Category: 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 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 4 Geometry Ex 4.7

Multiple Choice Questions :

Question 1.
The exterior angle of a triangle is equal to the sum of two

(1) Exterior angles
(2) Interior opposite angles
(3) Alternate angles
(4) Interior angles
Hint: Exterior angle = 180°- Interior angle = sum of interior opposite angle
Solution:
(2) Interior opposite angles

Question 2.
In the quadrilateral ABCD, AB = BC and AD = DC Measure of ∠BCD is

(1) 150°
(2) 30°
(3) 105°
(4) 72°
Hint:

Solution:
(3) 105°

Question 3.
ABCD is a square, diagonals AC and BD meet at O. The number of pairs of congruent triangles are

(1) 6
(2) 8
(3) 4
(4) 12
Solution:
(1) 6

Question 4.
In the given figure CE || DB then the value of x° is

(1) 45°
(2) 30°
(3) 75°
(4) 85°
Hint: 35° + x°+ 60° = 180° ⇒ x = 85°
Solution:
(4) 85°

Question 5.
The correct statement out of the following is

(1) ∆ABC ≅ ∆DEF
(2) ∆ABC ≅ ∆DFE
(3) ∆ABC ≅ ∆FDE
(4) ∆ABC ≅ ∆FED
Hint: ∠C = ∠D; ∠B = E; ∠A = ∠F
Solution:
(4) ∆ABC = ∆FED

Question 6.
If the diagonal of a rhombus are equal, then the rhombus is a
(1) Parallelogram but not a rectangle
(2) Rectangle but not a square
(3) Square
(4) Parallelogram but not a square
Solution:
(3) Square

Question 7.
If bisectors of ∠A and ∠B of a quadrilateral ABCD meet at O, then ∠AOB is

Hint:

Solution:
(2) \(\frac{1}{2}(\angle \mathbf{C}+\angle \mathbf{D})\)

Question 8.
The interior angle made by the side in a parallelogram is 90° then the parallelogram
is a
(1) rhombus
(2) rectangle
(3) trapezium
(4) Kite
Hint:
If one angle of a parallelogram is 90°, then it is a rectangle
Solution:
(2) rectangle

Question 9.
Which of the following statement is correct?
(1) Opposite angles of a parallelogram are not equal.
(2) Adjacent angles of a parallelogram are complementary.
(3) Diagonals of a parallelogram are always equal.
(4) Both pairs of opposite side of a parallelogram are always equal.
Hint:
Opposite sides of a parallelogram are equal.
Solution:
(4) Both pairs of opposite side of a parallelogram are always equal

Question 10.
The angles of the triangle are 3x – 40, x + 20 and 2x – 10 then the value of x is
(1) 40°
(2) 35°
(3) 50°
(4) 45°
Hint: 3x – 40 + x + 20 + 2x- 10 – 180° ⇒ 6x = 210 ⇒ x = 35°
Solution:
(2) 35

Question 11.
PQ and RS are two equal chords of a circle with centre O such that ∠POQ = 70°, then ORS =
(1) 60°
(2) 70°
(3) 55°
(4) 80°
Solution:
(3) 55°
Hint:

Question 12.
A chord is at a distance of 15cm from the centre of the circle of radius 25cm. The length of the chord is
(1) 25 cm
(2) 20 cm
(3) 40 cm
(4) 18 cm
Hint:

Solution:
(3) 40 cm

Question 13.
In the figure, O is the centre of the circle and ∠ACB = 40° then ∠AOB =
(1) 80°
(2) 85°
(3) 70°
(4) 65°

Hint:

Solution:
(1) 80°

Question 14.
In a cyclic quadrilaterals ABCD, ∠A = 4x, ∠C = 2x the value of x is

(1) 30°
(2) 20°
(3) 15°
(4) 25°
Hint:

Solution:
(1) 30°

Question 15.
In the figure, O is the centre of a circle and diameter AB bisects the chord CD at a point E such that CE = ED = 8 cm and EB = 4 cm. The radius of the circle is
(1) 8 cm
(2) 4 cm
(3) 6 cm
(4) 10 cm
Hint:

Solution:
(4) 10 cm

Question 16.
In the figure, PQRS and PTVS are two cyclic quadrilaterals, if ∠QRS = 80°, then ∠TVS =
(1) 80°
(2) 100°
(3) 70°
(4) 90°

Hint:

Solution:
(1) 80°

Question 17.
If one angle of a cyclic quadrilateral is 75°, then the opposite angle is
(1) 100°
(2) 105°
(3) 85°
(4) 90°
Hint: 180° – 75° =105°
Solution:
(2) 105°

Question 18.
In the figure, ABCD is a cyclic quadrilateral in which DC produced to E and CF is drawn parallel to AB such that ∠ADC = 80° and ∠ECF = 20°, then ∠BAD = ?
(1) 100°
(2) 20°
(3) 120°
(4) 110°

Hint:

Solution:
(3) 120°

Question 19.
AD is a diameter of a circle and AB is a chord. If AD = 30 cm and AB = 24 cm then the distance of AB from the centre of the circle is
(1) 10 cm
(2) 9 cm
(3) 8 cm
(4) 6 cm
Hint:

Question 20.
In the given figure, If OP = 17cm PQ = 30 cm and OS is perpendicular to PQ, then RS is
(1) 10 cm
(2) 6 cm
(3) 7 cm
(4) 9 cm

Hint:

Solution:
(4) 9 cm

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 3.14

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 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 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 5 Coordinate Geometry Ex 5.5

Question 1.
Find the centroid of the triangle whose vertices are
(i) (2, -4 ), (-3, -7) and (7, 2)
(ii) (-5, -5), (1, -4) and (-4, -2)
Solution:


Question 2.
If the centroid of a triangle is at (4,-2) and two of its vertices are (3, -2) and (5, 2) then find the third vertex of the triangle.
Solution:
Centroid G (x, y) = (4, -2)

Question 3.
Find the length of median through Aof a triangle whose vertices are A(-1, 3), B(1, -1) and C(5, 1).
Solution:

Question 4.
The vertices of a triangle are (1, 2), (h, -3) and (-4, k). If the centroid of the triangle is at the point (5, -1) then find the value of \(\sqrt{(h+k)^{2}+(h+3 k)^{2}}\)
Solution:

Question 5.
Orthocentre and centroid of a triangle are A(-3, 5) and B(3, 3) respectively. If C is the circumcentre and AC is the diameter of this circle, then find the radius of the circle.
Solution:

Question 6.
ABC is a triangle whose vertices are A(3, 4), B(-2, -1) and C(5, 3). If G is the centroid and BDCG is a parallelogram then find the coordinates of the vertex D.
Solution:

∴ The co-ordinates of the vertex D(x, y) = (1, 0)

Question 7.
If \(\) \(\) and \(\) are mid points of the sides of a triangle, then find the centroid of the triangle.
Solution:
“The centroid of the triangle obtained by joining the mid points of the sides of a triangle is the same as the centroid of the original triangle.”
∴ The mid points of the sides of the triangle are given as

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 Ex 5.4

Question 1.
Find the coordinates of the point which divides the line segment joining the points
A (4, -3) and B (9, 7) in the ratio 3 : 2.
Solution:

Question 2.
In what ratio does the point P (2, -5) divide the line segment joining A (-3, 5) and B (4, -9).
Solution:

Question 3.
Find the coordinates of a point P on the line segment joining A(1, 2) and B (6, 7) in such a way that AP = \(\frac{2}{5}\)AB.
Solution:

Question 4.
Find the coordinates of the points of trisection of the line segment joining the points A (-5, 6) and B (4, -3).
Solution:

Question 5.
The line segment joining A(6, 3) and B(-1, -4) is doubled in length by adding half of AB to each end. Find the coordinates of the new end points.
Solution:


Similarly by Mid point of DM = A(6, 3)

Question 6.
Using section formula, show that the points A(7, -5), B(9, -3) and C(13, 1) are collinear.
Solution:
∴ AB + BC = AC, Here B is the common Point. ∴ A, B, C are collinear

Question 7.
A line segment AB is increased along its length by 25% by producing it to C on the side of B. If A and B have the coordinates(-2, -3) and (2, 1) respectively, then find the coordinates of C.
Solution:

P is at 25% distance from B on its left and C is at 25% distance from B on its right.
∴ B is the mid point of PC

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 Ex 5.1

Question 1.
Plot the following points in the coordinate system and identify the quadrants P(-7, 6), Q(7, -2), R(-6, -7), S(3, 5) and T(3, 9)
Solution:

(i) P(-7, 6) = II Quadrant
(ii) Q(7, -2) = IV Quadrant
(iii) R(-6, -7) = III Quadrant
(iv) S(3, 5) = I Quadrant
(v) T(3, 9) = I Quadrant

Question 2.
Write down the abscissa and ordinate of the following
(i) P
(ii) Q
(iii) R
(iv) S

Solution:
(i) P(-4, 4)
Abscissa is -4
Ordinate is 4.

(ii) Q(3, 3)
Abscissa is 3
Ordinate is 3

(iii) R(4, -2)
Abscissa is 4
Ordinate is -2

(iv) S(-5, -3)
Abscissa is -5
Ordinate is -3

Question 3.
Plot the following points in the coordinate plane and join them. What is your conclusion about the resulting figure?
(i) (-5, 3) (-1, 3) (0, 3) (5, 3)
(ii) (0, -4) (0, -2) (0, 4) (0, 5)
Solution:
(i)

When we join the points, we see that they lie on a line which is parallel to x-axis.

(ii)

When we join the points, we see that they lie on a straight line which is y-axis.

Question 4.
Plot the following points in the coordinate plane. Join them in order. What type of geometrical shape is formed?
(i) (0, 0) (-4, 0) (-4, -4) (0, -4)
(ii) (-3, 3) (2, 3) (-6, -1) (5, -1)
Solution:
(i) (0, 0) (-4, 0) (-4, -4) (0, -4)

The type of geometrical shape is square.

(ii) (-3, 3) (2, 3) (-6, -1) (5, -1)

The type of geometrical shape is Trapezium.