Class 9

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Additional Questions

Exercise 3.1

Question 1.
Classify the following polynomials based on number of terms.
(i) x³ – x²
(ii) 5x
(iii) 4x⁴ + 2x³ + 1
(iv) 4.x³
(v) x + 2
(vi) 3x²
(vii) y⁴ + 1
(viii) y²⁰ + y¹⁸ + y²
(ix) 6
(x) 2u³ + u² + 3
(xi) u²³ – u⁴
(xii) y
Solution:
5x, 3x², 4x³, y and 6 are monomials because they have only one term.
x³ – x², x + 2, y⁴ + 1 and u²³ – u⁴ are binomials as they contain only two terms.
4x⁴ + 2x³ + 1 , y²⁰ + y¹⁸ + y² and 2u³ + u² + 3 are trinomials as they contain only three terms.

Question 2.
Classify the following polynomials based on their degree.
Solution:
p(x) = 3, p(x) = -7, p(x) = \(\frac{3}{2}\) are constant polynomials
p(x) = x + 3, p(x) = 4x, p(x) = \(\sqrt{3}\)x + 1 are linear polynomials, since the highest degree of the variable x is one.
p(x) = 5x² – 3x + 2, p(y) = \(\frac{5}{2}\) y² + 1, p(x) = 3x² are quadratic polynomials, since the highest degree of the variable is two.
p(x) = 2x³ – x² + 4x + 1, p(x) = x³ + 1, p(y) = y³ + 3y are cubic polynomials, since the highest degree of the variable is three.

Question 3.
Find the product of given polynomials p(x) = 3x³+ 2x – x² + 8 and q (x) = 7x + 2.

Solution:
(7x +2) (3x³ + 2x – x² + 8) = 7x(3x³ + 2x – x² + 8) + 2x
(3x³ + 2x – x² + 8) = 21x⁴ + 14x² – 7x³ + 56x + 6x³ + 4x – 2x²+ 16 = 21x⁴ – x³ + 12x⁴ + 60x + 16

Question 4.
Let P(x) = 4x⁴ – 3x + 2x³ + 5 and q(x) = x² + 2x + 4 find p (x) – q(x).

Solution:

Exercise 3.2

Question 1.
If p(x) = 5x³ – 3x² + 7x – 9, find
(i) p(-1)
(ii) p(2).
Solution:
Given that p(x) = 5x³ – 3x² + 7x – 9
(i) p(-1) = 5(-1)³ – 3(-1)² + 7(-1) – 9 = -5 – 3 – 7 – 9
= -24
(ii) p(2) = 5(2)³ – 3(2)³ + 7(2) – 9 = 40 – 12 + 14 – 9
∴ p(2) = 33

Question 2.
Find the zeros of the following polynomials.
(i) p(x) = 2x – 3
(ii) p(x) = x – 2
Solution:

Exercise 3.3

Question 1.
Find the remainder using remainder theorem, when
(i) 4x³ – 5x² + 6x – 2 is divided by x – 1.
(ii) x³ – 7x² – x + 6 is divided by x + 2.
Solution:
(i) Let p(x) = 4x³ – 5x² + 6x – 2. The zero of (x – 1) is 1.
When p(x) is divided by (x – 1) the remainder is p( 1). Now,
p(1) = 4(1)³ – 5(1)² + 6(1) – 2 = 4 – 5 + 6 – 2 = 3
∴ The remainder is 3.

(ii) Let p(x) = x³ – 7x² – x + 6. The zero of x + 2 is -2.
When p(x) is divided by x + 2, the remainder is p(-2). Now,
p(-2) = (-2)³ – 7(-2)² – (-2) +6 = – 8 – 7(4) + 2 + 6
= – 8 – 28 + 2 + 6 = -28.
∴ The remainder is -28.

Question 2.
Find the value of a if 2x³ – 6x² + 5ax – 9 leaves the remainder 13 when it is divided by x – 2.
Solution:
Let p(x) = 2x³ – 6x² + 5ax – 9
When p(x) is divided by (x – 2) the remainder is p(2).
Given that p(2) = 13
⇒ 2(2)³ – 6(2)² + 5a(2) – 9 = 13
⇒ 2(8) – 6(4) + 10a – 9 = 13
⇒ 16 – 24 + 10a – 9 = 13
⇒ 10a – 17 = 13
⇒ 10a = 30
⇒ ∴ a = 3

Question 3.
If the polynomials f(x) = ax³+ 4x² + 3x – 4 and g (x) = x³ – 4x + a leave the same remainder when divided by x – 3. Find the value of a. Also find the remainder.

Solution:
Let f(x) = ax³ + 4x² + 3x – 4 and g(x) = x³ – 4x + a,When f(x) is divided by (x – 3), the remainder is f(3).
Now f(3) = a(3)³4(3)² + 3(3) – 4 = 27a + 36 + 9 – 4
f (3) = 27a + 41 …(1)
When g(x) is divided by (x – 3), the remainder is g(3).
Now g(3) = 33 – 4(3) + a = 27 – 12 + a = 15 + a …(2)
Since the remainders are same, (1) = (2)
Given that, f(3) = g(3)
That is 27a + 41 = 15 +a
27a – a = 15 – 41
26a = -26

Sustituting a = -1, in f(3), we get
f(3) = 27(-1) + 41 = -27 + 41
f(3) = 14
∴ The remainder is 14.

Question 4.
Show that x + 4 is a factor of x³ + 6x² – 7x – 60.
Solution:

Question 5.
In (5x + 4) a factor of 5x³ + 14x² – 32x – 32
Solution:

Question 6.
Find the value of k, if (x – 3) is a factor of polynomial x³ – 9x² + 26x + k.
Solution:
Let p (x) = x³ – 9x² + 26x + k
By factor theorem, (x – 3) is a factor of p(x), if p(3) = 0
p(3) = 0
3³ – 9(3)² + 26 (3) + k = 0
27 – 81 + 78 + k = 0
k = -24

Question 7.
Show that (x – 3) is a factor of x³ + 9x² – x – 105.
Solution:
Let p(x) = x³ + 9x² – x – 105
By factor theorem, x – 3 is a factor of p(x), if p(3) = 0
p (3) = 3³ + 9(3)³ – 3 – 105
= 27 + 81 – 3 – 105
= 108 – 108
p(3) = 0

Therefore, x – 3 is a factor of x³ + 9x² – x – 105.

Question 8.
Show that (x + 2) is a factor of x³ – 4x² – 2x + 20.
Solution:
Let p(x) = x³ – 4x² – 2x + 20
By factor theorem,
(x + 2) is factor of p(x), if p(-2) = 0
p(-2) = (-2)³ – 4(-2)² – 2(-2) + 20 = -8 – 4(4) + 4 + 20
P(- 2) = 0
Therefore, (x + 2) is a factor of x³ – 4x² – 2x + 20

Exercise 3.4

Question 1.
Expand the following using identities :
(i) (7x + 2y)²
(ii) (4m – 3m)²
(iii) (4a + 3b) (4a – 3b)
(iv) (k + 2)(k – 3)
Solution:
(i) (7x + 2y)² = (7x)² + 2 (7x) (2y) + (2y)² = 49x² + 28xy + 4y²
(ii) (4m – 3m)² = (4m)² – 2(4m) (3m) + (3m)² = 16m² – 24mn + 9n²
(iii) (4a + 3b) (4a – 3b) [We have (a + b ) (a – b) = a² – b² ]
Put [a = 4a, b = 3b]
(4a + 3b) (4a – 3b) = (4a)² – (3b)² = 16a² – 9b²
(iv) (k + 2)(k – 3) [We have (x + a) (x – b) = x² + (a – b)x – ab]
Put [x = k, a = 2, b = 3]
(k + 2) (k – 3) = k² + (2 – 3)x – 2 × 3 = k² – x – 6

Question 2.
Expand : (a + b – c)²
Solution:
Replacing ‘c’ by ‘-c’ in the expansion of
(a + b + c)² = a² + b² + c² + 2 ab+ 2 bc + 2ca
(a + b + (-c))² = a² + b² + (-c)² + 2ab + 2b(-c) + 2(-c)a
= a² + b² + c² + 2ab – 2bc – 2ca

Question 3.
Expand : (x + 2y + 3z)²
Solution:
We know that,
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Substituting, a = x, b = 2y and c = 3z
(x + 2y + 3 z)² = x² + (2y)² + (3z)² + 2(x)(2y) + 2(2y)(3z) + 2(3z)(x)
= x² + 4y² + 9z² + 4xy + 12y² + 6zx

Question 4.
Find the area of square whose side length is m + n – q.
Solution:
Area of square = side × side = (m + n – q)²
We know that,
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
[m + n + (-q)]² = m² + n² + (-q)² + 2mn + 2n(-q) + 2(-q)m
= m² + n² + q² + 2mn – 2nq – 2qm
Therefore, Area of square = m² + n² + q² + 2mn – 2nq – 2qm = [m² + n² + q² + 2mn – 2nq – 2qm] sq. units.

EXERCISE 3.5

Question 1.
Factorise the following
(i) 25m^-2 – 16n²
(ii) x⁴ – 9x²
Solution:
(i) 25m² – 16n² = (5m)² – (4n)²
= (5m – 4n) (5m + 4n) [∵ a² – b² = (a – b)(a + b)]
(ii) x⁴ – 9x² = x²(x² – 9) = x²(x² – 3²) = x²(x – 3)(x + 3)

Question 2.
Factorise the following.
(i) 64m³ + 27n³
(ii) 729x³ – 343y³
Solution:
(i) 64m³ + 27n³ = (4m)³ + (3n)³
= (4m + 3n) ((4m)² – (4m) (3n) + (3n)²) [∵ a² + b² = (a + b) (a² – ab + b²)]
= (4m + 3n) (16m² – 12mn + 9n²)
(ii) 729x³ – 343y³ = (9x)³ – (7y)³ = (9x – 7y) ((9x)² + (9x) (7y) + (7y)²
= (9x – 7y) (81x² + 63xy + 49y²)
[∵ a³ – b³ = (a – b) (a² + ab + b²)]

Question 3.
Factorise 81x² + 90x + 25
Solution:
81x² + 90x + 25 = (9x)² + 2 (9x) (5) + 5²
= (9x + 5)² [ ∵ a² + 2ab + b² = (a + b)²]

Question 4.
Find a³ + b² if a + b = 6, ab = 5.
Solution:
Given, a + b = 6, ab = 5
a² + b² = (a + b)² – 3ab(a + b) = (6)³ – 3(5)(6) = 126

Question 5.
Factorise (a – b)² + 7(a – b) + 10
Solution:
Let a – b = p, we get p² + 7p + 10,

p² + 7p + 10 = p² + 5p + 2p + 10
= p(p + 5) + 2(p + 5) = (p + 5)(p + 2)
Put p = a – b we get,
(a – b)² + 7(a – b) + 10 = (a – b + 5)(a – b + 2)

Question 6.
Factorise 6x² + 17x + 12
Solution:

Exercise 3.6

Question 1.
Factorise x² + 4x – 96
Solution:

Question 2.
Factorise x² + 7xy – 12y²
Solution:

Question 3.
Find the quotient and remainder when 5x³ – 9x² + 10x + 2 is divided by x + 2 using synthetic division.
Solution:

Hence the quotient is 5x² – 19x + 48 and remainder is -94.

Question 4.
Find the quotient and remainder when -7 + 3x – 2x² + 5x³ is divided by -1 + 4x using synthetic division.
Solution:

Question 5.
If the quotient on dividing 5x⁴ + 4x³ + 2x + 1 by x + 3 is 5x³ + 9x² + bx – 97 then find the values of a, b and also remainder.
Solution:

Quotient 5x³ + 11x² + 33x – 97 is compared with given quotient 5x³ + ax² + bx – 97
co-efficient of x² is – 11 = a and co-efficient of x is 33 = b.
Therefore a = -11, b = 33 and remainder = 292.

Exercise 3.7

Question 1.
Find the quotient and the remainder when 10 – 4x + 3x² is divided by x – 2.
Solution:
Let us first write the terms of each polynomial in descending order (or ascending
order). Thus the given problem becomes (3x² – 4x + 10) ÷ (x – 2)

∴ Quotient = 3x + 2 and Remainder = 14, i.e 3x² – 4x + 10 = (x – 2) (3x + 2) + 14 and is in the form
Dividend = (Divisor × Quotient) + Remainder

Question 2.
If 8x³ – 14x² – 19x – 8 is divided by 4x + 3 then find the quotient and the remainder.
Solution:

Exercise 3.8

Question 1.
Factorise 2x³ – x² – 12x – 9 into linear factors.
Solution:
Let p(x) = 2x³ – x² – 12x – 9
Sum of the co-efficients = 2 – 1 – 12 – 9 = -20 ≠ 0
Hence x – 1 is not a factor
Sum of co-efficients of even powers with constant = -1 – 9 = -10
Sum of co-efficients of odd powers = 2 – 12 = -10
Hence x + 1 is a factor of x.
Now we use synthetic division to find the other factors.


Then p (x) = (x + 1)(2x² – 3x – 9)
Now 2x² – 3x – 9 = 2x² – 6x + 3x – 9 = 2x(x – 3) + 3(x – 3)
= (x – 3)(2x + 3)
Hence 2x³ – x² – 12x – 9 = (x + 1) (x – 3) (2x + 3)

Question 2.
Factorize x³ + 3x² – 13x – 15.
Solution:
Let p(x) = x³ + 3x² – 13x – 15
Sum of all the co-efficients = 1 + 3 – 13 – 15 = -24 + 0
Hence x – 1 is not a factor
Sum of co-efficients of even powers with constant = 3 – 15 = – 12
Sum of coefficients of odd powers = 1 – 13 = – 12
Hence x + 1 is a factor of p (x)
Now use synthetic division to find the other factors.

Hence p(x) = (x + 1) (x² + 2x – 15)
Now x³ + 3x² – 13x – 15 = (x + 1)(x² + 2x – 15)
Now x² + 2x – 15 = x² + 5x – 3x – 15 = x(x + 5) – 3 (x + 5) = (x + 5)(x – 3)
Hence x³ + 3x² – 13x – 15 = (x + 1)(x + 5)(x – 3)

Question 3.
Factorize x³ + 13x² + 32x + 20 into linear factors.

Solution:
Let, p(x) = x³ + 13x² + 32x + 20
Sum of all the coefficients = 1 + 13 + 32 + 20 = 66 ≠ 0
Hence, (x – 1) is not a factor.
Sum of coefficients of even powers with constant = 13 + 20 = 33
Sum of coefficients of odd powers = 1 + 32 = 33
Hence, (x + 1) is a factor of p(x)
Now we use synthetic division to find the other factors

Exercise 3.9

Question 1.
Find GCD of 25x³y² z, 45x²y⁴z³b
Solution:

Question 2.
Find the GCD of (y³ – 1) and (y – 1).
Solution:
y³ – 1 = (y – 1)(y³ + y + 1)
y – 1 = y – 1
Therefore, GCD = y – 1

Question 3.
Find the GCD of 3x² – 48 and x² – 7x + 12.
Solution:
3x² – 48 = 3(x² – 16) = 3(x² – 4³) = 3(x + 4)(x – 4)
x² – 7x + 12 = x² – 3x – 4x + 12 = x(x – 3) – 4 (x – 3) = (x – 3) (x – 4)
Therefore, GCD = x – 4

Question 4.
Find the GCD of a^x , a^{x + y}, a^{x + y + z}.
Solution:

Exercise 3.10

Question 5.
Solve graphically. x – y = 3 ; 2x – y = 11
Solution:

Exercise 3.11

Question 1.
Solve using the method of substitution. 5x – y = 5, 3x + y = 11
Solution:

Exercise 3.12

Question 2.
Solve by the method of elimination. 2x + y = 10; 5x – y = 11
Solution:

Exercise 3.13

Question 3.
Solve 5x – 2y = 10 ; 3x + y = 17 by the method of cross multiplication.
Solution:

Exercise 3.14

Question 1.
The age of Arjun is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father.
Solution:
Let the age of the father be x
Let the sum of the age of his sons be y
At present x = 2y ⇒ x – 2y = 0 …. (1)
After 20 years x + 20 = y + 40
x – y = 40 – 20

Exercise 3.15

Multiple Choice Questions :

Question 1.
The polynomial 3x – 2 is a
(1) linear polynomial
(2) quadratic polynomial
(3) cubic polynomial
(4) constant polynomial
Hint: A polynomial is linear if its degree is 1
Solution:
(1) linear polynomial

Question 2.
The polynomial 4x² + 2x – 2 is a
(1) linear polynomial
(2) quadratic polynomial
(3) cubic polynomial
(4) constant polynomial
Hint: A polynomial is quadratic of its highest power of x is 2
Solution:
(2) quadratic polynomial]

Question 3.
The zero of the polynomial 2x – 5 is

Hint:

Solution:
(1) \(\frac{5}{2}\)

Question 4.
The root of the polynomial equation 3x – 1 = 0 is

Hint:

Solution:
(2) x = \(\frac{1}{3}\)

Question 5.
Zero of (7 + 4x) is ___

Solution:
(2) \(\frac{-7}{4}\)

Question 6.
Which of the following has as a factor?
(1) x² + 2x
(2) (x – 1)²
(3) (x + 1)²
(4) (x² – 2²)
Solution:
(1) x² + 2x

Question 7.
If x – 2 is a factor of q (x), then the remainder is _____
(1) q (-2)
(2) x – 2
(3) 0
(4) -2
Solution:
(3) 0

Question 8.
(a – b) (a² + ab + b²) =
Solution:
(1) a³ + b³ + c³ – 3abc
(2) a² – b²
(3) a³ + b³
(4) a³ – b³
Solution:
(4) a³ – b³

Question 9.
The polynomial whose factors are (x + 2) (x + 3) is
(1) x² + 5x + 6
(2) x² – 4
(3) x² – 9
(4) x² + 6x + 5
Solution:
(1) x² + 5x + 6

Question 10.
(-a – b – c)² is equal to
(1) (a – b + c)²
(2) (a + b – c)²
(3) (-a + b + c)²
(4) (a + b + c)²
Solution:
(4) (a + b + c)²

Text Book Activities

Activity – 1
Write the Variable, Coefficient and Constant in the given algebraic expression,
Solution:

Activity – 2
Write the following polynomials in standard form.

Activity – 3

Question 1.
Find the value of k for the given system of linear equations satisfying the condition below:
(i) 2x + ky = 1; 3x – 5y = 7 has a unique solution
(ii) kx + 3y = 3; 12x + ky = 6 has no solution
(iii) (k – 3)x + 3y = k; kx + ky = 12 has infinite number of solution
Solution:

Question 2.
Find the value of a and b for which the given system of linear equation has infinite number of solutions 3x – (a + 1)y = 2b – 1, 5x + (1 – 2a)y = 3b
Solution:

Activity – 4
For the given linear equations, find another linear equation satisfying each of the given condition
Solution:

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.15

MULTIPLE CHOICE QUESTIONS :
Question 1.
If x³ + 6x² + kx + 6 is exactly divisible by (x + 2), then k = ?
(1) -6
(2) -7
(3) -8
(4) 11
Solution:
(4) 11
Hint: P(-2) = (-2)³ + 6 (-2)² + k (-2) + 6 = 0
⇒ – 8 + 24 – 2k +6 = 0
⇒ 22 = 2k
⇒ k = 11

Question 2.
The root of the polynomial equation 2x + 3 = 0 is
(1) \(\frac{1}{3}\)
(2) \(-\frac{1}{3}\)
(3) \(-\frac{3}{2}\)
(4) \(-\frac{2}{3}\)
Solution:
(3) \(-\frac{3}{2}\)

Question 3.
The type of the polynomial 4 – 3x³ is
(1) constant polynomial
(2) linear polynomial
(3) quadratic polynomial
(4) cubic polynomial.
Solution:
(4) cubic polynomial.
Hint: Polynomial of degree 3 is called cubic.

Question 4.
If x⁵¹ + 51 is divided by x + 1, then the remainder is
(1) 0
(2) 1
(3) 49
(4) 50
Solution:
(4) 50
Hint: P(-1 = (-1)⁵¹ + 51 = -1 + 51 = 50

Question 5.
The zero of the polynomial 2x + 5 is

Solution:
(2) \(-\frac{5}{2}\)

Question 6.
The sum of the polynomials p(x) = x³ – x² – 2, q(x) = x² – 3x + 1
(1) x³ – 3x – 1
(2) x³ + 2x² – 1
(3) x³ – 2x² – 3x
(4) x³ – 2x² + 3x – 1
Solution:
(1) x³ – 3x – 1
Hint:

Question 7.
Degree of the polynomial (y³ – 2) (y³ + 1) is
(1) 9
(2) 2
(3) 3
(4) 6
Solution:
(4) 6
Hint: (y³ – 2)(y³ + 1) = (y³ – 2)(y³ – 2) × 1 = y⁶ – 2y³ + y³ – 2 = y⁶ – y³ – 2

Question 8.
Let the polynomials be
(A) -13q⁵ + 4q² + 12q
(B) (x² + 4 ) (x² + 9)
(C) 4q⁸ – q⁶ + q²
(D) \(-\frac{5}{7}\)y¹² + y³ + y⁵
Then ascending order of their degree is
(1) A, B, D, C
(2) A, B, C, D
(3) B, C, D, A
(4) B, A, C, D
Solution:
(4) B, A, C, D
Hint: Degree of (A), (B) (C) & (D) are respectively be 5, 4, 8, 12

Question 9.
If p (a) = 0 then (x – a) is a ________ of p(x)
(1) divisor
(2) quotient
(3) remainder
(4) factor
Solution:
(4) factor

Question 10.
Zeros of (2 – 3x) is _____
(1) 3
(2) 2
(3) \(\frac{2}{3}\)
(4) \(\frac{3}{2}\)
Solution:
(3) \(\frac{2}{3}\)
Hint: 2 – 3x =0
-3x = – 2
x = \(\frac{2}{3}\)

Question 11.
Which of the following has x – 1 as a factor?
(1) 2x – 1
(2) 3x – 3
(3) 4x – 3
(4) 3x – 4
Solution:
(2) 3x – 3
Hint: p(x) = 3x – 3
P( 1) = 3(1) – 3 = 0
∴ (x – 1) is a factor of p(x)

Question 12.
If x – 3 is a factor of p (x), then the remainder is
(1) 3
(2) -3
(3) p(3)
(4) p(-3)
Solution:
(3) p(3)

Question 13.
(x + y)(x² – xy + y²) is equal to
(1) (x + y)³
(2) (x – y)³
(3) x³ + y³
(4) x³ – y³
Solution:
(3) x³ + y³

Question 14.
(a + b – c)² is equal to
(1) (a – b + c)²
(2) (-a – b + c)²
(3) (a + b + c)²
(4) (a – b – c)²
Solution:
(2) (-a – b + c)²
Hint: (a + b – c)² = [- (- a – b + c)]² = (-a – b + c)²

Question 15.
In an expression ax² + bx + c the sum and product of the factors respectively,
(1) a, bc
(2) b, ac
(3) ac, b
(4) bc, a
Solution:
(2) b, ac

Question 16.
If (x + 5) and (x – 3) are the factors of ax² + bx + c, then values of a, b and c are
(1) 1, 2, 3
(2) 1, 2, 15
(3) 1, 2, -15
(4) 1, -2, 15
Solution:
(3) 1, 2, -15
Hint: p(-5) = a (-5²) + b (-5) + c = 25a – 5b + c = 0 ……… (1)
p( 3) = a (3²) + bc + 3 + c = 9 + 3b + c = 0 …….. (2)
25a – 5b = 9a + 3b
25a – 9a = 3b + 5b
16a = 8 b
\(\frac{a}{b}=\frac{8}{16}=\frac{1}{2}\)
Substitute a = 1,b = 2 in (1)
25(1) – 5 (2) = -c
25 – 10 = 15 = – c
c = -15

Question 17.
Cubic polynomial may have a maximum of
(1) 1
(2) 2
(3) 3
(4) 4
Solution:
(3) 3

Question 18.
Degree of the constant polynomial is
(1) 3
(2) 2
(3) 1
(4) 0
Solution:
(4) 0

Question 19.
Find the value of m from the equation 2x + 3y = m. If its one solution is x = 2 and y = – 2.
(1) 2
(2) -2
(3) 10
(4) 0
Solution:
(2) -2
Hint: 2x + 3y = m, x = 2, y = – 2
m = 2(2) + 3(-2)
= 4 – 6 = -2

Question 20.
Which of the following is a linear equation?
(1) x + \(\frac{1}{x}\) = 2 x
(2) x(x – 1) = 2
(3) 3x + 5 = \(\frac{2}{3}\)
(4) x³ – x = 5
Solution:
(3) 3x + 5 = \(\frac{2}{3}\)
Hint: x + \(\frac{1}{x}\) = 2 ⇒ x² – 2x + 1 = 0; x(x – 1) = 2 ⇒ x² -x – 2 = 0

Question 21.
Which of the following is a solution of the equation 2x – y = 6?
(1) (2, 4)
(2) (4, 2)
(3) (3, -1)
(4) (0, 6)
Solution:
(2) (4, 2)
Hint: 2x – y = 6
2(4) – 2 = 8 – 2 = 6 = RHS

Question 22.
If (2, 3) is a solution of linear equation 2x + 3y = k then, the value of k is
(1) 12
(2) 6
(3) 0
(4) 13
Solution:
(4) 13
Hint: 2x + 3y = k
2(2) + 3(3) = 4 + 9 = 13

Question 23.
Which condition does not satisfy the linear equation ax + by + c = 0
(1) a ≠ 0,b = 0
(2) a = 0, b ≠ 0
(3) a = 0, b = 0, c ≠ 0
(4) a ≠ 0, b ≠ 0
Solution:
(3) a = 0, b = 0, c ≠ 0
Hint: a = 0, b =0, c ≠ 0 ⇒ (0)x + (0) y + c = 0 False

Question 24.
Which of the following is not a linear equation in two variable
(1) ax + by + c = 0
(2) 0x + 0y + c = 0
(3) 0x + by + c = 0
(4) ax + 0y + c = 0
Solution:
(2) 0x + 0y + c = 0
Hint: a and b both can not be zero

Question 25.
The value of k for which the pair of linear equations 4x + 6y – 1 = 0 and 2x + ky – 7 = 0 represents parallel lines is
(1) k = 3
(2) k = 2
(3) k = 4
(4) k = -3
Solution:
(1) k = 3
Hint: 4x + 6y = 1
6y = -4x + 1
y = \(\frac{-4}{6} x+\frac{1}{6}\) ……….. (1)
2x + ky – 7 = 0
ky = -2x + 7
y = \(\frac{-2}{k} x+\frac{7}{k}\) ……….. (2)
Since the lines (1) and (2) parallel m₁ = m₂
\(\frac{-4}{6}=\frac{-2}{k}\)
k = -2 × \(\frac{-6}{4}\) = 3

Question 26.
A pair of linear equations has no solution then the graphical representation is

Solution:
(2)

Hint: Parallel lines have no solution

Question 27.
If \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\) where a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 then the given pair of linear equation has ______ solution(s)
(1) no solution
(2) two solutions
(3) unique
(4) infinite
Solution:
(3) unique
Hint: \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\); unique solution

Question 28.
If \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\) where a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 then the given pair of linear equation has
(1) no solution
(2) two solutions
(3) infinite
(4) unique
Solution:
(1) no solution
Hint: \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\): parallel

Question 29.
GCD of any two prime numbers is _______
(1) -1
(2) 0
(3) 1
(4) 2
Solution:
(3) 1

Question 30.
The GCD of x⁴ – y⁴ and x² – y² is
(1) x⁴ – y⁴
(2) x² – y²
(3) (x + y)²
(4) (x + y)⁴
Solution:
(2) x² – y²
Hint:
x⁴ – y⁴ = (x² )² – (y²)² = (x² + y²) (x² – y²)
x² – y² = x² – y²
G.C.D. is = x² – y²

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.13

Question 1.
Solve by cross-multiplication method
(i) 8x – 3y = 12; 5x = 2y + 7
(ii) 6x + 7y – 11 = 0; 5x + 2y = 13
(iii) \(\frac{2}{x}+\frac{3}{y}=5 ; \frac{3}{x}-\frac{1}{y}+9=0\)
Solution:
(i) 8x – 3y = 12 ……….. (1)
5x – 2y = 7 ………… (2)
8x – 3y – 12 = 0
5x – 2y – 7 = 0
For cross multiplication method, we write the co-efficients as

(ii) 6x + 7y – 11 = 0
5x + 2y – 13 = 0
For cross multiplication method, we write the co-efficients as

∴ Solutions: x = 3; y = -1

(iii) \(\frac{2}{x}+\frac{3}{y}\) – 5 = 0 ……….. (1)
\(\frac{3}{x}-\frac{1}{y}\) + 9 = 0 ………… (2)
In (1), (2) Put \(\frac{1}{x}\) = a, \(\frac{1}{y}\) = b
(1) ⇒ 2a + 3b – 5 = 0
(2) ⇒ 3a – b + 9 = 0
For cross multiplication method, we write the co-efficients as

Solution: x = \(-\frac{1}{2}\); y = \(\frac{1}{3}\)

Question 2.
Akshaya has 2 rupee coins and 5 rupee coins in her purse. If in all she has 80 coins totalling ₹ 220, how many coins of each kind does she have.
Solution:
Let the number of 2 rupee coins be x
Let the number of 5 rupee coins be y
x + y = 80
2x + 5y = 220
x + y – 80 = 0
2x + 5y – 220 = 0
For cross multiplication method, we write the co-efficients as

∴ No. of 2 rupee coins = 60
No. of 5 rupee coins = 20

Question 3.
It takes 24 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for 8 hours and the pipe of the smaller diameter is used for 18 hours. Only half of the pool is filled. How long would each pipe take to fill the swimming pool.
Solution:
Let the time taken by the larger pipe be x hours and Set the time taken by the smaller pipe be y hours.
∴ \(\frac{1}{x}+\frac{1}{y}=\frac{1}{24}\)
In 1 hour the larger pipe can fill it = \(\frac{1}{x}\)
In 1 hour the smaller pipe can fill it \(\frac{1}{y}\)

For cross multiplication method, we write the co-efficients as

∴ To fill the full tank the larger pipe takes 40hrs
To fill the full tank the smaller pipe takes 60hrs

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 4 Geometry Ex 4.3

Question 1.
The diameter of the circle is 52 cm and the length of one of its chord is 20 cm. Find the distance of the chord from the centre.
Solution:
The distance of the chord from the centre O

Question 2.
The chord of length 30 cm is drawn at the distance of 8cm from the centre of the circle. Find the radius of the circle.
Solution:

Question 3.
Find the length of the chord AC where AB and CD are the two diameters perpendicular to each other of a circle with radius 4 \(\sqrt{2}\) cm and also find ∠OAC and ∠OCA.
Solution:
∆OAC is an isoceles triangle with one angle 90°
∴ ∠OAC + ∠OCA = 180° – 90°

2∠OAC = 90°
∠OAC = 45°
∴ ∠OCA = 45°

Question 4.
A chord is 12cm away from the centre of the circle of radius 15 cm. Find the length of the chord.
Solution:

Question 5.
In a circle, AB and CD are two parallel chords with centre O and radius 10 cm such that AB = 16 cm and CD = 12 cm determine the distance between the two chords?

Solution:
The distance between the two chord FE = OE + OF

∴ Distance between the chords is 14 cm

Question 6.
Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Solution:

The length of the common chord AB = AD + BD = (3 + 3) cm = 6 cm

Question 7.
Find the value of x° in the following

Solution:

Question 8.
In the given figure, ∠CAB = 25°, find ∠BDC, ∠DBA and ∠COB

Solution:

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.14

Solve by any one of the methods
Question 1.
The sum of a two digit number and the number formed by interchanging the digits is 110. If 10 is subtracted from the first number, the new number is 4 more than 5 times the sums of the digits of the first number. Find the first number.
Solution:
Let the two digit number be x y
Its place value = 10x + y
After interchanging the digits the number will be y x
Its place value = 10y + x
Their sum = 10x + y + 10y + x = 110
11x + 11y = 110
x + y = 10 ………….. (1)
If 10 is subtracted from the first number, the new number is 10x + y – 10
The sums of the digits of the first number is x + y
Its 4 more than 5 times is = 5(x + y) + 4
∴ 10x + y – 10 = 5x + 5y + 4
10x + y – 5x – 5y = 4 + 10
5x – 4y = 14 ………. (2)

Substitute y = 4 in (1)
x + 4 = 10
x = 10 – 4
x = 6
The first number is 64

Question 2.
The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes \(\frac{1}{2}\). Find the fraction.
Solution:
Let the numerator be x
Denominator be y
x + y = 12 ………. (1)
\(\frac{x}{y+3}=\frac{1}{2}\)
2x = y + 3
2x – y = 3 …………. (2)

Substitute y = 7 in (1)
x + 7 = 12
x = 12 – 7
x = 5
∴ The fraction is \(\frac{x}{y}=\frac{5}{7}\)

Question 3.
ABCD is a cyclic quadrilateral such that ∠A = (4y + 20)°, ∠B = (3y – 5)°, ∠C =(4x)° and ∠D = (7x + 5)°. Find the four angles.
Solution:
In a cyclic quadrilateral the sum of the four angles is 360° and
the sum of the opposite angles is 180°.
∠A + ∠C = 180°
4y + 20 + 4x = 180°
4x + 4y = 180 – 20
x + y = \(\frac{160}{4}\)
x + y = 40 ………….. (1)
∠B + ∠D = 180°
3y – 5 + 7x + 5 = 180
7x + 3y = 180 …………. (2)

Substitute y = 25° in (1)
x + 25 = 40
x = 40 – 25
x = 15°
∠A = (4y + 20)° = (4 × 25 + 20) = 100 + 20 = 120°
∠B = (3y – 5)° = (3 × 25° – 5) = 75° – 5° = 70°
∠C = (4x)° = 4 × 15° = 60°
∠D = (7x + 5)° = (7 × 15 + 5) = 105° + 5° = 110°
∴ ∠A = 120°, ∠B = 70°, ∠C = 60°, ∠D = 110°

Question 4.
On selling a T.V. at 5% gain and a fridge at 10% gain, a shopkeeper gains ₹ 2000. But if he sells the T.V. at 10% gain and the fridge at 5% loss, he gains ₹ 1500 on the transaction. Find the actual price of the T.V. and the fridge.
Solution:
Let the actual price of a T.V. = x
Let the actual price of a Fridge = y
\(\frac{5}{100}\)x + \(\frac{10}{100}\)y = 2000
\(\frac{5}{100}\) (x + 2y) = 2000
x + 2y = 2000 × \(\frac{100}{5}\)
x + 2y = 40000 ………. (1)
\(\frac{10}{100}\) x – \(\frac{5}{100}\)y = 1500
2x – y = 1500 × \(\frac{100}{5}\)
2x – y = 30000 ……….. (2)

Substitute y = 10000 in (1)
x + 2(10000) = 40000
x + 20000 = 40000
x = 40000 – 20000
x = 20000
∴ Actual price of T.V. = ₹ 20000
Actual price of Fridge = ₹ 10000

Question 5.
Two numbers are in the ratio 5 : 6. If 8 is subtracted from each of the numbers, the ratio becomes 4 : 5. Find the numbers.
Solution:
Let the two numbers be x, y
\(\frac{x}{y}=\frac{5}{6}\)
⇒ 6x = 5y
6x – 5y = 0 ………… (1)
\(\frac{x-8}{y-8}=\frac{4}{5}\)
⇒ 5(x – 8) = 4(y – 8)
5x – 40 = 4y – 32
5x – 4y = 40 – 32
5x – 4y =8 ………….. (2)

Substitute y = 48 in (1)
6x – 5(48) = 0
6x – 240 = 0
6x = 240
x = \(\frac{240}{6}\) = 40
\(\frac{x}{y}=\frac{40}{48}=\frac{5}{6}\)
∴ The numbers are in the ratio 5 : 6

Question 6.
4 Indians and 4 Chinese can do a piece of work in 3 days. While 2 Indians and 5 Chinese can finish it in 4 days. How long would it take for 1 Indian to do it? How long would it take for 1 Chinese to do it?
Solution:
Let for one Indian the rate of working be \(\frac{1}{x}\)
Let for one Chinese the rate of working be \(\frac{1}{y}\)

For cross multiplication method, we write the co-efficients as

∴ 1 Chinese can do the same piece of work = \(\frac{1}{y}\) = b = \(\frac{1}{36}\) = 36 days
1 Indian can do the piece of work = \(\frac{1}{x}\) = a = \(\frac{1}{18}\) = 18 days

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

Exercise 4.1

Question 1.
Find the complement of each of the following angles
(i) 63°
(ii) 24°
(iii) 48°
Solution:
(i) The complement of 63° = 90° – 63° = 27°
(ii) The complement of 24° = 90° – 24° = 66°
(iii) The complement of 48° = 90° – 48° = 42°

Question 2.
Find the supplement of each of the following angles
(i) 58°
(ii) 148°
(iii) 120°
Solution:
(i) The supplement of 58° = 180° – 58° = 122°
(ii) The supplement of 148° = 180° – 148° = 32°
(iii) The supplement of 120° = 180° – 120° = 60°

Question 3.
Find the value of x

Solution:

Question 4.
Find the values of x, y in the following figures

Solution:

Question 5.
In the given figure at right, side BC of ∆ABC is produced to D. Find ∠A and ∠C.

Solution:
From the figure
Exterior angle = 120°
⇒ ∠C = 180° – 120° = 60° (linear pair)
∴ ∠A = 180° – (40° + 60°) = 80°

Exercise 4.2

Question 1.
If the measures of three angles of a quadrilateral are 100°, 84° and 76° then, find the measure of fourth angle.
Solution:
Let the measure of the fourth angle be x°.
The sum of the angles of a quadrilateral is 360°
So, 100° + 84° + 76° + x° = 360°
260° + x° = 360°
x = 360° – 260° = 100°
Hence, the measure of the fourth angle is 100°.

Question 2.
In the parallelogram ABCD if ∠A = 65°, find ∠B, ∠C and ∠D.
Solution:
Let ABCD be a parallelogram in which ∠A = 65°
Since AD || BC we can treat AB as a transversal. So
∠A+∠B = 180°

65° +∠B = 180°
∠B = 180°-65°
∠B = 115°
Since the opposite angles of a parallelogram are equal, we have
∠C = ∠A = 65° and ∠D = ∠B = 115°
Hence, ∠B = 115°, ∠C = 65° and ∠D = 115°

Question 3.
If ABCD is a rhombus and if ∠A = 76°, find ∠CDB.
Solution:
∠A = ∠C = 76° (Opposite angles of a rhombus)
Let ∠CDB = x°. In ∆CDB, CD = CB
∠CDB + ∠CBD + ∠DCB = 180° (Angles of a triangle)

2x° + 76° = 180° ⇒ 2x° = 104°
x° = 52°
∴ ∠CDB = 52°

Question 4.
In a parallelogram, opposite sides are equal
Solution:
Given ABCD is a parallelogram
To Prove ABCD and DA = BC
Construction Join AC
Proof
Since ABCD is a parallelogram

AD || BC and AC is the transversal
∠DAC = ∠BCA ➝ (1) (alternate angles are equal)
AB || DC and AC is the transversal
∠BAC = ∠DCA ➝ (2) (alternate angles are equal)
In ∆ADC and ∆CBA
∠DAC = ∠BCA from (1)
AC is common
∠DCA = ∠BAC from (2)
∆ADC ≅ ∆CBA (By ASA)
Hence AD = CB and DC = BA (Corresponding sides are equal)

Question 5.
The angles of a quadrilateral are ¡n the ratio 1 : 2 : 3 : 4. Find all the angles. Let each ratio be x.

Solution:
Then the angles are x°, 2x°, 3x°, 4x°
x° + 2x° + 3x° + 4x° = 360°

Exercise 4.3

Question 1.
The radius of a circle 15 cm and the length of one of its chord is 24 cm. Find the distance of the chord from the centre.
Solution:
Distance of the chord from the centre.

Question 2.
The chord of length 32 cm is drawn at the distance of 12 cm from the centre of the circle. Find the radius of the circle.
Solution:

Question 3.
In a circle, AB and CD are two parallel chords with centre O and radius 5 cm such that AB = 8 cm and CD = 6 cm determine the distance between the chords?
Solution:

Question 4.
Find the value of x°

Solution:

Question 5.
Find the value of x°

Solution:

Exercise 4.4

Question 1.
Find the value of x in the figure.
Solution:

In the cyclic quadrilateral ABCD
∠ABC – 180°- 140° = 40°
∠BCA = 90°
∴ x = ∠BAC = 180°- (90° + 40°) = 50°

Question 2.
Find all the angles of the given cyclic quadrilateral ABCD in the figure.

Solution:

Question 3.
AB and CD are two parallel sides of a cyclic quadrilateral ABCD in the figure. such that AB = 12 cm, CD = 16 cm and the radius of the circle is 10cm. Find the shortest distance between the two sides AB and CD.
Solution:
In this figure,

The shortest distance between the two sides = 8 + 6 = 14 cm

Question 4.
In the given figure, AB and CD are the parallel chords of a circle with centre O, such that AB = 30 cm and CD = 40 cm. If OM ⊥ AB and OL ⊥ CD distance between LM is 35 cm. Find the radius of the circle?

Solution:

Exercise 4.5

Question 1.
Construct an equilateral triangle of sides 6 cm and locate its orthocentre.
Solution:

Construction:
(1) Draw the ∆ABC with the given measurements.
(2) Construct the altitudes from any two vertices (A and B) to their opposite sides BC and AC respectively.
(3) The point of intersection of the altitudes H is the orthocentre of the given ∆ABC.

Question 2.
Draw and locate the orthocentre of a right triangle PQR right angled at Q, with PQ = 4.5 cm and QR = 6 cm.
Solution:

Construction:
(1) Draw the ∆PQR with the given measurements.
(2) Construct altitudes from any two vertices (Q and R) to their opposite sides PR and PQ respectively.
(3) The point of intersection of the altitudes H is the orthocentre of the given ∆PQR.

Question 3.
Construct the circumcentre of the ∆ABC with AB = 5 cm, ∠A = 60° and ∠B = 80°, also draw two circumcircle and find the circum radius of the ∆ABC.
Solution:



Solution:
Step 1: Draw the ∆ABC with the given measurements.
Step 2 : Construct the perpendicular bisector of any two sides (AC and BC) and let them meet at S which is the circumcentre.
Step 3 : S as centre and SA = SB = SC as radius, draw the Circumcircle to passes through A,B and C. Circumradius = 3.9 cm.

Exercise 4.6

Question 1.
Draw the circumcircle for an equilateral triangle of side 6 cm.
Solution:

Construction:
(1) Draw the ∆ABC with the given measurements.
(2) Construct the perpendicular bisectors of AC and BC and let them meet at S which is the circumcentre.
(3) With S as centre and SA = SB = SC as radius, draw the circumcircle to pass through A, B and C.

Question 2.
Construct the centroid of ∆PQR such that PQ = 9 cm, PQ = 7cm, RP = 8 cm.
Solution:
In ∆PQR,
PQ = 5 cm,
PR = 6 cm
∠QPR = 60°

Construction :
Step 1 : Draw ∆PQR using the given measurements PQ = 9 cm, QR = 7 cm and RP = 8 cm and construct the perpendicular bisector of any two sides (PQ and QR) to find the mid-points M and N of PQ and QR respectively.
Step 2 : Draw the medians PN and RM and let them meet at G. The point G is the centroid of the given ∆PQR.

Question 3.
Draw and locate the centroid of the triangle ABC where right angle at A, AB = 8 cm and AC = 6 cm.
Solution:

Step 1 : Draw ∆ABC with the given measurements AB = 8 cm, ∠A = 90° and AC = 6 cm and construct the perpendicular bisector of any two sides (AB and AC) to find the mid points M and N of AB and BC respectively.
Step 2 : Draw the medians (C and BN and let them meet at G. The point G is the centroid of the given ∆ABC.

Question 4.
Construct the centroid of Ø PQR whose sides are PQ = 8 cm, QR = 6 cm, RP = 7 cm.

Solution:
Side = 6.5 cm

Construction:
Step 1 : Draw ∆ABC with AB = BC = CA = 6.5 cm
Step 2 : Construct angle bisectors of any two angles (A and B) and let them meet at 1.1 is the incentre of ∆ABC.
Step 3 : Draw perpendicular from I to any one of the side (AB) to meet AB at D.
Step 4 : With I as centre, ID as radius draw the circle. This circle touches all the sides of triangle internally.
Step 5 : Measure in radius. In radius = 1.9 cm.

Exercise 4.7

Multiple Choice Questions :

Question 1.
If an angle is equal to one third of its supplement, its measure is equal to
(1) 40°
(2) 50°
(3) 45°
(4) 55°
Hint:

Solution:
(3) 45°

Question 2.
In the given figure, OP bisect ∠BOC and OQ bisect ∠AOC. Then ∠POQ is equal to
(1) 90°
(2) 120°
(3) 60°
(4) 100°
Hint:


Solution:
(1) 90°

Question 3.
The complement of an angle exceeds the angle by 60°. Then the angle is equal to
(1) 25°
(2) 30°
(3) 15°
(4) 35°
Hint:

Solution:
(3) 15°

Question 4.
ABCD is a parallelogram, E is the mid-point of AB and CE bisects ∠BCD. Then ∠DEC is
(1) 60°
(2) 90°
(3) 100°
(4) 120°
Solution:
(2) 90°

Question 5.
If the length of a chord decreases, then its distance from the centre.
(1) increases
(2) decreases
(3) same
(4) cannot say
Solution:
(1) increases

Question 6.
In the figure, O is the centre of the circle and ∠ACB = 60° then ∠AOB =
(1) 60°
(2) 90°
(3) 120°
(4) 180°

Solution:
(3) 120°

Question 7.
The angle subtend by a semicircle at the centre is.
(1) 60°
(2) 90°
(3) 120°
(4) 180°
Solution:
(4) 180°

Question 8.
The angle subtend by a semicircle at the remaining part of the circumference is ___.
(1) 60°
(2) 90°
(3) 120°
(4) 180°

Text Book Activities

Activity – 3

Angle sum for a polygon.
Draw any quadrilateral ABCD.
Mark a point P in its interior. Join the segments PA, PB, PC and PD.
You have 4 triangles now.

How much is the sum of all the angles of the 4 triangles?
How much is the sum of the angles at their vertex, now P?
Can you now find the ‘angle sum’ of the quadrilateral ABCD?
Can you extend this idea to any polygon?
Solution:
Sum of the angles of the 4 triangle = 180° × 4 = 720°
Sum of the angles at their vertex, now p = 360°
Angle sum of the quadrilateral ABCD = 720°- 360° = 360°
Yes we can extend this idea to any polygon.

Activity – 4

Procedure.

1. Draw a circle with centre O and with suitable radius.
2. Make it a semi-circle through folding. Consider the point A, B on it.

3. Make crease along AB in the semi circles and open it.
4. We get one more crease line on the another part of semi circle, name it as CD (observe AB = CD)

5. Join the radius to get the ∆OAB and ∆OCD.

6. Using trace paper, take the replicas of triangle ∆OAB and ∆OCD.
7. Place these triangles ∆OAB and ∆OCD one on the other.

Activity – 6
Procedure:
1. Draw a circle of any radius with centre O.
2. Mark any four points A, B, C and D on the boundary. Make a cyclic quadrilateral ABCD and name the angles as in figure.

3. Figure Make a replica of the cyclic quadrilateral ABCD with the help of tracing paper.
4. Make the cutout of the angles A, B, C and D
5. Paste the angle cutout ∠1, ∠2, ∠3 and ∠4 adjacent to the angles opposite to A, B, C and D as in Figure.
6. Measure the angles ∠1 + ∠3, and ∠2 + ∠4.
Solution:
∠1 + ∠3 = 180°
∠2 + ∠4 = 180°

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.12

Question 1.
Solve by the method of elimination
(i) 2x – y = 3; 3x + y = 7
(ii) x – y = 5; 3x + 2y = 25
(iii) \(\frac{x}{10}+\frac{y}{5}\) = 14; \(\frac{x}{8}+\frac{y}{6}\) = 15
(iv) 3(2x + y) = 7xy; 3(x + 3y) = 11xy
(v) \(\frac{4}{x}\) + 5y = 7; \(\frac{3}{x}\) + 4y = 5
(vi) 13x + 11y = 70; 11x + 13y = 74
Solution:
(i) 2x – y = 3 ………….. (1)
3x + y = 7 ………… (2)

Substitute x = 2 in (1)
2(2) – y = 3
4 – y = 3
-y = 3 – 4
-y = -1
∴ Solution: x = 2; y = 1
Verification:
Substitute x = 2, y = 1 in (2)
3(2) + 1 = 7 = RHS
∴ Verified.

Substitute y = 2 in (1)
x – 2 = 5
x = 5 + 2
x = 7
∴ Solution: x = 7, y = 2
Verification:
Substitute x = 7, y = 2 in (2)
3(7) + 2(2) = 21 + 4 = 25 = RHS
∴ Verified.

Substitute y = 30 in (1)
x + 2 (30) = 140
x + 60 = 140
x = 140 – 60
x = 80
∴ Solution: x = 80; y = 30
Verification:
Substitute x = 80, y = 30 in (2)
3(80) + 4(30) = 240 + 120 = 360 = RHS
∴ Verified.

(iv) 3(2x +y) = 7xy ⇒ 6x + 3y = 7xy ………. (1)
3(x + 3y) = 11xy ⇒ 3x + 9y = 11xy ………….. (2)


Substitute a = 1 in (5)
6b + 3(1) = 7
6b + 3 = 7
6b = 7 – 3
b = \(\frac{4}{6}=\frac{2}{3}\)
∴a = \(\frac{1}{x}\) = 1 ⇒ x = 1
b = \(\frac{1}{y}=\frac{2}{3}\) ⇒ y = \(\frac{3}{2}\)
∴ Solution: x = 1; y = \(\frac{3}{2}\)

Substitute y = 4 in (1)
13x + 11 (4) = 70
13x + 44 = 70
13x = 70 – 44 = 26
x = \(\frac{26}{13}\) = 2
∴ Solution: x = 2; y = 4

Question 2.
The monthly income of A and B are in the ratio 3:4 and their monthly expenditures are in the ratio 5 : 7. If each saves ₹ 5,000 per month, find the monthly income of each.
Solution:
Let the monthly income of A and B be 3x and 4x respectively.
Let the monthly expenditure of A and B be 5y and 7y respectively.
∴ 3x – 5y = 5000 ……… (1)
4x – 7y = 5000 ……….. (2)

Substitute y = 5000 in (1)
3x – 5 (5000) = 5000
3x – 25000 = 5000
3x = 5000 + 25000
3x = 30000
x = 10000
∴ Monthly income of A is 3x = 3 × 10000 = ₹ 30000
Monthly income of B is 4x = 4 × 10000 = ₹ 40000

Question 3.
Five years ago, a man was seven times as old as his son, while five year hence, the man will be four times as old as his son. Find their present age.
Solution:
Let the man’s present age = x
Five years ago his age is = x – 5
Let his son’s age be = y
5 years ago his son’s age = y – 5
∴ x – 5 = 7(y – 5)
x – 5 = 7y – 35
x – 7y = -35 + 5
x – 7y = – 30 ……….. (1)
After 5 years, man’s age will be = x + 5
His son’s age will be = y + 5
∴ x + 5 = 4(y + 5)
x + 5 = 4y + 20
x – 4y = 20 – 5
⇒ x – 4y = 15 ………….. (2)

Substitute y = 15 in (1)
x – 7 (15) = -30
x – 105 – 30
x = – 30 + 105
x = 75
∴ Man’s Age = 75, His son’s Age =15

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Additional Questions

Exercise 5.1

Question 1.
State whether the following statements are true/false.
(i) (5, 7) is a point in the IV quadrant.
(ii) (-2, -7) is a point in the III quadrant.
(iii) (8, -7) lies below the x-axis.
(iv) (-2, 3) lies in the II quadrant.
(v) For any point on the x-axis its y-coordinate is zero.
Solution:
(i) False
(ii) True
(iii) True
(iv) True
(v) True

Question 2.
Locate the points
(i) (3, 5) and (5, 3)
(ii) (-2, -5) and (-5, -2) in the rectangular coordinate system.
Solution:

Question 3.
In which quadrant does the following points lie?
(i) (5, 2)
(ii) (-5, -8)
(iii) (-7, 1)
(iv) (8, -3)
Solution:
(i) I quadrant
(ii) III quadrant
(iii) II quadrant
(iv) IV quadrant.

Question 4.
Write down the ordinate of the following points.
(i) (7, 5)
(ii) (2, 9)
(iii) (-5, 8)
(iv) (7, -4)
Solution:
(i) 5
(ii) 9
(iii) 8
(iv) -4 (ordinate is the y-coordinate)

Exercise 5.2

Question 1.
Find the distance between the following pairs of points.
(i) (-4, 0) and (3, 0)
(ii) (-7, 2) and (5, 2)
Solution:
(i) The points (-4, 0) and (3, 0) lie on the x-axis. Hence,

(ii) The points (5,2) and (-7,2) lie on a line parallel to the x-axis. Hence the distance

Question 2.
Show that the three points (4, 2), (7, 5) and (9, 7) lie on a straight line.
Solution:
Let the points be A(4, 2), B(7, 5) and C(9, 7). By the distance formula.

Hence the points A, B and C are collinear.

Question 3.
Determine whether the points are vertices of a right triangle A(-3, -4), B(2, 6) and C (-6, 10).
Solution:

Hence ABC is a right angled triangle since the square of one side is equal to sum of the squares of the other two sides.

Question 4.
Show that the points (a, a), (-a, -a) and (\(-a \sqrt{3}, a \sqrt{3}\)) form an equilateral triangle.
Solution:
Let the points be represented by A (a, a), B(-a, -a) and C(\(-a \sqrt{3}, a \sqrt{3}\)) using the distance formula.

Since all the sides are equal the points form an equilateral triangle.

Question 5.
Prove that the points (-7, -3), (5, 10), (15, 8) and (3, -5) taken in order are the corners of a parallelogram.
Solution:
Let A, B, C and D represent the points (-7, -3), (5, 10), (15, 8) and (3, -5) respectively.

i.e. The opposite sides are equal. Hence ABCD is a parallelogram.

Question 6.
Show that the following points A (3, 1) B(6, 4) and C(8, 6) lies on a straight line.
Solution:
Using the distance formula, we have

Therefore the points lie on a straight line.

Question 7.
If the distance between the points (5, -2), (1, a) is 5 units. Find the value of a.

Solution:

Exercise 5.3

Question 1.
A, B and C are vertices of ∆ ABC. D, E and F are mid points of sides AB, BC and AC respectively. If the coordinates of A, D and F are (-3, 5), (5, 1) and (-5, -1) respectively. Find the coordinates of B, C and E.
Solution:

Question 2.
If A(10, 11) and B(2, 3) are the coordinates of end points of diameter of circle. Then find the centre of the circle.
Solution:

Question 3.
Find the coordinates of the point which divides the line segment joining the points (3, 1) and (5, 13) internally in the ratio 3 : 5.
Solution:

Exercise 5.4

Question 1.
Using section formula, show that the points A(7, -5), B(9, -3) and C(13, 1) are collinear.
Solution:

Question 2.
A car travels at an uniform speed. At 2pm it is at a distance of 5 km at 6 pm it is at a distance of 120 km. Using section formula, find at what distance it will reach 2 mid night.
Solution:

Question 3.
Find the coordinates of the point which divides the line segment joining the point A(3, 7) and B(-11, -2) in the ratio 5 : 1.
Solution:

Exercise 5.5

Question 1.
Find the centroid of the triangle whose vertices are (2, -5), (5, 11) and (9, 9)
Solution:

Question 2.
If the centroid of a triangle is at (10, -1) and two of its vertices are (3, 2) and (5, -11). Find the third vertex of the triangle.
Solution:

Exercise 5.6

Multiple Choice Questions :

Question 1.
The point (-2, 7) lies is the quadrant
(1) I
(2) II
(3) III
(4) IV
Hint:
(-, +) lies in II^nd quadrant
Solution:
(2) II

Question 2.
The point (x, 0) where x < 0 lies on
(1) OX
(2) OY
(3) OX’
(4) OY’
Hint:
(-, 0) lies on OX’
Solution:
(3) OX’

Question 3.
For a point A(a, b) lying in quadrant III.
(1) a > 0, b < 0
(2) a < 0, b < 0
(3) a > 0, b > 0
(4) a < 0, b > 0
Hint:
(-, -) lies in III^rd quadrant
Solution:
(2) a < 0, b < 0

Question 4.
The diagonal of a square formed by the points (1, 0) (0, 1) and (-1, 0) is
(1) 2
(2) 4
(3) \(\sqrt{2}\)
(4) 8
Hint:

Solution:
(1) 2

Question 5.
The triangle obtained by joining the points A(-5, 0) B(5, 0) and C(0, 6) is
(1) an isosceles triangle
(2) right triangle
(3) scalene triangle
(4) an equilateral triangle
Hint:
Triangles having two sides equal are called isosceles.
Solution:
(a) an isosceles triangle

Text Book Activities

Activity 1.
Plot the following points on a graph sheet by taking the scale as 1cm = 1 unit. Find how far the points are from each other? A (1, 0) and D (4, 0). Find AD and also DA. Is AD = DA? You plot another set of points and verify your result.

Solution:
AD = DA is correct.

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.11

Question 1.
Solve, using the method of substitution
(i) 2x – 3y = 7; 5x + y = 9
(ii) 1.5x + 0.1y = 6.2; 3x – 0.4y = 11.2
(iii) 10% of x + 20% of y = 24; 3x – y = 20
(iv) \(\sqrt{2} x-\sqrt{3} y=1 ; \sqrt{3} x-\sqrt{8} y=0\)
Solution:
(i) 2x – 3y = 7 ………….. (1)
5x + y = 9 ………….. (2)
Step (1)
From the equation (2)
5x+ y = 9
y = -5x + 9
Step (2)
substitute (3) in (1)
2x – 3(-5x + 9) = 7
2x + 15x – 27 = 7
17x = 7 + 27
17x = 34
x = \(\frac{34}{17}\) = 2; x = 2
Step (3)
substitute x = 2 in (3)
y = – 5(2) + 9 = -10 + 9 = -1
Solution: x = 2; y = -1

(ii) 1.5x + 0.1y = 6.2 …………. (1)
3x – 0.1y = 11.2 ………….. (2)
Multiply (1 ) x 10 15x + y = 62 ……….(1)
(2) × 10 ⇒ 30x – 4y = 112 …………. (4)
Step (1)
From equation (3)
15x + y = 62
y = -15x + 62 …………. (5)
Step (2)
substitute (5) in (4)
30x – 4 (-15x + 62) = 112
30x + 60x – 248 = 112
90x = 112 + 248
90x = 360
x = \(\frac{360}{90}\)
x = 4
Step (3)
substitute x = 4 in (5)
y = -15(4) + 62
= -60 + 62
y = 2
Solution: x = 4; y = 2

x + 2y = 240 ………. (1)
3 x – y =20 ……….. (2)
Step (1)
From equation (2)
3x – y = 20
-y = 20 – 3x
y = 3x – 20 — (3)
Step (2)
substitute (3) in (1)
x + 2(3x – 20) = 240
x + 6x – 40 = 240
7x = 240 + 40
x = \(\frac{280}{7}\)
x = 40
Step (3)
substitute x = 40 in (3)
y = 3 (40) – 20
= 120 – 20 = 100
Solution : x = 40 and y = 100

(iv) \(\sqrt{2} x-\sqrt{3} y\) = 1 ………… (1)
\(\sqrt{3} x-\sqrt{8} y\) = 0 ……….. (2)
Step (1)
From the equation (2)


Solution: x = \(\sqrt{8}\) and y = \(\sqrt{3}\)

Question 2.
Raman’s age is three times the sum of the ages of his two sons. After 5 years his age will be twice the sum of the ages of his two sons. Find the age of Raman.
Solution:
Let Raman’s age = x
Let the sum of his two sons age = y
now x = 3y ⇒ x – 3y = 0 ……… (1)
After 5 years,
Step (3)
x + 5 = 2(y + 10)
x + 5 = 2y + 20
x – 2y = 20 – 5
x – 2y = 15
Step (1)
From equation (1) x = 3y
Step (2)
Substitute x = 3y in (2)
3y – 2y = 15
y = 15
Step (3)
Substitute y = 15 in (1)
x = 3y = 3 × 15
x = 45
∴ Raman’s age is 45 years.

Question 3.
The middle digit of a number between 100 and 1000 is zero and the sum of the other digit is 13. If the digits are reversed, the number so formed exceeds the original number by 495. Find the number.
Solution:
Let the number be x0y
x + y = 13 …………….. (1)
If the digits are reversed the number so formed is y0x
x0y = 100x + 10 × 0 + 1 × y
y0x = 100y + 10 × 0 + 1 × x
100y + x – (100x + y) = 495
100y + x – 100x – y = 495
-99x + 99y = 495 ………….. (2)

Substitute x = 4 in (1)
4 + y = 13 = 13 – 4 = 9
The number is 409.

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.2

Question 1.
Find the value of the polynomial f(y) = 6y – 3y² + 3 at (i) y = 1 (ii) y = -1 (iii) y = 0
Solution:
(i) At y = 1,
f(1) = 6(1) – 3(1)² + 3 = 6 – 3 + 3 = 6
(ii) At y = -1,
f(-1) = 6(-1) – 3(-1)² + 3 = -6 – 3 + 3 = -6
(iii) At y = 0,
f(0) = 6(0) – 3(0)² + 3 = 0 – 0 + 3 = 3

Question 2.
If f(x) = x² – \(2 \sqrt{2} x\) + 1, find p (\(2 \sqrt{2}\))
Solution:
p(\(2 \sqrt{2}\)) = (\(2 \sqrt{2}\))² – \(2 \sqrt{2}\)(\(2 \sqrt{2}\)) + 1
= 4 × 2 – 4 × 2 + 1
= 8 – 8 + 1 = 1

Question 3.
Find the zeros of the polynomial in each of the following :
(i) p(x) = x – 3
(ii) p(x) = 2x + 5
(iii) q(y) = 2y – 3,
(iv) f(z) = 8z
(v) p(x) = ax where a ≠ 0,
(vi) h(x) = ax + b, a ≠ 0, a, b ∈ R
Solution:
(i) x = 3.
p( 3) = 3 – 3 = 0
∴ The zero of the polynomial is x = 3.

(iv) f(z) = 8z,
If 8z = 0
z = \(\frac{0}{8}\) = 0
f(0) = 8(0) = 0
∴ z = 0 is the zero of the given polynomial.

Question 4.
Find the roots of the polynomial equations.
(i) 5x – 6 = 0
(ii) x + 3 = 0
(iii) 10x + 9 = 0
(iv) 9x – 4 = 0
Solution:
(i) 5x – 6 = 0
5x = 6
∴ x = \(\frac{6}{5}\)

(ii) x + 3 = 0
∴ x = -3

(iii) 10x + 9 = 0
10x = -9
∴ x = \(\frac{-9}{10}\)

(iv) 9x – 4 = 0
9x = 4
∴ x = \(\frac{4}{9}\)

Question 5.
Verify whether the following are zeros of the polynomial indicated against them, or not.
(i) p(x) = 2x – 1, x = \(\frac{1}{2}\)
(ii) p(x) = x³ – 1, x = 1,
(iii) p(x) = ax + b, x = \(\frac{-b}{a}\)
(iv) p(x) = (x + 3) (x – 4), x = 4, x = -3
Solution:
(i) p(x) = 2x – 1, x = \(\frac{1}{2}\)
p(\(\frac{1}{2}\)) = 2(\(\frac{1}{2}\)) – 1 = 1 – 1 = 0
∴ x = \(\frac{1}{2}\) is the zero of the given polynomial.

(ii) p(x) = x³ – 1, x = 1
p(1) = 1³ – 1 = 1 – 1 = 0
∴ x = 1 is the zero of the given polynomial

(iii) p(x) = ax + b, x = \(\frac{-b}{a}\)
p(\(\frac{-b}{a}\)) = a(\(\frac{-b}{a}\)) + b
= -b + b = 0
∴ x = \(\frac{-b}{a}\) is the zero of the given polynomial.

(iv) p(x) = (x + 3) (x – 4), x = 4, x = -3
p(-3) = (-3 + 3) (-3 – 4) = 0(-7) = 0
p(4) = (4 + 3) (4 – 4) = 7(0) = 0
∴ x = -3, x = 4 are the zeros of the given polynomial.

Question 6.
Find the number of zeros of the following polynomial represented by their graphs.

Solution:
(i) The curve cuts the x-axis at two points. ∴ The equation has 2 zeros.
(ii) Since the curve cuts the x-axis at 3 different points. The number of zeros of the given curve is three.
(iii) Since the curve doesn’t cut the x axis. The number of zeros of the given curve is zero.
(iv) The curve cut the x-axis at one point. ∴ The equation has one zero.
(v) The curve cut the x axis at one point. ∴ The equation has one zero.