2.
Find the value of θ if
(i) sin θ = 0.9975
(ii) cos θ = 0.6763
(iii) tan θ = 0.0720
(iv) cos θ = 0.0410
(v) tan θ = 7.5958
Solution:
(i) sin θ = 0.9975
From the table we get,
= 0.9974 + 0.0001
= 85° 54′ + 1′
= 85° 55′ (or) 85° 56′ (or) 85° 57

(ii) cos θ = 0.6763
= 0. 6769 – 0.0006′
= 47° 24′ + 3′
= 47° 27

(iii) tan θ = 0. 0720
= 0. 0717 + 0.0003
= 4° 6′ + 1′
= 4° 7′

(iv) cos θ = 0.0410
= 0.0419 – 0.0009
= 87° 36′ + 3′
= 87° 39′

(v) tan θ = 7. 5958
= 7. 5958 + 0 (from the natural table)
= 82° 30′ (tangent table)

Question 3.
Find the value of the following:
(i) sin 65° 39′ + cos 24° 57’ + tan 10° 10′
(ii) tan 70° 58′ + cos 15° 26′ – sin 84° 59′
Solution:
(i) sin 65° 39′ = 0.9111
cos 24° 57′ = 0.9066
tan 10° 10′ = 0.1793
sin 65° 39′ + cos 24° 57′ + tan 10° 10′
= 0.9111 + 0.9066 + 0.1793
= 1.9970
sin 65° 39′ + cos 24° 57′ + tan 10° 10′ = 1.9970

(ii) tan 70° 58′ = 2. 8982
cos 15° 26′ = 0. 9639
sin 84° 59′ = 0.9962
tan 70° 58′ + cos 15° 26′ – sin 84° 59′
= 2. 8982 + 0. 9639 – 0.9962
= 3.8621 – 0.9962
= 2.8659

Question 4.
Find the area of a right triangle whose hypotenuse is 10 cm and one of the acute angle is 24°24′.
Solution:
In the ΔABC,

sin 24° 24′ = \(\frac{AB}{AC}\)
0. 4131 = \(\frac{AB}{10}\)
∴ AB = 4.131 cm
In the ΔABC,
cos 24° 24′ = \(\frac{BC}{AC}\)
0.9107 = \(\frac{BC}{10}\)
∴ BC = 9.107 cm
Area of the right angle = \(\frac{1}{2}\) × BC × AB
= \(\frac{1}{2}\) × 9.107 × 4.131
= \(\frac{37.62}{2}\)
Area of the triangle = 18.81 cm²

Question 5.
Find the angle made by a ladder of length 5m with the ground, if one of its end is 4m away from the wall and the other end is on the wall.
Solution:
AC is the length of the ladder.

In the right ΔABC,
cos θ = \(\frac{BC}{AC}\)
cos θ = \(\frac{4}{5}\)
= 0.8
cos θ = 0.8000
θ = 36° 52′
Angle made by a ladder with the ground is 36° 52′

Question 6.
In the given figure, HT shows the height of a tree standing vertically. From a point P, the angle of elevation of the top of the tree (that is ∠P) measures 42° and the distance to the tree is 60 metres. Find the height of the tree.
Solution:
Let the height of the tree HT be “x”

In the ΔHTP,
tan 42° = \(\frac{HT}{PT}\)
0.9004 = \(\frac{x}{60}\)
x = 0.9004 × 60
= 54.024
The height of the tree = 54.02 m

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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 3.13 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 3.13

Question 1.
Determine the nature of the roots for the following quadratic equations
(i) 15x² + 11x + 2 = 0
Answer:
Here a = 15, b = 11, c = 2
∆ = b² – 4ac
∆ = 11² – 4(15) × 2
= 121 – 120
∆ = 1 > 0
So the equation will have real and unequal roots

(ii) x² – x – 1 = 0
Answer:
Here a = 1, b = -1, c = -1
∆ = b² – 4ac
= (-1)² – 4(1)(-1)
= 1 + 4 = 5
∆ = 1 > 0
So the equation will have real and unequal roots.

(iii) \(\sqrt { 2 }\) t² – 3t + 3\(\sqrt { 2 }\) = 0
Answer:
Here a = \(\sqrt { 2 }\) , b = -3, c = 3\(\sqrt { 2 }\)
∆ = b² – 4ac
= (-3)² – 4(\(\sqrt { 2 }\)) (3\(\sqrt { 2 }\))
= 9 – 24 = -15
∆ = -15 < 0
So the equation will have no real roots.

(iv) 9y² – 6\(\sqrt { 2y }\) + 2 = 0
Answer:
Here a = 9, b = -6\(\sqrt { 2 }\), c = 2
∆ = b² – 4ac
= (-6\(\sqrt { 2 }\))² – 4(9) (2)
= 72 – 72
= 0
So the equation will have real and equal roots.

(v) 9a²b²x² – 24abcdx + 16c²d² = 0, a ≠ 0, b ≠ 0
Answer:
Here a = 9a²b²; b = -24 abed, c = 16c²d²
∆ = b² – 4ac
= (-24abcd)² – 4(9a²b²) (16c²d²)
= 576a²b²c²d² – 576a²b²c²d²
∆ = 0
So the equation will have real and equal roots.

Question 2.
Find the value(s) of ‘k’ for which the roots of the following equations are real and equal.
(i) (5k – 6)² + 2kx + 1 = 0
(ii) kx² + (6k + 2)x + 16 = 0
Solution:
(5k – 6)x² + 2kx + 1 :
a = (5k – 6), b = 2, c = 1
Δ = b² – 4ac
⇒ (2k)² – 4 (5k – 6)(1)
⇒ 4k² – 20k + 24 = 0 [∵ Since the roots are real and equal)
⇒ k² – 5k + 6 = 0
⇒ (k – 3)(k – 2) = 0
k = 3, 2

(ii) kx² + (6k + 2)x + 16 = 0
a = k, b = (6k + 2), c = 16
Δ = b² – 4ac [∵ the roots are real and equal)
⇒ (6k + 2)² – 4 × k × 16 = 0
⇒ 36k² + 24k + 4 – 64k = 0
⇒ 36k² – 40k + 4 =0
⇒ 36k² – 36k – 4k + 4 =0
⇒ 36k (k – 1) – 4 (k – 1) = 0
⇒ 4 (k – 1) (9k – 1) =0
⇒ k = 1 or k = \(\frac{1}{9}\)

Question 3.
If the roots of (a – b)x² + (b – c)x + (c – a) = 0 are real and equal, then prove that b, a, c are in arithmetic progression.
Answer:
(a – b) x2 + (b – c) x + (c – a) = 0
Here a = (a – b);b = b – c ; c = c – a

Since the equation has real and equal roots ∆ = 0
∴ b² – 4ac = 0
(b – c)² – 4(a – b)(c – a) = 0
b² + c² – 2bc -4 (ac – a² – bc + ab) = 0
b² + c² – 2bc – 4ac + 4a² + 4bc – 4ab = 0
b² + c² + 2bc -4a (b + c) + 4a² = 0
(b + c)² – 4a (b + c) + 4a² = 0
[(b+c) – 2a]² = 0 [using a² – 2ab + b² = (a – b)²]
b + c – 2a = 0
b + c = 2a
b + c = a + a
c – a = a – b (t₂ – t₁ = t₃ – t₂)
b,a,c are in A.P.

Question 4.
If a, b are real then show that the roots of the equation (a – b)² – 6(a + b)x – 9(a – b) = 0 are real and unequal.
Solution:
(a – b)x² – 6(a + b)x – 9(a – b) = 0
Δ = b² – 4ac
= (-6(a + b)² – 4(a – b)(-9(a – b))
= 36(a + b)² + 36(a – b)²
= 36 (a² + 2ab + b²) + 36(a² – 2ab + b²)
= 72a² + 12b²
= 72(a² + b²) > 0
∴ The roots are real and unequal.

Question 5.
If the roots of the equation (c² – ab)x² – 2(a² – bc)x + b² – ac = 0 are real and equal prove that either a = 0 (or) a³ + b³ + c³ = 3abc
Answer:
(c² – ab)x² – 2(a² – bc)x + b² – ac = 0
Here a = c² – ab ; b = – 2 (a² – bc); c = b² – ac
Since the roots are real and equal
∆ = b² – 4ac
[-2 (a² – bc)]² – 4(c² – ab) (b² – ac) = 0
4(a² – bc)² – 4[c² b² – ac³ – ab³ + a²bc] = 0
Divided by 4 we get
(a² – bc)² – [c² b² – ac³ – ab³ + a²bc] = 0
a⁴ + b² c² – 2a² bc – c²b² + ac³ + ab³ – a²bc = 0
a⁴ + ab³ + ac³ – 3a²bc = 0
= a(a³ + b³ + c³) = 3a²bc
a³ + b³ + c³ = \(\frac{3 a^{2} b c}{a}\)
a³ + b³ + c³ = 3 abc
Hence it is proved