Class 10

Students often refer to SSLC Maths Textbook Solutions and Class 10 Maths Chapter 5 Second Degree Equations Important Extra Questions and Answers Kerala State Syllabus to clear their doubts.

SSLC Maths Chapter 5 Second Degree Equations Important Questions and Answers

Second Degree Equations Class 10 Extra Questions Kerala Syllabus

Second Degree Equations Class 10 Kerala Syllabus Extra Questions

Question 1.
Sum of a number and its reciprocal is \(\frac{26}{5}\). The number is
(a) 5
(b) 1
(c) 2
(d) 6
Answer:
(a) 5

Question 2.
Solutions of the equation (x – 1)² = 100 are
(a) 9, -10
(b) 10, -9
(c) 10, -10
(d) 11, -9
Answer:
(d) 11, -9

Question 3.
A small square is drawn inside a big square as shown in the figure.
The space between the squares is 2 and area of the inner square is 81

a) If the side of the outer square is x then write an equation.
Answer:
(x – 4)² = 81

b) Find the side of the outer square.
Answer:
x – 4 = 9, – 9
x – 4 = 9, x = 13
Side of the outer square is 13 cm

Question 4.
One side of a rectangle is 4 more than the other side. Area of the rectangle is 140 sq.cm
a) If the small side is x then write the equation.
Answer:
(x + 4)x = 140

b) Find the sides of the rectangle.
Answer:
x(x + 4) = 140
x² + 4x = 140, x² + 4x + 4 = 144,
(x + 2)² = 144, x + 2 = 12, x = 10
Sides are 10 cm and 14 cm

Question 5.
Two chords of a circle intersect at a point inside the circle.The intersecting point divides one chords into the segments of length 9 cm and 4 cm.The length of other chord is 15 cm
a) If one segment of the second chord of length 15 cm is x then what is the length of the other segment?
b) Form a second degree equation connecting length of the segments
c) Find the length of the segments on either side of the intersecting point on the second chord.
Answer:
a) 15 – x
b) x(15 – x) = 9 × 4, -x² + 15x = 36, x² – 15x = -36
c) Solving x = 12. Length of the segments are 12, 3

Question 6.
Sum of the area of two squares is 80.The sum of its perimetres is 48.
a) If the sides are x and y then what is x + y?
b) Form an equation representing the sum of the areas
c) Find the side of the the squares.
Answer:
a) Sides are x and y. 4x + 4y = 48, x + y = 12

b) Sides are x and 12 – x
x² +(12 – x)² = 80
x² + 144 – 24x + x² = 80
2x² – 24x = -64,
x² -12x = -32

c) x² – 12x + 36 = -32 + 36,
(x – 6)² = 4,
x – 6 = 2, x = 8 cm,
y = 12 – 8 = 4 cm

Question 7.
The perpendicular sides of a right triangle differ by 2. Area of the triangle is 24
a) Form an equation on x connecting the sides and area ?
b) Find the sides of the triangle
Answer:
a) Perpendicular sides are x and x + 2
\(\frac{1}{2}\) × x(x + 2) = 24, x² + 2x = 48, x² + 2x +1 = 49

b) (x + 1 )² = 49, x + 1 = 7, x = 6
Perpendicular sides are 6 and 8.
Hypotenuse is \(\sqrt{6^2+8^2}\) = 10

Question 8.
Manju’s age after 15 years will be the square of his age before 15 years.
a) If the present age is x then what is the age before and after 15 years.
b) Write the equation connecting the given conditions.
c) Find the present age
Answer:
a) x – 15 and x + 15

b) (x – 15)² = (x + 15), x² -30x + 225
= x + 15, x² -31x + 210 = 0

c) Solving x = 21. Present age is 21 years

Question 9.
Sum of the squares of three consecutive integers is 14. If the middle number is x then
a) What are the other numbers?
b) Form an equation
c) Find the numbers.
Answer:
a) x – 1, x + 1

b) (x – 1)² + x² + (x + 1)² = 14
3x² + 2 = 14, 3x² = 12, x² = 4, x = 2

c) Numbers are 1, 2, 3

Question 10.
A, B, C, D are the numbers on a calendar page as given below

a) If A = x write B,C and D
b) If A × C = 48 then form an equation
c) Find x
d) Find B, C and D
e) Can A + B+ C + D = 25 in any squares in the calendar? Justify your answer
Answer:
a) B = x + 1, C = x + 8, D = x + 1
b) x(x + 8) = 48, x² + 8x = 48
c) (x + 4)² = 64, x + 4 = 8, x = 4
d) Numbers are A = 4, B = 5, C = 12, D = 11
e) No. A + B + C + D always a multiple of 4.

Question 11.
Sum of a number and its square is 30. Which of the following is the number?
(a) -5
(b) 1
(c) 2
(d) -6
Answer:
(d) -6

Question 12.
x + √x = 6 in the form ax² + bx + c = 0 is
(a) x² + 12x + 6 = 0
(b) x² – 13x + 36 = 0
(c) x² – 10x – 12 = 0
(d) x² + 13x – 36 = 0
Answer:
(b) x² – 13x + 36 = 0

Question 13.
Sum of the areas of two squares is 130. Side of one square is 2 more than the side of the other square .
a) If the side of the small square is x then what is the side of the big square ?
b) Form a second degree equation using the condition.
Answer:
a) Side of the big square is x + 2
b) x² + (x + 2)² = 130
x² + x² + 4x + 4 = 130,
2x² + 4x + 4 – 130 = 0,
2x² + 4x – 126 = 0,
x² + 2x – 63 =0

Question 14.
The product of two consecutive even numbers is 360
a) If the odd number in between these numbers is x then write the numbers .
b) Form an equation using the given condition.
c) Find the numbers.
Answer:
a) Numbers are x – 1, x + 1
b) (x – 1)(x + 1) = 360, x² – 1 = 360 .
c) x² = 361, x = \(\sqrt{361}\) = 19.
Numbers 19 – 1 = 18, 19 + 1 = 20

Question 15.
Consider the arithmetic sequence 5, 9, 13, 17, 21…
a) Write the algebraic form of this sequence.
b) What is the position of the term in the sequence whose square is 625?
c) Is 36 a term of this sequence. How can you realize it ?
d) What is the position of 49 in this sequence ?
Answer:
a) x_n = dn + (f – d)
= 4n + (5 – 4)
= 4n + 1

b) (4n + 1)² = 625,
4n + 1 = \(\sqrt{625}\) = 25,
4n = 24,
n = 6

c) All terms are odd numbers. The even number 36cannot be a term of this sequence

d) 4n + 1 = 49,
4n = 48,
n = 12.
12th term is 49

Question 16.
Hypotenuse of a right angled triangle is 1 less than twice its small side.Third side is 1 more than its small side
a) If the small side is x what is the length of other two sides.
b) Form an equation connecting the length of the sides.
c) Calculate the length of the sides of the triangle.
Answer:
a) Hypotenuse = 2x – 1, Third side = x + 1

b) (2x – 1)² = x² + (x + 1)²,
4x² – 4x + 1 = x² + x² + 2x + 1
2x² – 6x = 0

c) x = 3. Sides are :
Hypotenuse, 2x – 1 = 6 – 1 = 5 cm.
Other two sides are 3cm, 4cm.

Question 17.
Length of a rectangle is 4 more than its breadth. Area of the rectangle is 357 sq.cm
a) If the breadth is x then what is its length?
b) Write an equation connecting length, breadth and area.
c) Find the lenghth and breadth of the rectangle.
Answer:
a) Length x + 4
b) x(x + 4) = 357, x² + 4x = 357
c) x² + 4x + 4 = 357 + 4 = 361,
(x + 2)² = 361,
x + 2 = \(\sqrt{361}\) = 19,
x = 19 – 2 = 17
Breadth 17cm, length 17 + 4 = 21 cm

Question 18.
Sum of the areas of two squares is 468sq.cm.The difference between the perimetres is 24cm.
a) If the small side is x then what is the length of the big side ?
b) What is the perimetre of the big square?
c) Write the length of the sides the squares in x
d) Form a second degree equation and find the length of the small square.
e) Find the length of the big square.
Answer:
a) Length of the big side be y, difference between the perimeter,
4y – 4x = 24,
4 (x – y) = 24,
x – y = 6,
y = x +6

b) 4x + 24

c) Side of the small square is x, Side of the big square is \(\frac{4 x+24}{4}\) = x + 6

d) x² +(x + 6)² = 468,
x² + x² + 12x + 36 = 468,
2x² +12x = 432
x² + 6x = 216,
x² + 6x + 9 = 225,
(x + 3)² = 225
(x + 3) = 15, x = 12
Side of the small square is 12 cm

e) Length of the big square is 12 + 6 = 18 cm

Question 19.
A two digit number is 4 times the sum of the digits. Also the number is 3 times the product of the digits.
a) Form an equation by taking x, y as the digits.
b) Make a second degree equation using the given condition.
c) Find the numbers.
Answer:
a) Digit in tens place x, digit in one’s place y Number is 10x + y
10x + y = 4(x + y)
10x + y = 3xy

b) 10x + y = 4x + 4y,
6x = 3y, y = 2x
10x + y = 3xy
⇒ 10x + 2x = 3x × 2x
12x = 6x²
6x² – 12x = 0

c) x = 0, x =2.
Tens place cannot be 0.
Tens place = 2,
one’s place 2x = 4
Number = 24

Question 20.
A number whose square is decreased by 119 equal to 10 times 8 less than that number.
a) If x is the number then write the equation
b) Find the number.
Answer:
a) x² – 119 = (x – 8) x 10
b) x² – 10x = 119 – 80, x² – 10x = 39 x² – 10x + 25 = 64
⇒ (x – 5)² = 64
x – 5 = 8 or x – 5 = -8
x = 13, x = – 3

Question 21.
A man is 5 times as old as his son. Sum of the squares of their ages is 2106
a) If x is son’s age then write the equation
b) Find the age of man and son?
Answer:
a) x² + (5x)² = 2106
x² + 25x² = 2106,
26x² = 2106, x² = 81,
x = 9

b) Son’s age 9 years , mans age 45years

Question 22.
Sum of a number and its square 9 times the next highest number.
a) If x is the number then write the equation
b) Find the number
Answer:
a) Let x be the number. x² + x = 9 (x + 1)
x² + x – 9x = 9,
x² – 8x = 9
x² – 8x + 16 = 9+ 16
⇒ (x – 4)² = 25

b) x – 4 = 5 or -5, x = 9,-1

Question 23.
Length of a rectangle is 8 more then its breadth.
a) If the breadth is x then what is its length?
b) If the area is 240 sq.cm form a second degree equation.
c) Calculate the length and breadth
Answer:
a) x + 8
b) x(x + 8) = 240
c) x² + 8x = 240,
x² + 8x + 16 = 240 + 16
(x + 4)² = 256, x + 4 = 16, x = 12
Sides are 12cm, 20cm

Question 24.
Consider the sequence of even numbers 2, 4, 6, 8
a) What is the algebraic form for the sum of first n even numbers?
b) How many terms from the beginning in the or-der makes the sum 210?
Answer:
a) This is the sequence of even numbers. Sum of the first n even numbers is n(n + 1)

b) n(n+ 1) = 210 × n² + n = 210,
n² + n + \(\frac{1}{4}\) = 210 + \(\frac{1}{4}\)
(n + \(\frac{1}{2}\))² = 210 + \(\frac{1}{4}\)
(n + \(\frac{1}{2}\))² = \(\frac{841}{4}\),
n + \(\frac{1}{2}\) = \(\frac{29}{2}\),
n = 14
Sum of first 14 terms is 210

Question 25.
The sum of the squares of two consecutive natural numbers is 313.
a) If one number is x then what is the other?
b) Form an equation using this condition.
c) Find the numbers.
Answer:
a) x + 1

b) x² +(x + 1 )² – 313
x² + x² + 2x + 1 = 313
2x² + 2x – 312 = 0
x² + x – 156 = 0

c) x = 12, -13, Numbers are 12, 13

Students often refer to Kerala Syllabus 10th Standard Maths Textbook Solutions Chapter 5 Second Degree Equations Questions and Answers Notes Pdf to clear their doubts.

SSLC Maths Chapter 5 Second Degree Equations Questions and Answers

Second Degree Equations Class 10 Questions and Answers Kerala State Syllabus

SCERT Class 10 Maths Chapter 5 Second Degree Equations Solutions

Class 10 Maths Chapter 5 Kerala Syllabus – Square Problems

(Textbook Page No.85)

Question 1.
When each side of a square was reduced by 2 metres to make a smaller square, its area became 49 square metres. What was the length of a side of the original square?
Answer:
Side of small square = \(\sqrt{49}\) = 7 m
Length of the side of original square is 9 m

Question 2.
There is a 2 metre wide path around a square ground. The area of the ground and path together is 1225 square metres. What is the area of the ground alone?
Answer:
Let x be the side
(x + 4)² = 1225
⇒ x + 4 = 35,
x = 31 m Area of the ground alone x² = 312 = 961 m²

(Textbook Page No.87)

Question 1.
1 added to the product of two consecutive even numbers gave 289. What are the numbers?
Answer:
If 1 is not added the product will be 288
Let x – 1 and x + 1 be the two consecutive even numbers.
(x – 1)(x + 1) = 288
⇒ x² – 1 = 288,
x² = 289,
x = 17 m
Numbers are 16,18

Question 2.
9 added to the product of two consecutive multiples of 6 gave 729. What are the numbers9
Answer:
x – 3 and x + 3 are two consecutive multiples of 6
If 9 is not added to the product (x – 3)(x + 3) = 720
x² – 32 = 720,
x² = 729,
x = 27
Numbers are 24 and 30

Question 3.
The first few terms of the arithmetic sequence 9, 11, 13, … were added and then 16 added to the sum, to get 256. How many terms were added?
Answer:
n^th term of the sequence is 2n + 1
2(1 + 2 + 3 + …….. + n) + 7n = 240,
2(\(\frac{n}{2}\)(n + 1)) + 7n + 16 = 256
This can be simplified as n² + 8n + 16 = 256
(n + 4)²= 162
⇒ n + 4 = 16,
n = 12
Sum of the first 12 terms and 16 is added we get 256

Class 10 Maths Kerala Syllabus Chapter 5 Solutions – Square Completion

(Textbook Page No.92)

Question 1.
An isosceles triangle like the one shown below must be made:

The height must be 2 metres less than the base and the area should be 12 square metres. What should be the lengths of the sides of the triangle?
Answer:
Base x , height x – 2
\(\frac{1}{2}\) × x(x – 2) – 12
⇒ x(x – 2) = 24
x² – 2x = 24, x² – 2x + 1 =25
(x- 1)² = 25, x – 1 = 5, x = 6,
Height = 6 – 2 = 4

Altitude , half of the base and other side of the triangle makes a right triangle.
Side of the triangle = 5 metre.
Sides are 5 m, 5 m and 6 m

Question 2.
One side of a right triangle is one centimetre shorter than twice the side perpendicular to it and the hypotenuse is one centimetre longer than twice the length of this side. What are the lengths of the sides?
Answer:
The shorter side is x . The side perpendicular to it is 2x – 1.
Hypotenuse = 2x + 1
(2x + 1)² = (2x- 1)² + x²
4x² + 4x + 1 = 4x² – 4x + 1 + x²
x² – 8x = 0
⇒ x = 8
Sides are 8 cm, 15 cm, 17 cm

Question 3.
A pole 2.6 metres long leans against a wall. The foot of the pole is 1 metre away from the wall. When the foot of the pole was pushed a little away, the top end slid down by the same length. How much was the foot moved?

Answer:
AC² + AB² = BC²
⇒ AC² + 1² = 2.62
AC = 2.62 – 1² = 5.76,
AC = 2.4

The foot and top of the pole slid the distance
x (2.4 – x)² + (1 + x)² = (2.62)²
⇒ x = 1.4m

Question 4.
The product of two consecutive odd numbers is 195. What are the numbers?
Answer:
Odd numbers are x, x + 2
x(x + 2) = 195, x² + 2x + 1 = 196
⇒ (x + 1)² = 14², x + 1 = 14, x = 13
Numbers are 13, 15

Question 5.
How many terms of the arithmetic sequence 5, 7, 9, … must be added to get 140 as the sum ?
Answer:
Algebraic form of the sequence is 2n + 3
Sum of the first n terms = (x₁ + x_n) × \(\frac{n}{2}\)
(5 + 2n + 3) × \(\frac{n}{2}\) = n² + 4n
n² + 4n = 140
⇒ n² + 4n + 4 = 144,
(n + 2)² = 144,
n + 2 = 12, n = 10
Sum of first 10 terms is 140.

SCERT Class 10 Maths Chapter 5 Solutions – Two Solutions

(Textbook Page No. 96)

Question 1.
The product of a number and 2 added to it is 168. What are the numbers?
Answer:
Let x be the number.
x(x + 2) = 168,
x² + 2x = 168
x² + 2x + 1 = 169
⇒ (x + 1)² = 169 x + 1 = 13, or – 13, x = 12 or – 14

Question 2.
Find two numbers with sum 4 and product 2
Answer:
Let the numbers be 2 – x and 2 + x. The sum is 4
(2 – x)(2 + x) = 2
⇒ 2² – x² = 2,
x² = 2, x = √2 or -√2
Numbers are 2 + √2, 2 – √2

Question 3.
Consider the arithmetic sequence 55, 45, 35, ……
(i) How many terms of this, starting from the first, must be added to get 175 as the sum?
(ii) How many terms of this, starting from the first, must be added to get 180 as the sum?
(iii) How many terms of this, starting from the first, must be added to get 0 as the sum?
Answer:
x_n = -10n + 65
(i)Number of terms added is an odd number.
\(\left(\frac{n+1}{2}\right)^{t h}\)
term is the middle term. It is,
55 + (\(\frac{n+1}{2}\) – 1)d = 55 + \(\frac{n+1}{2}\) × – 10
= 55 – 5(n – 1)
= 60 – 5n
Then, 60 – 5n = \(\frac{175}{n}\)
60n – 5n² =175
5n² – 60n = -175
n² – 12n = -35
n² – 12n + 36 = 1
(n – 6)² = 1
n – 6 = 1
n = 7
So 7 terms must be added to get sum 175.

(ii) 1th term = (-10 × 7) + 65 = -70 + 65 = -5
Sum of six terms = 175 – (-5) =180
So 6 terms must be added to get sum 180.

(iii) To get the sum 0,
Middle term must be zero,
60 – 5n = 0
5n = 60
n = 12
So 12 terms must be added to get sum 0.

Second Degree Equations Class 10 Notes Pdf

Class 10 Maths Chapter 5 Second Degree Equations Notes Kerala Syllabus

Introduction
The unit is introduced non algebraic approach of solving problems related to squaring of numbers. Three types of questions are discussed in the unit. First one is the problems which involves square directly. For example, square of a number is 1. The numbers are 1 and -1.
This can be written as x² = 1 and solutions of this equation are 1 and -1.

• Secondly we discuss the problems whose algebraic form can be written in the square for by doing some basic operations on it. For example, the sum of the area and perimeter of a square is 140 cm². If x is the side then we can write x² + 4x = 140
• We add square the half of the coefficient of x on both sides. Now we get x² + 4x + 4 = 140 + 4. This makes the equation (x + 4)² = 122
• Third part of the unit is the problems involving two solutions. While discussing the section we recall the polynomial pictures for the better understanding of the situation.

→ The algebraic approach on problem solving is essential when the simple logic and mental calculation become difficult.

→ Square problems are the beginning in the study of second degree equations.For example, area of a square is 100 cm². What is the length of the side of the square?
By simple reasoning, the side is 10 cm
If the square is zoomed by increasing its side by a small amount area of the new square becomes 121 cm2. How much the side increases?
By a simple logic we can see the increment is 1 cm
Algebraically, let it be x. (10 + x)² = 121 , 10 + x = 11, x = 1cm

→ Some problems arise in slightly different way. For example : Sum of the area and perimeter of a square is 140 cm². If x is the side then we can make an equation as x² + 4x = 140
Adding 4 on both sides, the equation becomes x² + 4x + 4 = 144 and it can be written as (x + 2)² = 144, x + 2 = 12, x = 10 cm. This process is generally known as completing square.

Generally a second degree equation will have two solutions. In certain situations both solutions are acceptable.
For example, A boy’s age after 15 years will be square of his age 15 years ago. What is his present age?
This problem can be stated algebraically as (x – 15)² = x + 15 where x is the present age. On solving this problem algebraically or otherwise we get two solutions. These are x = 10, 21.
x = 10 cannot be acceptable. So answer is 21. That is present age is 21 years
Two solutions of a second degree equation is illustrated graphically by using the graph of a second degree polynomial p(x) = x² – 4x + 3

As we discussed in 9th standard ‘graphs of polynomials’ the graph cut horizontal line at x = 1 and x = 3. The length of verical line which cut the graph from a point (x) on the horizontal line will be the value of p(x) at x.
Here p(1) and p(3) are 0. That is x² – 4x + 3 = 0 has solutions x = 1 and x = 3

→ A second degree equation is of the form, ax² + bx + c = 0, where a ≠ 0, and a, b. c are real numbers.

→ Solution of a Quadratic Equation

• The solution is a value of x that satisfies the equation.
• A quadratic equation generally has two solutions.

→ Square Completion Method

• A method to solve quadratics by expressing in the form, (x + p)² = q
• Useful in deriving the quadratic formula and solving geometrical problems.

→ Applications:

• Geometry (area and perimeter)
• Age problems
• Motion and speed
• Real-life word problems

Students often refer to Kerala Syllabus 10th Standard Maths Textbook Solutions Chapter 6 Trigonometry Questions and Answers Notes Pdf to clear their doubts.

SSLC Maths Chapter 6 Trigonometry Questions and Answers

Trigonometry Class 10 Questions and Answers Kerala State Syllabus

SCERT Class 10 Maths Chapter 6 Trigonometry Solutions

Class 10 Maths Chapter 6 Kerala Syllabus – Angles And Sides

(Textbook Page No. 102)

Question 1.
Calculate the area of parallelograms shown below

Answer:
In the first diagram triangle AMD is a 45° – 45° – 90° triangle

Draw perpendicular to AB. Since AD = 2 cm , DM = \(\frac{2}{\sqrt{2}}=\) = √2 cm
Area of the parallelogram is 4 × √2 = 4√2 cm²
In the second diagram triangle AMD is a 30° – 60° – 90° triangle
AD = 2 cm , DM = √3 cm.
Area of the parallogram = 4 × √3 = 4√3 cm²

Question 2.
Calculate the area of triangles shown below:

Answer:
Draw CM perpendicular to AB in all diagrams

AMC is a 30° – 60° – 90° triangle
CM = 1 cm
Area of the triangle is = \(\frac{1}{2}\) × 3 × 1 = \(\frac{3}{2}\) cm²

In the second diagram AMC is a 30° – 60° – 90° triangle CM = √3 cm.
Area of the triangle is= \(\frac{1}{2}\) x 3 x √3 = \(\frac{3}{2}\)√3 cm²

In the third diagram AMC is a 45° – 45° – 90°
In third triangle, CM = √2
Area of the triangle is \(\frac{1}{2}\) × 3 × √2 = \(\frac{3}{2}\) √2 cm²

Class 10 Maths Kerala Syllabus Chapter 6 Solutions – New Measures Of Angles

(Textbook Page No. 110)

Question 1.
The lengths of two sides of a triangle are 8 centimetres and 10 centimetres and the angle between them 40°
(i) What is the area of this triangle?
(ii) What is the area of the triangle with lengths of two sides the same, but the angle between them 140°?
Answer:
(i) In the figure

sin 40° = \(\frac{h}{8}\), 0.6428 = \(\frac{h}{8}\), h = 5.14 cm
Area = \(\frac{1}{2}\) × 10 × 5.14
= 25.70 cm²

(ii) In the second figure
h = 5.14 and area is 25.70 cm²

Question 2.
The length of sides of a rhombus is 5 centimetres and one of its angles is 100°. Calculate its area.
Answer:
In the figure ∠A = 180 – 100 = 80°

sin 80° = \(\frac{h}{2}\), 0.9848 = \(\frac{h}{5}\) ⇒ h = 4.9240
Area = 5 × 4.9240
= 24.62 cm²

Question 3.
The lengths of the diagonals of a parallelogram are 8 centimetres and 12 centimetres and the angle between them 50°. Calculate its area.
Answer:

\(\frac{AM}{6}\) = sin 50° ⇒ AM = 6 × 0.76 = 4.56
Area of triangle ABD = \(\frac{1}{2}\) × 8 × 4.56 = 18.24 cm²
Area of parallelogram = 2 × 18.24 = 36.48 cm²

Question 4.
A triangle is to be drawn with one side 8 centimetres long and one of the angles on it 40°. What is the minimum length of the side opposite this angle?
Answer:
We know that shortest distance from a point to a line is the perpendicular distance
BC is the perpendicular distance. Triangle ACB is a right triangle

See the diagram

\(\frac{BC}{8}\) = sin 40 = 0.6427
⇒ BC = 8 × 0.6427 = 5.14cm
Minimum length of the side opposite to 40° angle is 5.14 cm

Question 5.
The length of the sides of a rhombus is 5 centimetres and one of its angles is 70°. Calculate the length of its diagonals.
Answer:
See the diagram

sin 35° = \(\frac{b}{5}\) ⇒ b = 5 × 0.5735 = 2.86 cm
cos 35° = \(\frac{a}{5}\) ⇒ a = 5 × 0.8191 = 4.09 cm
Length of diagonals = 2 × 2.86 = 5.72 cm ,
= 2 × 4.09 = 8.18 cm

SCERT Class 10 Maths Chapter 6 Solutions – Triangles And Circles

(Textbook Page No. 114)

Question 1.
The pictures below show two triangles and their circumcircles:

Calculate the radius of each circle.
Answer:
Draw a rough diagram and mark the measurements.

Draw the diameter BP and join PC Triangle
BPC is a right triangle. ∠P = 30°
By the property of 30° – 60° – 90° triangle, Diameter 4.
Radius 2 cm

Draw a diameter AP, join PC
APCB is a cyclic quadrilateral.
∠P= 180 – 135 = 45°
∆ACP is a 45°, 45°, 90°
∆AC = PC = 3cm
AP = 3√2
Radius of the circle \(\frac{3}{2}\)√2cm

Question 2.
A circle is to be drawn, passing through the ends of a line 5 centimetres long; and the angle in one of the segments made by the line should be 80°. What should be the radius of the circle?
Answer:
See the diagram

sin 80° = \(\frac{5}{2r}\), 0.9848 = \(\frac{5}{2r}\)
r = \(\frac{5}{2 \times 0.9848}\) = 2.53 cm

Question 3.
The picture shows part of a circle:

What is the radius of the circle?
Answer:
See the digram

In the diagram the circle is completed by drawing the alternate arc.
Angle made by the chord in the alternate arc is 180 – 140 = 40°
Now we can complete a right triangle.
sin 40° = \(\frac{8}{2r}\) where r is the radius of the circle.
0.64 = \(\frac{8}{2r}\) ⇒ r = 6.25 cm
Note that, if a chord makes an angle c at the centre then length of chord will be 2r sin (\(\frac{c}{2}\))

SSLC Maths Chapter 6 Questions and Answers – Ratio Of Sides

(Textbook Page No. 118)

Question 1.
Two triangles and their circumcircles are shown in the figure.

Use the sine table to calculate the diameters of the circles and the other two sides of the triangles correct to a millimetre
Answer:

The length of chord d which makes a° angle on the arc. It can be found that d = 2r sin a°
4 = 2r sin 85°
⇒ r = \(\frac{2}{\sin 85^{\circ}}\) ≈ 2.01cm sin 85°
In the second diagram, angle is 80°
r = \(\frac{2}{0.98}\) = 2.04 cm

Kerala Syllabus Class 10 Maths Chapter 6 Solutions – Another Measure

(Textbook Page No. 121,122)

Question 1.
One angle of a rhombus is 50° and the shorter diagonal is 6 centimetres. What is its area?
Answer:
In the diagram

Diagonals of a rhombus bisect perpendicularly. In triangle APB, ∠BAP = 25°.
tan 25° = \(\frac{3}{AP}\)
AP = 6.433 cm
Area of triangle ABD = \(\frac{1}{2}\) × 6 × 6.43
= 19.29 cm²
Area of rhombus is half the product of diagonals. You can follow this method to area.

Question 2.
A ladder leans against a wall with its foot 2 metres away from the wall. The angle between the ladder and the ground is 40°. How high is the top of the ladder from the ground?
Answer:

tan 40° = \(\frac{h}{2}\)
⇒ h = 2 × tan 40°
= 2 × 0.83
= 1.66 m

Question 3.
Three rectangles are to be cut along the diagonals to make triangles which are then rearranged to form a regular pentagon as shown below

The sides of the pentagon must be 30 centimetres. What should be the lengths of the sides of the rectangles?
Answer:

AQ, CQ are the sides of large rectangle. BP, AP are the sides of small rectangles.
sin 54° = \(\frac{AP}{30}\) ⇒ AP = 24.27, AC = 48.54 cm
cos54° = \(\frac{BP}{30}\) ⇒ BP = cos 54 × 30 = 17.63 cm
sin 18° = \(\frac{CQ}{48.54}\) ⇒ CO = 14.99 cm
cos 18° = \(\frac{AQ}{48.54}\) ⇒ AQ = 46.16cm

Question 4.
The perpendiculars in the picture below are drawn 1 centimetre apart:

Prove that their heights are in an arithmetic sequence. What is the common difference?
Answer:
In the first right triangle, let x be the base and h h
the height. \(\frac{h}{x}\) = tan 40° ⇒ h = x tan 40°
For the second right triangle height = (x + 1) tan 40°
For the third right triangle height = (x + 2) tan 40°
xtan 40°, x tan 40° + tan 40°,
x tan 40° + 2 tan 40° ………………..
is the resulting sequence. It is an arithmetic sequence.
Common difference is tan 40°

Question 5.
Calculate the area of a regular pentagon of sides 10 centimetres.
Answer:
\(\frac{AP}{10}\) = sin 54°
⇒ AP = 8.09, AC = 16.18 cm

\(\frac{BP}{10}\) = cos 54°
⇒ BP = 10 × cos 54° = 5.8 cm

Area of triangle ABC is
\(\frac{1}{2}\) × 16.18 × 5.8 = 46.922 cm²

Area of triangle AED is 46.922 cm²
cos 18° = \(\frac{AQ}{AC}\) ⇒ AQ = 15.388 cm

Area of triangle ACD is
\(\frac{1}{2}\) × 10 × 15.38 = 76.94 cm²
Area of pentagon = 46.92 + 46.92 + 76.94
= 170.78 cm²

Class 10 Maths Chapter 6 Kerala Syllabus – Heights And Distance

(Textbook Page No. 125)

Question 1.
When the sun is seen at an angle of elevation of 40°, the length of a tree’s shadow is 18 metres
(i) What is the height of the tree?
(ii) What would be the length of the shadow, when the sun is at an elevation of 80°
Answer:
Diagram

(i) tan 40° = \(\frac{h}{18}\) ⇒ h = 15.10 m
(ii) tan 80°= \(\frac{15.10}{x}\) ⇒ x = 2.66 m

Question 2.
From top of a building, a person sees the foot of a shop 30 metres away at an angle of depression 25°. What is the height of the building?
Answer:
Diagram

tan 25° = \(\frac{h}{30}\)
⇒ h = 30 × tan 25
⇒ h = 13.98 m

Question 3.
From the top of an electric post, two wires are stretched to either side and attached to the ground, making angles 55° and 45° with the ground. The distance between the feet of the wires is 25 metres. What is the height of the post?
Answer:
Diagram

tan 55° = \(\frac{x}{25-x}\)
⇒ 1.42 = \(\frac{x}{25 – x}\)
⇒ 35.5 = 2.4x
⇒ x = \(\frac{35.5}{2.4}\)
⇒ x = 14.79 m

Question 4.
A person, 1.75 metres tall, standing at the foot of a tower sees the top of a hill 40 metres away at an elevation of 60°. From the top of the tower, he sees it at an elevation of 50°. Calculate the heights of the tower and the hill.
Answer:

\(\frac{DE}{CE}\) = tan 50° ⇒ DF = 40 × tan 50° = 47.67 m
\(\frac{DF}{BE}\) = tan 60° ⇒ DF = 40 × tan 60° = 69.28 m
EF = 69.28 – 47.67
= 21.61m
Height of the tower AC = 21.61 + 1.75
= 23.36 m
Height of the hill DG = 47.67 + 23.36
= 71.03 m

Question 5.
A boy, 15 metres tall, standing at the edge of a canal sees the top of a tree on the other edge at an angle of elevation of 80°. Stepping 15 metres back, he saw it at an angle of elevation of 40°. How wide is the canal and how tall is the tree?
Answer:
See the diagram

\(\frac{y}{x}\) = tan 80°, \(\frac{y}{x+15}\) = tan40°
y = x tan80°, Therefore \(\frac{x \tan 80}{x+15}\) = tan 40
x = \(\frac{15 \tan 40^{\circ}}{\tan 80^{\circ}-\tan 40^{\circ}}=\frac{12.58}{4.8}\) = 2.62 m

Width of the canal = 2.62
Height of the tree y + 1.5 = x tan 80 +1.5
= 16.35m

Trigonometry Class 10 Notes Pdf

Class 10 Maths Chapter 6 Trigonometry Notes Kerala Syllabus

Introduction
In our high school class we introduce trigonometry as an extension of the similarity of triangles. Whenever a triangle is scaled its angles and ratio of sides remain unchanged. That means angle can be measured using ratio of sides. The measurement of angles using ratio of sides are called trigonometric measure of angles.

The unit starts with properties of two special triangles. First one is 45°- 45°- 90° triangles. Such a triangle can be obtained by drawing diagonal to a square. Second special triangle is 30° – 60° – 90° triangle. The altitude of a equilateral triangle divides the triangle into two such triangles.

The relation between sides and angles of these triangles give an insight into the trigonometric measures of the angle. Sine measure, Cosine measure and Tangent measure are three basic trigonometric measures of an angle.

Problems related to circles and triangles are solved in the unit with the help of trigonometry. Calculation of heights and distances are another area of discussion.
A trigonometric table is given at the end of the unit. This is helpful for solving problems which contain angles between 1° and 89°.

→ The diagonal divides a square into two right triangles.These triangles are equal and have angles 45°, 45°, 90°

→ On scaling a triangle angles remains same. The sides opposite to equal angles change proportion-ally.
If the sides opposite to 45° angles is a then the side opposite to 90° angle will be a √2a

→ Altitude of an equilateral triangle divides the triangle into two right triangles. These are equal triangles
Each right triangle have angles 30°, 60° and 90°
If the side of the equilateral triangle is 2, clearly it will be the side opposite to 90° angle of the right triangle
The side opposite to 30° angle is 1 , the half of the side opposite to 90° angle.
Side opposite to 60° angle is √3.

→ If the side opposite to 90° angle is ‘a ’ then the side opposite to 30° will be \(\frac{a}{2}\), side opposite to 60° is \(\frac{a}{2}\)√3

→ The sides of a right triangle change proportionally on scaling the triangle. For an acute angle a° of the right triangle the ratio of opposite side to the hypotenuse is the sine measure of that angle.
The ratio of adjacent side to the hypotenuse is the cosine measure of that angle .

→ In the figure

sin a° = \(\frac{p}{d}\), cos a° = \(\frac{q}{d}\)

→ Note that, in this proportion sin a° is the proporionality constant in the change of height of the perpendicular with respect to the distance of the point from the vertex.
Also, cos a° is the proportionality constant in the change of distance from the vertex of the angle to the foot of perpendicular with respect to the distance of the point from the vertex.

→ Note that, sin 30° = cos 60° = \(\frac{1}{2}\)
sin 60° = cos 30° = \(\frac{\sqrt{3}}{2}\)
sin 45° = cos 45° = \(\frac{1}{\sqrt{2}}\)

→ In a triangle, p and q he the two sides and a0 is the angle between these sides then area of the triangle is \(\frac{1}{2}\)pq sin a°

→ The length of an arc of a circle can be calculated using the central angle of the arc.
Apart from degree measure Radian is another measure of angle. This related to central angle of the arc of a circle and its radius

If s is the arc length and r is the radius then \(\frac{s}{r}\) will be the central angle of the arc in radian measure

→ The length of any chord of a circle is twice the product of the radius and the sine of half the central angle

→ Consider a point on one side of an angle of a0,and the perpendicular from this point to the other side; if the height of the perpendicular is p and the dis-tance of the foot of the perpendicular from the vertex is q, then.
tan a° = \(\frac{p}{q}\)

→ Trigonometric measures of angles ranging from 1° to 89° is given in the table.
When angle increases from 0 to 90° sin measure increases from 0 to 1 and cos measure decreases from 1 to 0
If the sum of two angles is 90° then sine of one angle is equal to cosine of other angle.
sin 1° = cos 89°, sin 10° = cos 80°, sin 45° = cos 45°

→ In any triangle with angles 45°, 45°, 90° both the shorter skies have the same length and the length of the longest side is √2 times this length

→ In any triangle with angles 30°, 60°, 90° the length of the longest side is 2 times the length of the shortest side; and the length of the side of medium size is √3 times the length of the shortest

→ In an isosceles triangle the perpendicular from the top vertex to the bottom side bisects the top angle.

→ Consider a point on one side of an angle of o°, at distance d from the vertex, and the perpendicular from this point to the other side; if the height of the perpendicular is p and the the distance of the foot of the perpendicular from the vertex is q, then.

sin a° = \(\frac{p}{d}\)
cos a° = \(\frac{q}{d}\)

→ The length of any chord of a circle is twice the product of the radius and the sine of half the central angle

→ The central angle of each chord is twice the angle opposite to it in the triangle.

→ Consider a point on one side of an angle of a°, and the perpendicular from this point to the other side; if the height of the perpendicular is p and the distance of the foot of the perpendicular from the vertex is q, then.

tan a° = \(\frac{p}{q}\)

→ Line of sight: It’s the imaginary line drawn from the eye of an observer to the object being viewed.

→ Angle of Elevation: The angle formed upwards between the horizontal line and the line of sight, when the observer looks up at an object.

→ Angle of Depression: The angle formed downwards between the horizontal line and the line of sight, when the observer looks down at an object.

Students often refer to SSLC Maths Textbook Solutions and Class 10 Maths Chapter 4 Mathematics of Chance Important Extra Questions and Answers Kerala State Syllabus to clear their doubts.

SSLC Maths Chapter 4 Mathematics of Chance Important Questions and Answers

Mathematics of Chance Class 10 Extra Questions Kerala Syllabus

Mathematics of Chance Class 10 Kerala Syllabus Extra Questions

Question 1.
Each letter of the word MALAYALAM are written in small paper pieces and kept in a box. One is taken from the box without looking. The probability of getting A is
(a) \(\frac{2}{9}\)
(b) \(\frac{3}{9}\)
(c) \(\frac{4}{9}\)
(d) \(\frac{1}{9}\)
Answer:
(c) \(\frac{4}{9}\)

Question 2.
The factors of 24 are written in small paper pieces separately and kept in a box. One is taken from the box without looking. The probability of getting even
(a) \(\frac{1}{8}\)
(b) \(\frac{7}{8}\)
(c) \(\frac{6}{8}\)
(d) \(\frac{3}{8}\)
Answer:
(c) \(\frac{6}{8}\)

Question 3.
First 10 terms of the arithmetic sequence 3n + 1 are written in small paper pieces separately and kept in a box.One is taken from the box without looking. What is the probability of getting odd number.
(a) \(\frac{1}{2}\)
(b) \(\frac{3}{5}\)
(c) \(\frac{5}{11}\)
(d) \(\frac{3}{8}\)
Answer:
(a) \(\frac{1}{2}\)

Question 4.
Radius of the big circle is equal to the diameter of small circle.

A fine dot is placed into the figure without looking The probability of falling the dot in the unshaded part?
Answer:
\(\frac{3}{4}\)

Question 5.
The numbers 2,3,4 are written in small paper pieces kept in one box. Fractions \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}\) are written in paper pieces and kept in another box. One is taken from each box and find the product
a) What are the outcome pairs
Answer:

Products are 1, \(\frac{2}{3}, \frac{1}{2}, \frac{3}{2}\), 1\(\frac{3}{4}\), 2, \(\frac{4}{3}\), 1

b) What is the probability of getting an integer as the product?
Answer:
\(\frac{4}{9}\)

Question 6.
Numbers 1, 2, 3,…25 are written in small paper pieces and put in a box.One is taken from it without looking into the box.
a) What is the probability of getting an even number?
Answer:
\(\frac{12}{25}\)

b) What is the probability of getting an odd number?
Answer:
\(\frac{13}{25}\)

c) What is the probability of getting a prime number?
Answer:
\(\frac{9}{25}\)

Question 7.
A die in which the numbers 1 to 6 are written on the faces is thrown
a) What is the probability of falling an even numbered face?
Answer:
Probability of falling even face = \(\frac{3}{6}=\frac{1}{2}\)

b) What is the probability of getting an odd numbered face?
Answer:
Probability of falling odd face = \(\frac{3}{6}=\frac{1}{2}\)

c) What is the probability of getting a prime numbered face?
Answer:
Probability of falling prime numbered face = \(\frac{3}{6}=\frac{1}{2}\)

Question 8.
Two digit numbers are written in small paper pieces and placed in a box. One is taken from the box at random
a) How many multiples of 5 are there in the box?
Answer:
10, 11, 12, ……….. 99 are the two digit numbers.
Number of two digit numbers is 90
Multiples of five are 10,15, 20…95
Number of numbers = 18

b) What is the probability of getting a multiple of 5?
Answer:
Probability of getting a multiple of five = \(\frac{18}{90}\)

c) What is the probability of not getting a multiple of 5 ?
Answer:
Probability of not getting a multiple of 5
= 1 – \(\frac{18}{90}=\frac{72}{90}=\frac{8}{10}\)

Question 9.
Two dice numbered 1 to 6 are thrown at together.
a) Write the outcomes as pairs
Answer:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1) , (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1) , (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1) , (5, 2), (5, 3), (5,4), (5, 5), (5, 6)
(6, 1) , (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

b) What is the probability of the occurence of equal numbers?
Answer:
\(\frac{6}{36}=\frac{1}{6}\)

c) What is the probability of the occurence of perfect squares ?
Answer:
(1, 1), (1, 4), (4, 1), (4, 4).
Probability \(\frac{4}{36}\)

d) What is the probability of the occurence of multiple of 2 in one die and multiple of 3 in other die ?
Answer:
(2, 3), (4, 3), (6, 3), (2, 6), (4, 6), (6, 6), (3, 2), (3, 4), (3, 6), (6, 2), (6, 4)
Probability \(\frac{11}{36}\)

Question 10.
What is the probability of getting 5 Sundays in the month December?
Answer:
There are 31 days in December. 28 days decide 4weeks, so four Mondays.
The combinations are (Sunday, Monday, Tuesday), (Monday, Tuesday, Wednesday), (Tuesday, Wednesday, Thursday), (Wednesday, Thusday, Friday), (Thusday, Friday, Saturday), (Friday, Saturday, Sunday), (Saturday, Sunday, Monday).
There are three combinations in which Sundays occur.Probability of occuring five Sundays is \(\frac{3}{7}\)

Question 11.
The numbers 21, 22, 23… 250 are written in small paper pieces and placed in a box.
a) Write the sequence of numbers comes in the right end of these numbers ?
Answer:
2, 4, 8, 6, 2,4, 8, 6…

b) If one is taken from the box at random, then what is the probability of getting a number with 4 in ones place?
Answer:
Up to 248, the set of digits 2,4,8,6 repeats 12 times. One’s place of 249 is 2 and one’s place of 250 is 4
Probability of getting 4 in one’s place is = \(\frac{13}{50}\)

c) What is the probability of getting a number with 8 in ones place?
Answer:
Probability of getting 8 in one’s place is = \(\frac{12}{50}\)

d) What is the probability of getting a number with 2 in ones place?
Answer:
Probability of getting 2 in one’s place is = \(\frac{13}{50}\)

e) What is the probability of not getting a number with 2 in ones place?
Answer:
Probability of not getting 2 in one’s place is = 1 – \(\frac{13}{50}\)
= \(\frac{37}{50}\)

Question 12.
Two digit numbers are written in small paper pieces and placed in a box.
a) How many paper slips are there in the box?
b) If one is taken from the box, what is probability of getting a number with digits same?
c) If one is taken from the box, what is probability of getting a number in which the product of the digits a prime number.
d) What is the probability of getting a prime number?
Answer:
a) 10, 11, 12…99 are the numbers. There are 90 numbers

b) Numbers with same digits are 11, 22, 33, 44, 55, 66, 77, 88, 99
Total number of these numbers is 9 Probability = \(\frac{9}{90}\) = \(\frac{1}{10}\)

b) In the two digit numbers with product of the digits a prime, one digit is 1 and other digit is one of the numbers 2, 3, 5, 7

c) Numbers are 12, 13, 15, 17, 21,31,51,71.
Probability = \(\frac{8}{90}=\frac{4}{45}\)

d) There are 25 prime numbers below 100.4 of them are one digit primes and the rest of the 21 numbers are two digit primes. Probability is
= \(\frac{21}{90}=\frac{7}{30}\)

Question 13.
There are two circles in the picture.One is inside other.Radius of the small circle is half of the radius of the big circle.

a) If the radius of the small circle is r then what is the area of the small circle and big circle ?
b) If a fine dot is placed into the figure, what is the probability of falling the dot in the small circle?
c) What is the probability of falling the dot the
yellow shaded part in the figure.
Answer:
a) Area of small circle πr²
Area of big circle π × (2r)² = 4πr²
b) Chance of getting black \(\frac{\pi r^2}{4 \pi r^2}=\frac{1}{4}\)
c) Probability of falling the dot in the yellow shade is 1 – \(\frac{1}{4}=\frac{3}{4}\)

Question 15.
Triangle PQR is drawn by joining the midpoints of the sides of triangle ABC.

a) How many equal triangles are there in the figure?
Answer:
4 triangles.
∆PQR, ∆APQ, ∆PCR, ∆QRB are equal triangles

b) A fine dot is placed into the figure. What is the probability of falling the dot in triangle PQR?
Answer:
\(\frac{1}{4}\)

c) How many parallelograms are there in the picture?
Answer:
3 Parallelograms
PQRC, PQBR, PRQA are equal

d) A fine dot is placed into the figure. What is the probability of falling the dot in the parallelogram PQRC ?
Answer:
To fall the dot in PQRC it is necessary to fall either in triangle PCR or triangle PQR
Probability is \(\frac{2}{4}=\frac{1}{2}\)

Question 15.
ACP is drawn in the square ABCD and shaded. P is the mid point of the side of the square

a) If the side of the square is ‘a’ then what is the altitude to the side PC of the shaded triangle.
Answer:
AB = a

b) If the side of the square is V then what is the area of the shaded triangle?
Answer:
Height of triangle APC is = \(\frac{a}{2}\)
Height = a
Area = \(\frac{1}{2} \times \frac{a}{2}\) × a = \(\frac{\mathrm{a}^2}{4}\)

c) If a fine dot is placed into the figure then what is the probability of falling the dot in the shaded triangle?
Answer:
Probability = \(\frac{\mathrm{a}^2}{4}\) ÷ a = \(\frac{1}{4}\)

Question 16.
There are two squares in the figure.The perimetre of the outer square is 28 cm , the perimetre of the inner square is 20 cm

a) What is the area of the outer square?
Answer:
Side of outer square = \(\frac{28}{4}\) = 7 cm
Area = 7² = 49 sq.cm

b) What is the area of inner square?
Answer:
Side of inner square \(\frac{20}{4}\) = 5 cm
Area = 5² = 25 sq.cm

c) What is the area of the shaded triangle ?
Answer:
Area of the region in between the squares
= 49 – 25 = 24 sq.cm
Area of shaded part = \(\frac{24}{4}\) = 6 sq.cm

d) If a fine dot is placed into the figure then what is the probability of falling the dot in the shaded triangle?
Answer:
Probability = \(\frac{6}{49}\)

Question 17.
The mid points of the two sides and one vertex of a square are joined in such a way as to get a triangle which is coloured in the picture.

a) If the side of the square is a, what is area of the unshaded triangles ?
Answer:
Unshaded part = (\(\frac{1}{2}\) × a × \(\frac{a}{2}\)) × 2 + \(\frac{1}{2} \times \frac{a}{2} \times \frac{a}{2}\)
= \(\frac{a^2}{2}+\frac{a^2}{8}=\frac{5 a^2}{8}\)

b) What is the area of the shaded triangle?
Answer:
Area of shaded part = a – \(\frac{5 a^2}{8}=\frac{3 a^2}{8}\)

c) If a fine dot is placed into the figure then what is the probability of falling the dot in the coloured traingle?
Answer:
Probability = \(\frac{3 a^2}{8}\) ÷ a² = \(\frac{3}{8}\)

Question 18.
Manju has three ornaments :Green, Red and Blue ear rings and chains.She ware it in different ways.
a) How many ways she can ware the ornaments?
b) What is the probability of waring ornaments of same colour?
c) What is the probability of wearing the orna-ments of different colours?
Answer:
a) Number of pairs 3 × 3 = 9, (Green, Green), (Green , Red), (Green, Blus) (Blue, Green), (Blue, Red), (Blue, Blue) (Red, Green), (Red, Red), (Red, Blue)
b) (Green, Green), (Red, Red), (Blue, Blue) Probability = \(\frac{3}{9}=\frac{1}{3}\)
c) Probability of wearing different colours is 1 – \(\frac{1}{3}=\frac{2}{3}\)

Question 19.
A box contains 4 black balls and 3 white balls. Another box contains 5 black balls and 3 white balls. One from each box is taken at random.
a) How many pair of balls are possible ?
b) What is the probability of getting both balls black?
c) What is the probability of getting both balls white?
d) What is the probability of getting balls of dif-ferent colours?
Answer:
a) Total number of possible selections = (3 + 4) × (5 + 3) = 7 × 8 = 56

b) Probability of getting both black
\(\frac{4 \times 5}{56}=\frac{20}{56}=\frac{5}{14}\)

c) Probability of getting both white \(\frac{3 \times 3}{56}=\frac{9}{56}\)

d) Probability of getting balls of different colours \(\frac{(4 \times 3)+(3 \times 5)}{56}=\frac{27}{56}\)

Question 20.
A box contains four paper slips carrying numbers 1,2,3,4. Another box contains paper slips carrying numbers 1, 2, 3. One from each box is taken at random and entered as pairs.
a) How many pairs are possible ?
b) What is the probability of getting a pair with the product of the digits odd?
c) What is the probability of getting a pair with the product of the digits even?
Answer:
a) Number of pairs 4 × 3 = 12
(1, 1) , (1, 2), (1, 3)
(2, 1), (2, 2), (2, 3)
(3, 1), (3, 2), (3, 3)
(4, 1), (4, 2), (4, 3)

b) Pairs of getting a product odd are
(1, 1), (1, 3), (3, 1), (3, 3)
Probability \(\frac{4}{12}=\frac{1}{3}\)

c) Probability of getting product even = 1 – \(\frac{1}{3}=\frac{2}{3}\)

Question 21.
A bag contains 4 black beads and 3 white beads.Another bag contains 4 black beads and 5 white beads. One is taken from each bag
a) What is the probability of getting both white ?
b) What is the probability of getting both black ?
c) What is the probability of getting one white and one black?
d) What is the probability of getting atleast one black?
Answer:
a) First bag contains 7 beads and second bag con-tains 9 beads. When one from each box are taken we get 7×9 = 63 pairs.
Number of pairs with both white is, 3 × 5 = 15
Probability of getting both white is \(\frac{15}{63}=\frac{5}{21}\)

b) Number of pairs with both black is, 4 × 4 = 16
Probability of getting both black is \(\frac{16}{63}\)

c) Probability of getting one white and one black is \(\frac{4 \times 5+3 \times 4}{63}=\frac{32}{63}\)

d) Probability of getting atlest one black is
\(\frac{4 \times 5+3 \times 4+4 \times 4}{63}=\frac{48}{63}=\frac{16}{21}\)

Question 22.
A box contains four paper strips on which the numbers 1,2,3,4 are written. Another box contains the strips 1, 2, 3. One is taken from each box
a) What are the possible outcomes
Answer:
(1, 1), (1, 2), (1, 3)
(2, 1), (2, 2), (2, 3)
(3, 1), (3, 2), (3, 3)
(4, 1), (4, 2), (4, 3)

b) What is the probability of getting the sum a multiple of 3 ?
Answer:
Favourable outcomes are
(1, 2) , (2, 1), (3, 3), (4, 2)
Probability is = \(\frac{4}{12}=\frac{1}{3}\)

c) What is the probability of getting the sum a multiple of 2 ?
Answer:
Favourable outcomes are
(1, 1), (3, 1), (1, 3), (2, 2), (3, 3), (4, 2)
Probability is \(\frac{6}{12}=\frac{1}{2}\)

d) What is the probability of getting the product a multiple of 6
Answer:
Favourable outcomes are (2,3), (3,2), (4,3)
Probability is \(\frac{3}{12}=\frac{1}{4}\)

Students often refer to SSLC Maths Textbook Solutions and Class 10 Maths Chapter 3 Arithmetic Sequences and Algebra Important Extra Questions and Answers Kerala State Syllabus to clear their doubts.

SSLC Maths Chapter 3 Arithmetic Sequences and Algebra Important Questions and Answers

Arithmetic Sequences and Algebra Class 10 Extra Questions Kerala Syllabus

Arithmetic Sequences and Algebra Class 10 Kerala Syllabus Extra Questions

Question 1.
Algebraic form of an arithmetic sequence is x_n = 3n +1. Common difference is
(a) 3
(b) 4
(c) -3
(d) 5
Answer:
(a) 3

Question 2.
Sum of the first n terms of an arithmetic sequence n² + n. What is its common difference .
(a) 3
(b) 4
(c) 2
(d) 5
Answer:
(c) 2

Question 3.
Sum of the first n terms of an arithmetic sequence is n² + n. What is its 13th term?
(a) 31
(b) 26
(c) 20
(d) 50
Answer:
(b) 26

Question 4.
a, a – 1, a – 2 ……. is an arithmetic sequence. What is its n^th term?
(a) a + n + 1
(b) a + n – 1
(c) a – n – 1
(d) a – n + 1
Answer:
(d) a – n + 1

Question 5.
The algebraic form of the sequence \(\frac{1}{7}, \frac{3}{7}, \frac{5}{7}\) ……….?
(a) \(\frac{n}{7}\)
(b) \(\frac{2 n+1}{7}\)
(c) \(\frac{2 n-1}{7}\)
(d) None of these
Answer:
(c) \(\frac{2 n-1}{7}\)

Question 6.
Algebraic form of an arithmetic sequence is
(a) What is its common difference ?
Answer:
\(\frac{1}{3}\)

(b) At what position 15 becomes a term of the sequence?
Answer:
44^th term is 15

Question 7.
Sum of the first n terms of an arithmetic sequence is 3n² + 2n
(a) What is its common difference?
(b) Find the sum of first 20 terms
Answer:
(a) 6
(b) 3 × 20² + 2 × 20 = 1240

Question 8.
\(\frac{1}{11}, \frac{2}{11}, \frac{3}{11} \ldots\) is an arithmetic sequence.
(a) Write the algebraic form of this sequence?
(b) Find the sum of first 10 terms of this sequence?
Answer:
(a) \(\frac{n}{11}\)
(b) \(\frac{1+2+3+\cdots 10}{11}=\frac{55}{11}\) = 5

Question 9.
n^th term of an arithmetic sequence is 1 – 4n
(a) What is the common difference ?
(b) Find the sum of first 25 terms of this sequence.
Answer:
(a) – 4
(b) x₂₅ = 1 – 4 x 25 = -99, x₁ = 1 – 4 × 1 = -3
Sum =(x₁ + x₂₅) × \(\frac{25}{2}\)
= (-3 + -99) × \(\frac{25}{2}\) = -1275

Question 10.
Sum of the first«terms of an arithmetic sequence is 5«2 + 3n .
(a) What is the common difference ?
(b) Write the algebraic form of this sequence.
Answer:
(a) d = 10
(b) f = 8, d = 10
⇒ x_n = 10n – 2

Question 11.
Consider the arithmetic sequence 1, 4, 7, 10….
(a) Write the n^th term of the sequence?
(b) Find the expresson for the sum of first terms of this sequence.
(c) Calculate the sum of first 20 terms of this sequence.
Answer:
(a) x_n = 3n – 2
(b) Sum = (x₁ + x_n) × \(\frac{n}{2}\) = (1 + 3n – 2) × \(\frac{n}{2}\)
= \(\frac{3 n^2}{2}-\frac{n}{2}\)
(c) \(\frac{3 \times 20^2}{2}-\frac{20}{2}\) = 590

Question 12.
Consider the arithmetic sequence 6, 10, 14,…
(a) What is the common difference of this sequence?
(b) Find the sum of first n terms of this sequence
(c) Can sum of some terms of this sequence 1225? How do you know this?
Answer:
(a) d = 4

(b) x_n = 4n + 2
Sum =(6 + 4n + 2) × \(\frac{n}{2}\) = 2n² +4 n

(c) All terms are even numbers. Sum of even numbers cannot be an odd number. 1225 cannot be the sum

Question 13.
First term of an arithmetic sequence is \(\frac{1}{2}\) and common difference \(\frac{3}{4}\)
(a) What is the algebra of this sequence?
(b) At what position 50 becomes a term of the se-quence
(c) What is the sum of first 11 terms?
Answer:
(a) x_n = dn + (f – d)
= \(\frac{3}{4}\)n + \(\left(\frac{1}{2}-\frac{3}{4}\right)=\frac{3 n-1}{4}\)

(b) \(\frac{3 n-1}{4}\) = 50
⇒ 3n – 1 = 200,
3n = 201,
n = 67

(c) Sum = \(\frac{3(1+2+3+\cdots+11)-11}{4}=\frac{187}{4}\)

Question 14.
a + 1, a + 2, a + 3 ……….. is an arithmetic sequence
(a) What is its algebraic form?
(b) What is the sum of first 20 terms?
Answer:
(a) x_n = a + n
(b) Sum = 20a + \(\frac{20 \times 21}{2}\)
= 20a + 210

Question 15.
Algebraic form ofan arithmetic sequence is 3n + 2
(a) What is the common difference?
(b) What is its 15th term? ‘
(c) What is the sum of first 29 terms?
Answer:
(a) d = 3
(b) x₁₅ = 3 × 15 + 2 = 47
(c) Since x₁₅^th term is the middle term the first 29 terms sum will be x₁₅ × 29
= 47 × 29
= 1363

Question 16.
Consider the arithmetic sequence -1, 3, 7, …………
(a) What is the common difference?
Answer:
d = 4

(b) Write the n^th term of this sequence.
Answer:
4n – 5

(c) Is 95 a term of the sequence ?
Answer:
4n – 5 = 95
⇒ 4n = 100,
n = 25

(d) Calculate the sum of the terms upto 95.
Answer:
Sum = 4(1 + 2 + 3 + ……… + 25) – 5 × 25
= 4 × 325 – 125
= 1175

Question 17.
Sum of the first n terms of an arithmetic sequence is n² +5n
(a) Find the sum of first 15 terms
Answer:
15² + 5 × 15 = 300

(b) WTiat is its 8th term?
Answer:
x₈ = \(\frac{300}{15}\) = 20

(c) Write the algebraic form of the sequence.
Answer:
f = 6, d= 2
⇒ x_n = 2n + 4

Question 18.
Sum of the first 3 terms of an arithmetic sequence is 15 . Sum of the first 4 terms is 28
(a) What is the 4th term?
Answer:
x₄ = 28 – 15 = 13

(b) What is the sum of first 7 terms of the sequence?
Answer:
x₄ × 7 = 91

(c) What is the third term ?
Answer:
2d = 13 – 5 = 8,
d = 4,
x₃ = 13 – 4 = 9

(d) Express the sequence algebraically
Answer:
x_n = 4n – 3

Question 19.
Sum of the first n terms of an arithmetic sequence is n2
(a) What is the first term?
Answer:
1² = 1

(b) Find the common difference
Answer:
1² + 2² = 5, x₂ = 3, d = 2

(c) Write the sequence algebraically
Answer:
2n – 1

Question 20.
Sum of the first n terms of an arithmetic sequence is n² + n
(a) Find the common difference of the arithmetic sequence.
Answer:
x₁ = 1² + 1 = 2, x₁ + x₂ = 6
⇒ x₂ = 4
d = 4 – 2 = 2

(b) Write the sequence argebraically
Answer:
x_n = 2n

(c) Find the sum : 3 + 5 + 7 + …. + 51
Answer:
25th term is 50. Each term of 3, 5, 1…. is one more than 2, 4, 6….
3 + 5 + 7 + ……….. + 51 = 25² + 25 + 25 = 675

Question 21.
If \(\frac{1+3+5+\cdots(2 n-1)}{2+5+8+\cdots 23}\) = 9 then what is n ?
Answer:
\(\frac{n^2}{2+5+8+\cdots+23}\) = 9
2, 5, 8….23 has 8 terms. Sum is 100
\(\frac{n^2}{100}\) = 9
⇒ n² = 900,
n = 30

Question 22.
n^th term of an arithmetic sequence is 2n +1.
(a) Write the sequence in the numerical form
(b) What is its 100th term?
(c) Calculate the sum of first 100 terms
(d) Find the sum of first 100 terms of the arith-metic sequence 4, 6, 8
Answer:
(a) 3, 5, 7….
(b) 2 × 100+ = 201
(c) 2(1 +2 + 3 +….+100) + 100 = 10200
(d) The terms of 4, 6, 8,… are one more than the terms of 2n + 1 Sum is 10200 + 100 = 10300

Question 23.
Let x₁, x₂, x₃ ……… be the terms of an arithmetic sequence.
Answer:
Given that x₁ + x₅ + x₁₀ + x₁₅ + x₂₀ + x₂₄ = 225
Calculate the sum of first 24 terms of this sequence.
x₁ + (x₁ + 4d) + (x₁ + 9d) + (x₁ + 14d) + (x₁ + 19d) + (x₁ + 23d) = 225
Answer:
6x₁ + 69d = 225, 2x₁ + 23d = 75 24
Sum = (x₁ + x₂₄) × \(\frac{24}{2}\)
= (x₁ + x₁ + 23d) × 12
= (2x₁ + 23d) × 12
= 75 × 12 = 900

Question 24.
Algebraic form of an arithmetic sequence is \(\frac{3}{7}\) n +1
(a) What is the common difference?
(b) Obtain the expression for the sum of first n terms of this sequence
(c) Find the sum of first 21 terms of the sequence.
Answer:
(a) \(\frac{3}{7}\)
(b) Sum = (x₁ + x_n) × \(\frac{n}{2}=\frac{3 n^2+17 n}{14}\)
(c) \(\frac{3 \times 21^2+17 \times 21}{14}\) = 120

Question 25.
-117, -114, -111… is an arithmetic sequence.
(a) What is the common difference ?
(b) Find the nth term of the sequence
(c) At what position 0 becomes a term of the se-quence?
(d) What is the sum of first 79 terms of this se-quence?
Answer:
(a) 3
(b) 3n – 120
(c) 3n – 120 = 0 × n = 40
(d) When we consider 79 terms, 40th term will be the middle term. It is 0
Sum of 79 terms is 0

Question 26.
The sum of first 10 terms is 230 . The sum of first 15 terms is 495.
(a) If pn² + qn is the sum of first n terms then write a pair of equations
Answer:
100p + 10q = 230
⇒ 10p + q = 23

225 + 15q = 495
⇒ 15p + q = 33

5p = 10, p = 2, q = 3

(b) Find p and q and write the expression of the sum of first n terms
Answer:
2n² +3n

(c) Find the algebraic form of the sequence
Answer:
x_n = 4n +1

Question 27.
10 times 10th term of an arithmetic sequence is 20 times its 20th term.
(a) Find the 30th term of the sequence.
(b) What is the product of first 30 terms ?
Answer:
(a) 10(f + 9d) = 20(f + 19d)
⇒ f + 29d = 0, x₃₀ = 0

(b) 0

Question 28.
The sums of the first n terms of three arithmetic sequences are y₁, y₂ and y₃. The first term of each is 1 and common differences are 1,2 and 3 in the order. Prove that y₁ + y₃ = 2y₂
Answer:
s₁ = 1 + 2 + 3 … n = \(\frac{n(n+1)}{2}\)
s₂ = 1 + 3 + 5 + …….. + 2n – 1 = n²
For the third sequence, f = 1, d = 3
n^thth term is 3n – 2 Sum of first n terms = \(\frac{3 n^2}{2}-\frac{n}{2}\)
s₁ + s₃ = \(\frac{n(n+1)}{2}+\frac{3 n^2}{2}-\frac{n}{2}\)
On simplifying s₁ + s₃ = 2n²= 2s₂

Question 29.
Sum of the first n odd numbers is k
(a) What is the sum of first n even numbers?
Answer:
k + √k

(b) What is the sum of first n natural numbers ?
Answer:
\(\frac{k+\sqrt{k}}{2}\)

(c) Find \(\frac{1+2+3+4+\cdots+15}{16+17+18+\cdots+30}\)
Answer:
Find \(\frac{8}{23}\)

Question 30.
Consider the sequence of numbers which leaves the remainder 4 on dividing by 7 .
(a) Write the algebraic form of the sum of the terms of this sequence.
(b) Calculate the sum of first 20 terms of this se-quence
Answer:
(a) 4, 11, 18…. is the sequence
Sum of the first n terms = pn² + qn
p + q = 4, 4p + 2q = 15
Solving q = \(\frac{1}{2}\), p = \(\frac{7}{2}\)
Sum of first n terms = \(\frac{7}{2}\) n² + \(\frac{1}{2}\)n

(b) \(\frac{7}{2}\) × 20² + \(\frac{20}{2}\)
= 1400 + 10
= 1410

Students often refer to Kerala Syllabus 10th Standard Maths Textbook Solutions Chapter 4 Mathematics of Chance Questions and Answers Notes Pdf to clear their doubts.

SSLC Maths Chapter 4 Mathematics of Chance Questions and Answers

Mathematics of Chance Class 10 Questions and Answers Kerala State Syllabus

SCERT Class 10 Maths Chapter 4 Mathematics of Chance Solutions

Class 10 Maths Chapter 4 Kerala Syllabus – Chances As Numbers

(Textbook Page No. 74, 75)

Question 1.
A box contains 6 black and 4 white balls. If a ball is picked from it, what is the probability that it is black? And the probability that it is white?
Answer:
(a) \(\frac{6}{10}=\frac{3}{5}\)
(b) \(\frac{4}{10}=\frac{2}{5}\)

Question 2.
A bag contains 3 red balls and 7 green balls. Another contains 8 red and 7 green balls
(a) If a ball is drawn from the first bag what is the probability that it is red?
Answer:
\(\frac{3}{10}\)

(b) From the second bag?
Answer:
\(\frac{8}{15}\)

(c) The balls in both bags are put together in a single bag. If a ball is drawn from this, what is the probability that it is red?
Answer:
\(\frac{11}{25}\)

Question 3.
A bag contains 3 red beads and 7 green beads. Another bag contains one more of each. The probability of getting a red from which bag is greater?
Answer:
From the first bag, probability of getting red is \(\frac{3}{10}\)
In second bag, Number of red balls 4 , number of green balls 8
Probability of getting red \(\frac{4}{12}=\frac{2}{6}\)
\(\frac{3}{10}=\frac{18}{60}\), \(\frac{2}{6}=\frac{20}{60}\)
\(\frac{20}{60}>\frac{18}{60}\)
Taking ball from the second bag is more probable to get red

Class 10 Maths Kerala Syllabus Chapter 4 Solutions – Number Probability

(Textbook Page No. 76)

Question 1.
In each of the problems below, compute the probability as a fraction and then write it in decimal form and as a percent
(i) If one number from 1, 2, 3, …, 10 is chosen, what is the probability that it is a prime number?
(ii) If one number from 1, 2, 3, …, 100 is chosen, what is the probability that it is a two-digit number?
(iii) Every three digit number is written in a slip of paper, and all the slips are put in a box. If one slip is drawn, what is the probability that it is a palindrome?
Answer:
(i) \(\frac{4}{10}\) 0.4 40%
(ii) \(\frac{90}{100}=\frac{9}{10}\), 0.9,90%\frac{90}{100}=\frac{9}{10}
(iii) \(\frac{10}{100}=\frac{1}{10}\), 0.1, 10%

Question 2.
A person is asked to say a two-digit number. What is the probability that is a perfect square?
Answer:
\(\frac{6}{90}=\frac{1}{15}\)

Question 3.
A person is asked to say a three-digit number.
(i) What is the probability that all three digits of this number are the same?
Answer:
\(\frac{9}{900}=\frac{1}{100}\)

(ii) What is the probability that the digit in one’s place of this number is zero?
Answer:
\(\frac{90}{900}=\frac{1}{10}\)

(iii) What is the probability that this number is a multiple of 3?
Answer:
\(\frac{300}{900}=\frac{1}{3}\)

Question 4.
Four cards with numbers 1, 2, 3, 4 on them, are joined to make a four-digit number.
(i) What is the probability that the number is greater than four thousand?
Answer:
There are 24 such numbers. 6 of them are more than 4000 Probability of getting more than 4000 is \(\frac{6}{24}=\frac{1}{4}\)

(ii) What is the probability that the number is less than four thousand?
Answer:
\(\frac{18}{24}=\frac{3}{4}\)

SCERT Class 10 Maths Chapter 4 Solutions – Geometrical Probability

(Textbook Page No. 78)

In each picture below, the description of the green region inside the yellow one is given. In each, find the probability that a dot put in the picture, without looking, falls within the green region

Question 1.
The square got by joining the midpoints of a larger square

Answer:
The square got by joining the midpoints of a larger square

Draw horizontal and vertical lines. It divides the figure into 8 equal parts. Four of them are shaded.So probability is \(\frac{1}{2}\).

Question 2.
The triangle got by joining alternate vertices of a regular hexagon

Answer:
The triangle got by joining alternate vertices of a regular hexagon

Divide the regular hexagon into six equal triangles.
Three of them are green probability is \(\frac{3}{6}=\frac{1}{2}\)

Question 3.
The regular hexagon formed between two equal equilateral triangle.

Answer:
Regular hexagons formed by two equilateral triangles

Divide the green hexagon into six equal triangles.
These are 6 among 12 equal triangles.Probability is \(\frac{6}{12}=\frac{1}{2}\)

Question 4.
A square drawn with vertices on a circle

Answer:
A square drawn with vertices on the circle Let a be the side of the square. Diagonal √2 a
Radius of the circle \(\frac{a}{\sqrt{2}}\)
Area of the circle π\(\frac{a^2}{2}\)
Probability is \(\frac{a^2}{\frac{\pi a^2}{2}}=\frac{2}{\pi}\)

Question 5.
The circle that just fits within a square

Answer:
The circle that just fits within a square
Radius of the circle r.
Area = πr²
Side of the square 2r. Area of the square 4r²;
Probability = \(\frac{\pi}{4}\)

SSLC Maths Chapter 4 Questions and Answers – Pairs

(Textbook Page No. 80, 81)

Question 1.
Rajani has three necklaces and three earrings of green, blue and red stones. In how many different ways can she wear them ? What is the probability of her wearing a necklace and earrings of the same colour ? Of different colours?
Answer:
Different ways: 3 × 3 = 9
Probability of wearing a necklace and carring of same colour = \(\frac{3}{9}=\frac{1}{3}\)
Probability of wearing a necklace and carring of different colour = \(\frac{6}{9}=\frac{2}{3}\)

Question 2.
A box contains four slips numbered 1, 2, 3, 4 and another box contains two slips numbered 1 and 2. If one slip is drawn from each box, what is the probability of the sum of the numbers being odd? What is the probability of the sum being even?
Answer:
(1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1), (4, 2) Pairs having sum odd are (1, 2), (2, 1), (3, 2) (4, 1)
Probability of getting sum odd = \(\frac{4}{8}=\frac{1}{2}\)
Probability of getting sum even = \(\frac{4}{8}=\frac{1}{2}\)

Question 3.
A box contains four slips numbered 1, 2, 3, 4, and another box contains three slips numbered 1, 2, 3. If one slip is drawn from each box, what is the probability of the product of the numbers being odd? What is the probability of the product being even?
Answer:
Possible outcomes are
(1.1), (1, 2), (1, 3)
(2.1), (2, 2), (2, 3)
(3.1), (3, 2), (3, 3)
(4, 1), (4, 2), (4, 3)
Probability of getting product odd = \(\frac{1}{3}\)
Probability of getting product even = \(\frac{2}{3}\)

Question 4.
From all two-digit numbers using only the digits 1, 2, and 3, one number is chosen
(i) What is the probability of both digits being the same?
(ii) What is the probability of the sum of the digits being 4?
Answer:
(i) Outcomes are 11, 12, 13, 21, 22, 23, 31, 32, 33
Probability is \(\frac{3}{9}=\frac{1}{3}\)

(ii) \(\frac{3}{9}=\frac{1}{3}\)

Question 5.
A game for two players. Before starting, each player has to decide whether he wants an odd number or even number. Then both raise some fingers of one hand at the same time. If the sum of numbers of fingers is odd, the one who chose odd number at the beginning wins; if even, the one who chose even number wins. Which is the better choice at the beginning, odd or even?
Answer:
Pairs are
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5)

Sums are
2, 3, 4, 5, 6
3, 4, 5, 6, 7
4, 5, 6, 7, 8
5, 6, 7, 8, 9
6, 7, 8, 9, 10
Probability of getting the sum even = \(\frac{13}{25}\)
Probability of getting the sum odd = \(\frac{12}{25}\)
Choosing even at the beginning is better.

Kerala Syllabus Class 10 Maths Chapter 4 Solutions – More Pairs

(Textbook Page No. 84)

Question 1.
There are 30 boys and 20 girls in Class 10 A and 15 boys and 25 girls in Class 10 B. One student is to be chosen from each Class.
(i) What is the probability of both being girls ?
Answer:
There are 30 boys and 20 girls in Class 10 A and 15 boys and 25 girls in Class 10 B. One student is to be chosen from each Class.
\(\frac{20 \times 25}{2000}=\frac{500}{2000}=\frac{1}{4}\)

(ii) What is the probability of both being boys ?
Answer:
\(\frac{30 \times 15}{2000}=\frac{450}{2000}=\frac{9}{40}\)

(iii) What is the probability of one being a boy and one being a girl?
Answer:
\(\frac{30 \times 25+20 \times 15}{2000}=\frac{105}{200}=\frac{21}{40}\)

(iv) What is the probability of at least one being a boy?
Answer:
\(\frac{1500}{2000}=\frac{15}{20}=\frac{3}{4}\)

Question 2.
One is asked to say a two-digit number
(i) What is the probability of both digits being the same ?
Answer:
\(\frac{9}{90}=\frac{1}{10}\)

(ii) What is the probability of the first digit being greater than the second?
Answer:
Without actual listing
10, (20, 21), (30, 31, 32)…. (90, 91, 92….98)
Total numbers = 1 + 2 + 3 + …. + 9 = 45
Probability is \(\frac{45}{90}\)

(iii) What is the probability of the first digit being less than the second?
Answer:
8 + 7 + 6 + 5 + 4 + 3 + 2+1=36
Probability is \(\frac{36}{90}\)

Question 3.
Two dice with faces numbered from 1 to 6 are rolled together. What are the possible sums that can be got ? Which sum has the maximum possibility?
Answer:
Sums are listed below
2, 3, 4, 5, 6, 7
3, 4, 5, 6, 7, 8
4, 5, 6, 7, 8, 9
5, 6, 7, 8, 9, 10
6, 7, 8, 9, 10, 11
7, 8, 9, 10, 11, 12

Question 4.
One box contains 10 slips numbered 1 to 10 and another contains slips numbered with multiples of 5, up to 25. One slip is drawn from each box.
(i) What is the probability of both numbers being odd?
(ii) What is the probability of the product of the numbers being odd ?
(iii) What is the probability of the sum of the numbers being odd?
Answer:
(i) 5 odd numbers in the first box and 3 odds in the second box. Probability is \(\frac{15}{50}=\frac{3}{10}\)
(ii) To get product odd, number from first box is odd and number from second box odd
Probability is \(\frac{15}{50}=\frac{3}{10}\)
(iii) To get sum odd, numbers from the first box odd from the second box even or number from the first box even and number from second box is odd.
Probability is \(\frac{5 \times 2+5 \times 3}{50}=\frac{25}{50}=\frac{1}{2}\)

Mathematics of Chance Class 10 Notes Pdf

Class 10 Maths Chapter 4 Mathematics of Chance Notes Kerala Syllabus

Introduction
The unit ‘Mathematics of Chance’ or ‘Probability’ discusses the numerical measurement of chance. There are experiments whose outputs cannot be predicted. Such experiments are generally known as probability experiments.
Tossing a coin and throwing a dice are probability experiments. The outcomes in the tossing a coin are Head H and Tail T. We define probability of getting H on a single toss as \(\frac{1}{2}\) .
There is a section ‘Geometric probability’ in the unit. It measures probability as the ratio of areas.
The third section of the unit is related to fundamental principle of counting. If an activity can be performed in m ways, another activity by n ways then both the activities can be performed in m × n ways. Questions related to this situation is a part of the unit.

→ The experiments whose results cannot be predicted are called probability experiments. Tossing a coin, throwing a dice are the examples of probability experiments

→ On tossing a coin the expected outcomes are Head and Tail. The probability of getting head on tossing is \(\frac{1}{2}\)

→ Throwing a dice is another experiment. Dice is a cubical object numbered 1 to 6 on its faces. The expected outcomes are 1, 2, 3, 4, 5, 6. On throwing, the probability of getting a prime numbered face is \(\frac{3}{6}\)

→ Probability is the measure of chance. Sometimes by comparing area of geometric shapes also comes as the part of probability experiments.
For example, we know that the diagonal of a rectangle divides the rectangle into two equal triangles.Let one of them be shaded. A fine dot is placed into the rectangle without looking in it. The probability of falling the dot in the shaded part is the ratio of area of shaded part to the area of the rectangle. It is \(\frac{1}{2}\).

→ There are situations to measure probability by pairing of objects.
A box contains 3 black balls and 3 white balls. Another box contains 4 black balls and 2 white balls One ball is taken from each box and write as pairs
Number of pairs is (3 + 3) × (4 + 2) = 6 × 6 = 36
Probability of getting both white = \(\frac{3 \times 2}{36}\)
Probability of getting both black = \(\frac{3 \times 4}{36}\)
Probability of getting one black and one white = \(\frac{3 \times 2 + 3 \times 4}{36}\)

→ Probability Experiments

• Experiments where results cannot be predicted in advance are called probability experiments.
• Examples: Tossing a coin, throwing a dice.

→ Tossing a Coin

• Possible outcomes: Head (H) or Tail (T).
• Probability of getting a Head = \(\frac{1}{2}\), and Tail = \(\frac{1}{2}\)

→ Throwing a Dice

• A standard dice has six faces numbered from 1 to 6
• Possible outcomes: 1, 2, 3, 4, 5, 6.
• Prime numbers among them: 2, 3, 5

Students often refer to SSLC Maths Textbook Solutions and Class 10 Maths Chapter 2 Circles and Angles Important Extra Questions and Answers Kerala State Syllabus to clear their doubts.

SSLC Maths Chapter 2 Circles and Angles Important Questions and Answers

Circles and Angles Class 10 Extra Questions Kerala Syllabus

Circles and Angles Class 10 Kerala Syllabus Extra Questions

Question 1.
In the given figure BC is the diameter of a circle and AB = AC. Then ∠ABC is

(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer:
45°

Question 2.
O is the centre of the circle and ∠ACB = 30°. What is ∠AOB?

(a) 30°
(b) 15°
(c) 60°
(d) 90°
Answer:
60°

Question 3.
O is the center of a circle. If ∠OAB = 40° and C is a point on the circle then Z ACB is

(a) 40°
(b) 50°
(c) 80°
(d) 100°
Answer:
50°

Question 4.
A and B are the points on a circle. Pisa point on the alternate segment of arc AB. If ∠ACB = 30° then what is the central angle of small arc AB.

(a) 40°
(b) 50°
(c) 30°
(d) 60°
Answer:
60°

Question 5.
In the figure AB = AC = AD, ∠BAC = 40°. What is ∠BDC?

Answer:
20°

Question 6.
In the figure radius OA is perpendicular to radius OB. Radius of the circle is 10 cm

(a) What is the measure of ∠ACB
(b) What is the length AB?
Answer:
(a) 45°
(b) 10√2

Question 7.
In the figure AB is the diameter of a circle. ∠Q is \(\frac{1}{3}\) of of ∠APB.

(a) What is the measure of ∠APB?
(b) What is the length AQB?
Answer:
(a) 90°
(b) \(\frac{1}{3}\) × 90 = 30°

Question 8.
In triangle ABC. angles are in the ratio 1: 2: 3. Angle B is the largest and A is the smallest angle.
(a) What are the angles?
(b) What is the position of A based on the circle with diameter BC?
Answer:
(a) Smallest angle = \(\frac{1}{6}\) × 180 = 30°
∠A =30°, ∠B = 90°, ∠C = 60°

(b) On the circle

Question 9.
The vertices of a quadrilateral ABCD are on a circle. ∠A = x, ∠B = 3y, ∠C = 4x, ∠D = 2y

(a) Find x, y
(b) Find the angles of ABCD
Answer:
(a) x + 4x = 180 ⇒ 5x = 180, x = 36
Angles are 36°, 144°

(b) 2y + 3y = 180 ⇒ 5y = 180,y = 36°
Angles are 72°, 108°

Question 10.
ABCD is a trapezium in which AB is parallel to CD and AD = BC.
Prove that ABCD is a cyclic quadrilateral

Answer:
AD = BC ⇒ ∠A = ∠B
Since AB is parallel to CD, ∠B + ∠C =180°
Therefore ∠A + ∠C = 180°
ABCD is cyclic

Question 11.
The comers of a trapezium are on a circle.
Prove that it is an isosceles trapezium .
Answer:
ABCD is a trapezium in which AB is parallel to CD and vertices are on the circle.
∠B + ∠C = 180°, ∠A + ∠C = 180°
That is ∠A = ∠B. Base angles of a trapezium are equal. ABCD is isosceles.

Question 12.
Can all parallelogram be cyclic ?Justify your answer
Answer:
No. For a parallelogram opposite angles are equal
If it is cyclic opposite angle sum is 180°.
That is, opposite angles are 90° each.That is it is a rectangle.
That is, cyclic parallelogram is a rectangle.

Question 13.
O is the centre of the circle, PQ is parallel to OA, PR is parallel to OB

(a) If the central angle of arc ASB is 40° then what is the measure of ∠QPR?
(b) What is the central angle of arc QSR?
Answer:
(a) ∠QPR= 40°.(Angle between two lines is equal to angle between the lines parallel to them)
(b) Central angle of arc QSR is 80°

Question 14.
Draw a right triangle having longest side 7 cm and one of the other sides is 3 cm
Answer:
See the rough diagram

Question 15.
Angles of the quadrilateral ABCD are in the ratio 1: 2: 4: 3 in the order.
a) What are the angles?
b) Is this a cyclic quadrilateral? Why?
Answer:
(a) Take the angles as x, 2x, 4x and 3x
x + 2x + 4x + 3x = 360,
10x = 360,
x = 36
Angles are 36°, 72°, 144°

(b) 36 + 144 = 360, 72 + 108 = 180.
Opposite angle sum is 180°. So it is cyclic

Question 16.
O is the centre of the circle with diametre AB. Another circle is drawn with AO as the diametre

(a) What are the measure of ∠APO, ∠ACB
(b) Outer circle has radius 5 cm and BC = 8 cm. What is the length OP?
(c) Is AP = PC? Why?
(d) What is the length of AC?
Answer:
(a) 90°. Reason ∠APO, ∠ACB are the angles in the semicircle

(b) Triangle APO, triangle ACB are similar.
\(\frac{A O}{A B}=\frac{O P}{B C}\)
\(\frac{5}{10}=\frac{O P}{8}\), OP = 4 cm

(c) AC is the chord of big circle. OP is perpendicu-lar from centre to this chord. OP bisect AC. Therefore
AP = PC

(d) AC = \(\sqrt{10-8}\) = 6 cm.

Question 17.
Sides of triangle ABC are AB = 5 cm, AC = 12 cm, BC = 13 cm
(a) What kind of triangle is this ?
(b) What is the position of A based on the circle with diametre BC?
(c) What is the position of C based on the circle with diametre ZB?
(d) What is the position of B based on the circle with diametre AC?
Answer:
(a) Right triangle
(b) On the circle
(c) Outside the circle
(d) Outside the circle

Question 18.
In the figure AB is the diametre of a semicircle.Three angles x, y, z are marked outside. on the semicircle and inside the semicircle.

(a) What is the value of y?
(b) If x, y, z are in an arithmetic sequence,then what is x + z ?
(c) If the common difference of the sequence is 50 then find x and z
Answer:
(a) y = 90°
(b) x + z = 2 × 90 = 180° (Refer the properly of arithmetic sequence)
(c) y = 90, x = 90 – 50 = 40°, z = 90 + 50 = 140°

Question 19.
Draw a circle of radius 3 cm .Construct the angles 30° and 150° with vertices on the circle using compasses and scale only.
Answer:
• Draw a circle of radius 3 cm .Mark the center of the circle as O
• Mark a point A on the circle. Draw the radius OA.
• With A as the center and OA as radius, draw an arc which cut the circle at B. Join OB, ∠AOB = 60°
• Mark a point P on the complement of the arc AB, which makes 60° at the center. ∠APB = \(\frac{1}{2}\) × 60 = 30°
• Mark a point Q on the arc AB. ∆AQB = 180 – 30 = 150°

Question 20.
In the figure ∆ABC, ∆AOC, ∆ADC are in an arithmetic sequence
(a) What is the relation between angle ABC and angle AOC
(b) What is the relation between angle ABC and ADC
(c) Find the measure of these angles
Answer:
(a) ∆AOC = 2 × ∆ABC
(b) ∆ABC + ∆ADC = 180°
(c) Let ∆ABC = x, ∆AOC = y, ∆ADC = z
x, y, z are in an arithmetic sequence. Therefore 2y = x + z

From the relations noted above
y = 2x,
x + z = 180
2y = x + z
2 × 2x = x + y = 180
4x = 180 x = 45
x = 45°,y = 90°, z = 135°
∠ABC = 45°, ∠AOC = 90°, ∠ADC = 135°

Question 21.
ABCD is a square. The diagonals AC and BD intersect at O.

(a) What is the measure of angle AOD?
(b) What is the measure of angle APD?
(c) What is the measure of angle AQD
Answer:
(a) Diagonals of a square are perpendicular to each other. ∠AOD = 90°
(b) ∠APD = 45°
(c) ∠AQD = 180 – 45 = 135°

Question 22.
In the figure O is the centre of the circle, ∠BAO = 20°, ∠BCO = 10°
(a) What is the measure of angle ABC?
(b) What is the measure of angle AOC?
(c) What is the measure of angle ADC?
(d) Find the angles of triangle AOC
(e) If the diametre of the circle is 10 cm then find the length of the chord AC
Answer:
(a) In triangle OAB. OA = OB. Angles opposite to the equal sides are equal. Similarly in the case of triangle OBC also.
∠ABC = 20+ 10 = 30°
(b) ∠AOC = 2 × 30 = 60°
(c) ∠ADC= 180 – 30 = 150°
(d) Triangle AOC, OA = OC. ∠OAC = ∠OCA = 180 – 60 = 60°
∆OAC is an equilateral triangle. Angles are 60° each.
(e) OA = AC = OC = 5cm, radius 5cm.

Question 23.
In the figure O is the centre of the circle. If angle ADC = 140°, angle 4EC= 60°then B

(a) What is the measeure of ∠APC and ∠AQC
(b) What is the measure of angle AOC?
(c) Fnd the angles of the quadrlateral PEQB
Answer:
(a) ∠APC= 180-140 = 40°, ∠AQC = 40°
(b) ∠AOC = 2 × 40 = 80°
(c) In the quadrilateral ∠AEQ = ∠AEC = 60°, ∠EPB = 180 – 40 = 140°, ∠EQB = 140° ∠PBQ = 360 – (140 + 140 + 60) = 20.
Angles are 140°, 60°, 140°, 20°

Question 24.
In the figure O is the centre of the circle, then

(a) What kind of triangle is OAC?
(b) What is the measure of angle ABC?
(c) What is the measure of angle ADC?
(d) If the radius of the circle is 6 cm then what is the length of the chord AC.
Answer:
(a) OA = OC, ∠OAC = ∠OCA = 45°, ∠AOC = 90° ∆OAC is an isosceles right triangle
(b) ∠ABC = – AOC = 45°
(c) ∠ADC = 180 – 45 = 135°
(d) AC = \(\sqrt{6^2+6^2}=6 \sqrt{2}\) cm

Question 25.
ABC is an isosceles triangle with AB = AC, ∠ABC= 50°.

(a) Name two cyclic quadrilaterals in this picture.
(b) What is the measure of angle D?
(c) What is the measure of ∠BEC?
Answer:
(a) Quadrilateral ABEC and quadrilateral DBESare cyclic.
(b) ∠ABC = ∠ACB = 50“
∴ ∠A = 180 – 100 = 80°
∠D = 80°
(c) ∠BEC = 180 – 80 = 100°

Question 26.
ABCD is a cyclic quadrilateral. AB is the diametre of the circle. AD = CD and ∠ADC = 130°.

(a) What is the measure of ∠ACB?
(b) What is the measure of ∠ABC?
(c) Find ∠DCB.
(d) What is the measure of ∠BAD?
Answer:
(a) ∠ACB = 90° (Angle in the semicircle)
(b) ∠ABC = 180 – 130 = 50°
(c) Since CD = AD,ihe angles opposite to the equal sides of triangle ADC are equal.
∠DCA = 25°, ∠DCS = 90 + 25 = 115°
(d) ∠BAD = 180 – 115 = 65°

Question 27.
Prove that any cyclic parallelogram is a rectangle.
Answer:
ABCD is a parallelogram .(Draw rough figure)

Opposite angles are equal.
∠A = ∠C, ∠B = ∠D

Sum of the opposite angles is 180°
∠A + ∠C = 180°, ∠A – ∠C
∴ ∠A = 90°, ∠C = 90°
∠B + ∠D = 180°, ∠B = ∠D
∠B = 90°, ∠D = 90°
ABCD is a square

Question 28.
In triangle ABC, AB = AC. P and Q are the mid points of the side AB and AC.
(a) Draw a rough diagram and join the points P and Q.
(b) Prove that BPQC is a cyclic quadrilateral.
(c) If ∠A in triangle ABC is 20°, find the angles of the trapezium BPQC
Answer:
(a) Figure

(b) Since AB = AC, ∠B = ∠C .
Line joining the mid points of two sides of a triangle is parallel to the third side. PQ is parallel to SC.
In PBCQ, ∠B + ∠P = 180° (co interior angles)
Since ∠C = ∠S, ∠C + ∠P = 180°
PQCB is a cyclic quadrilateral

(c) ∠A = 20°
∠B = ∠C = \(\frac{180-20}{2}\) = 80°
∠B + ∠P = 180°, ∠S = 100°, ∠Q = 100°
Angles are ∠P = 100°, ∠Q = 100°,
∠B = 80°, ∠C = 80°

Question 29.
In the figure SD = CD, ∠DSC = 25°

(a) What is the measure of ∠SDC?
(b) What is the measure of ∠BAC?
(c) What is the measure of ∠EBC?
Answer:
(a) In triangle SDC, SD = CD.
Angle opposite to these sides are equal. ∠BCD = 25°
∠BOC = 180 – (25 + 25) = 130°
(b) ∠BAC= 180 – 130 = 50°
(c) ∠EBC = ∠BAC = 50 °, ∠ESC
= 180 – (90 + 50)
= 180 – 140
= 40°

Question 30.
Two circles intersect at S and E as in the figure. The points A – B – C are along a line. Also the points D – E – F are also on a line

(a) Prove that AD is parallel to CF
(b) If AC = DF suggest a suitable name to the quadrilateral ADFC
(c) Prove that ADFC is a cyclic quadrilateral.
Answer:

(a) Draw BE. ABED is a cyclic quadrilateral.
If ∠DAB = x then ∠BED = 180 – x,
∠BEF = 180 – (180 – x) = x.
BEFC is cyclic. ∠C = 180 – x.
In quadrilateral ADFC,
∠A + ∠C = x + 180 – x = 180°
Co interior angle sum is 180°.
AD is parallel to CF

(b) ADFC is a trapezium.
Since AC = DF is an isosceles trapezium.

(c) Angles at the ends of parallel sides of an isosceles trapezium are equal.
Since ∠A = ∠D and ∠A + ∠C = 180° then ∠D + ∠C = 180°
ADFC is a cyclic quadrilateral.

Question 31.
AB is the diametre of the circle. CD is a chord of length equal to radius of the circle.
(a) What is the measure of ∠COD?
(b) What is the measure of ∠CBD?
(c) What is the measure of ∠BCF?
(d) Find the measure of ∠CPD
Answer:
(a) Draw OC, OD, OCD is an equilateral traingle.
∠COD = 60°

(b) ∠CBD = \(\frac{1}{2}\) × 60 = 30°

(c) ∠BCA = 90° (angle in the semicircle).
∴ BCP = 90°.

(d) In traingle BCP,
∠CPD = ∠CPB
= 180-(90 + 30)
= 60°

Students often refer to SSLC Maths Textbook Solutions and Class 10 Maths Chapter 1 Arithmetic Sequences Important Extra Questions and Answers Kerala State Syllabus to clear their doubts.

SSLC Maths Chapter 1 Arithmetic Sequences Important Questions and Answers

Arithmetic Sequences Class 10 Extra Questions Kerala Syllabus

Arithmetic Sequences Class 10 Kerala Syllabus Extra Questions

Question 1.
x, y, z are the three consecutive terms of an arithmetic sequence. Which of the following is correct?
(a) y = x + z
(b) 2y = x + z
(c) y = x – z
(d ) x = y + z
Answer:
(b) 2y = x + z

Question 2.
The first three digit term of the arithmetic sequence 1, 6, 11, 16 …….. is
(a) 101
(b) 102
(c) 100
(d) 103
Answer:
(a) 101

Question 3.
13^th term of an arithmetic sequence is 20. What is the sum of Is’ and 25s’ term?
(a) 30
(b) 50
(c) 40
(d) 20
Answer:
(c) 40

Question 4.
Distance from 10” term to 15:,: term is 11. What is the distance from 10;- term to 20,r- term?
(a) 33
(b) 10
(c) 22
(d) 20
Answer:
(c) 22

Question 5.
The position at which 100 becomes a term of the sequence \(\frac{1}{7}, \frac{2}{7}, \frac{3}{7}\)
(a) 100
(b) 300
(c) 700
(d) 200
Answer:
(c) 700

Question 6.
a, a – 1, a – 2, a – 3 is an arithmetic sequence.
a) What is the common difference of the sequence ?
b) What is its 10th term?
Answer:
a) Common difference = – 1
b) a – 9

Question 7.
10^th term of an arithmetic sequence is 25 and its common difference 4
a) What is the 15th term?
b) What is the 5th term?
Answer:
a) 15^th term = 10^th term + 5 × common difference
= 25 + 5 × 4 = 45

b) 5th term = 10^th term – 5 × common difference = 25 – 20 = 5

Question 8.
Sum of the 7 consecutive terms of an arithmetic sequence is 119.
a) Which term comes in the middle of these 7 terms?
b) What is the sum of two terms on either side of the middle term just closer to it?
Answer:
a) -y = 17
b) 2 × 17 = 34

Question 9.
Common difference of an arithmetic sequence is \(\frac{1}{2}\)
a) What is the difference between 5th term and 15th term?
b) If 5th term is 20 then what is the first term of the sequence?
Answer:
a) Difference is 10 times common difference. It is 10 × \(\frac{1}{2}\) = 5
b) First term = 5th term – 4 times common difference = 20 – 4 × \(\frac{1}{2}\) = 18

Question 10.
□ , 10, □ , □ , 31 are five consecutive terms of an arithmetic sequence.
a) What is the common difference ?
b) Write the terms in the boxes.
Answer:
a) Difference between second term and fifth term is three times common difference
3 × common difference = 31 – 10 = 21
Common difference is \(\frac{21}{3}\) = 7

b) 3, 10, 17, 24, 31

Question 11.
Angles of a right triangle are in arithmetic sequence.
a) Which angle comes in the middle?
b) What are the angles?
Answer:
a) Middle term = \(\frac{180}{3}\) = 60
b) Largest term (third angle )= 90°. So common difference is 30°
Angles are 30°, 60°, 90°

Question 12.
Angle sum of a pentagon is 540? .The angles when arranged in the ascending order makes an arithmetic sequence.
a) Which angle comes as the middle term of the sequence?
b) If the first term is 42° then what is the difference between two adjacent terms?
Answer:
a) Middle term of the arithmetic sequence is \(\frac{540}{5}\) = 108
Middle angle is 108°

b) Difference between first term and third term is two times common difference It is 108 – 42 = 66.
Common difference is 33.

Question 13.
Common difference of an arithmetic sequence is 10
a) What is the difference between sixth term and first term?
b) Write two term positions having the same difference.
Answer:
a) Difference between sixth term and first term is five times common difference.
It is 5 × 10 = 50

b) Difference 7^th term – 2nd term, 8^th term – 3^rd term gives the same difference.

Question 14.
Sum of the first five terms of an arithmetic sequence and sum of the first eight terms of the same arithmetic sequence are equal.
a) What is the sum of sixth, seventh and eighth terms of the sequence ?
b) What is the seventh term?
Answer:
a) 70 – 70 = 0
b) Sum of sixth , seventh , eighth term is 0. So 7^th term is \(\frac{0}{3}\) = 0

Question 15.
√2, √8, √18… is an arithmetic sequence
a) What is the common difference?
b) What is the difference between first term and eleventh term?
Answer:
a) Common difference is √8 – √2 = 2√2 – √2 = √2
b) Difference is ten times common difference. It is 10√2

Question 16.
Look at the sequence of squares

a) Write the number of sticks used for making squares as sequence
b) Describe the sequence in your own words
c) How many sticks are needed to make 1O’* term?
Answer:
a) 4, 7, 10…
b) This is the sequence of numbers 1 more than multiple of 3
c) 3 × 10 + 1 = 31

Question 17.
This the sequence of triangles made by sticks

a) Write the number of sticks used for making triangles as a sequence
b) Describe this sequence in your own words.
c) How many sticks are needed to make 50th term of the sequence?
Answer:
a) 3, 5, 7…
b) This is the sequence of numbers 1 more than multiple of 2
c) 2 × 50 + 1 = 101

Question 18.
Third term of an arithmetic sequence is 12 and its fifth term20
a) What is the common difference?
b) What is the first term ?
c) Write the sequence numerically
Answer:
a) Difference between fifth term and third term is two times common difference.
2 × common difference = 20 – 12 = 8, Common difference = 4
b) First term = third term – 2 common difference = 12 – 8 = 4
c) 4, 8, 12…

Question 19.
The 4th term of an.arithmetic sequence is 54 and its 9^th term is 99.
(a) What is the common difference of the sequence ?
(b) Will the difference between any two terms of this sequence be 900 ? Why ?
Answer:
a) Difference between ninth term and fourth term is five times common difference.
5 times common difference = 99 – 54 = 45, common difference is 9

(b) We know that difference between two terms of an arithmetic sequence is a multiple of common difference.
Here 900 is the multiple of 9. So 900 can be the difference between two terms.

Question 20.
The sum of 1st and 19th terms of an arithmetic se-quence is 50. The sum of its 1st and 20th terms is 54.
a) What is the common difference of the sequence ?
b) What is the sum of its 7^th and 13^th terms ?
c) Find its 10^th term.
Answer:
a) Common difference = 54 – 50 = 4
b) Sum of 7^th and 13^th term is 50
c) Tenth term is = \(\frac{50}{2}\) = 25

Question 21.
700 rupees is used for purchasing 7 prizes . Cost of each prize is 20 rupees more than the cost of the prize just below it.
a) What is the cost of fourth prize?
b) What is the cost of first prize?
c) Write costs of prizes in the decending order.
Answer:
a) Fourth term = 100
b) First prize is 3 × 20 more than fourth term. It is 100 + 3 × 30= 16
c) 160, 140, 120, 100, 80, 60, 40

Question 22.
5^th term of an arithmetic sequence is 10 and 10^th term is 5.
a) What is the common difference of the sequence?
b) Find the first term?
c) What is its 15^th term?
Answer:
a) Difference between 10^th term and 5^th term is five times common difference.
5 – 10 = 5 common difference
common difference = \(\frac{-5}{5}\) = -1 5
b) First term = 10 – 4 × -1 = 14
c) Fifteenth term = tenth term + 5 × common difference = 5 + -1 × 5 = 0

Question 23.
7^th term of an arithmetic Sequence is 30 and common difference 3
a) What is its 13^th term ?
b) What is its first term?
c) What is the sum of first 13 terms?
Answer:
7th term of an arithmetic Sequence is 30 and common difference 3
a) 13^th term = seventh term + six times common difference
13^th term = 30 + 6 × 3 = 48

b) First term + thireenth term = 2 × 7 ^th term = 60
First term = 60 – 48 = 12
OR
First term = seventh term -6 times common difference = 12

c) Sum of the first 13 terms = 390

Question 24.
Angles of a pentagon are in arithmetic sequence when arranged in the ascending order. First term is 42°
a) What is its middle term?
b) What is the difference between two adjacent terms?
c) Find the largest term of the pentagon
Answer:
a) Angle sum of a pentagon (n – 2) × 180
= 3 × 180 = 540
Middle term = \(\frac{540}{5}\) = 108

b) Difference between third term and first term is two times common difference
2 × common difference = 108 – 42 = 66,
Common difference = 33
c) Fifth term = 108 + 2 × 33 = 108 + 66 = 174
Angles are 42°, 75°, 108°, 141°, 174°

Question 25.
Eight consecutive terms of an arithmetic sequence are written as pairs taking terms equidistant from both ends (9, 58), (16, 51), (23,44), (30, 37)
a) What is the common difference of the sequence ?
b) What is the sum of these terms?
Answer:
a) 6th and 5th terms are paired as (30, 37). Common difference is 3
b) Pair sum 67. Sum of terms = 4 × 67 = 268

Question 26.
Consider the arithmetic sequence 4, 7, 10…
(a) What is its 13^th term ?
(b) What is the sum of its first 25 terms ?
(c) Find the sum of first 25 terms of the arithmetic sequence 8, 14, 20
Answer:
Consider the arithmetic sequence 4, 7, 10…
(a) Thirteenth term = first term + 14 times common difference
13^th term = 4 + 12 × 3 = 40
(b) Sum of first 25 terms = thireenth term × 25 = 40 × 25 = 1000
(c) Each term is two times the terms of the first sequence
Sum = 2 × 1000 = 2000

Question 27.
The sum of the first 5 terms of an arithmetic sequence is 65 and the sum of the first 9 terms is 189.
a) What is the 3^rd term of the sequence ?
b) What is the 5^th term of the sequence ?
c) What is the common difference of the sequence ?
Answer:
a) Third term = \(\frac{65}{5}\) = 13
b) Fifth term = \(\frac{189}{9}\) = 21
c) 2 × common difference = 21 – 13 = 8, Common difference is 4

Question 28.
7^th term of an arithmetic sequence is 18 and its 18^th term is 7.
a) What is the common difference of the sequence ?
b) Find the 25^th term of the sequence.
c) What is the sum of first 49 terms of the sequence ?
d) What is the sum of first 50 terms of the sequence ?
Answer:
7 th term of an arithmetic sequence is 18 and its 18^th term is 7.
a) 18^th term – 7^th term = 11 times common difference 7 – 18 = 11 × common difference,
Common difference = – 1

b) 25^th term = 18^th term + 7 times common difference. It is 7 + 7 × – 1 = 0

c) 25^th term is 0. Sum of first 49 terms = 0

d) 50^th term = 25^th term + 25 times common difference. It is 0 + 25 × -1 = -25
Sum of 50 terms = sum of 49 terms + 50^th term = 0 + -25 = -25

Question 29.
There are 20 terms in an arithmetic sequence. Sum of the first and last terms is 67.
a) What is the sum of 2^nd and 19^th term ?
b) If the 10^th term is 32, What is the 11^th term ?
c) What is the common difference of the sequence ?
d) What is the first term ?
Answer:
a) 67
b)35
c) 3
d) 5

Question 30.
The sum of first 9 terms of an arithmetic sequence is 144 and the sum of next 6 terms is 231.
a) What is the 5^th term of the sequence ?
b) What is its 81^th term ?
c) What is the common difference of the sequence?
Answer:
Solve the problem yourself.
Sum of first 9 terms is 144 . From this find its 5^th term Sum of first 15 terms will be 144 + 231 = 375. Find 8^th term.
Using 5^th term and 8^th term find the common difference

Students can use SSLC Malayalam Adisthana Padavali Notes Unit 2 Chapter 3 കെത്തളു Kethalu Summary in Malayalam Pdf to grasp the key points of a lengthy text.

Class 10 Malayalam Kethalu Summary

Kethalu Class 10 Summary

Class 10 Malayalam Adisthana Padavali Unit 2 Chapter 3 കെത്തളു Summary

ഗ്രന്ഥകാരപരിചയം

സുകുമാരൻ ചാലിഗദ്ധ : വയനാട്ടിലെ ആദിവാസി സമൂഹത്തിൽ നിന്നുയർന്നുവന്ന ശ്രദ്ധേയ നായ കവി. റാവുള ഭാഷയിലും മലയാളത്തിലും കവിതയെഴുതുന്നു. കേരളസാഹിത്യ അക്കാ ദമി ജനറൽ കൗൺസിൽ അംഗമാണ്. മലയാള കവിതയിലേക്ക് ഗോത്രഭാഷയുടെ ചൂടും ചൂരും കൊണ്ടുവന്ന കവിയാണിദ്ദേഹം. കാട്ടുപക്ഷി, വളവിൽമാരൻ, കൊക്കര, ഇപ്പം പേരില്ല, തെറ് കുതിര, ഇടിമുട്ടി, കൂ………. കൂ……….. കൂ …… ഏട്ടക്കൂരി മുതലായവയാണ് പ്രസിദ്ധകവിതകൾ.

സുകുമാരൻ ചാലിദ്ധയും ഗോത്രകവിതകളും

വരികളിലൂടെ വർണാഭമായ ഒരു ജീവിതം തേടിയാണ് സുകുമാരൻ ചാലിഗദ്ധയാത്ര തുടങ്ങുന്ന ത്. ആദിവാസികൾക്ക് കവികളില്ലെങ്കിലും കവിതകളുണ്ട്. ഗായകരില്ലെങ്കിലും പാട്ടുകളുണ്ട്. ഈ കവിതകളും പാട്ടുകളുമാണ് സുകുമാരനെ ആകർഷിച്ചത്. പട്ടിണിയിലും, ലഹരിയിലും തച്ചുത കർക്കപ്പെട്ട ജീവിതങ്ങൾ ചുറ്റുപാടും വെന്തുരുകുമ്പോൾ കവിതയിലൂടെ ജീവിതത്തിന് വർണ ങ്ങൾ പ്രദാനം ചെയ്യുകയാണ് സുകുമാരൻ. കലയും സാഹിത്യവും ആദിവാസികൾക്ക് അത പഥ്യമുള്ള കാര്യങ്ങളല്ല. കവിതയെഴുതിയാലോ പാട്ടുപാടിയാലോ ഊരിലെ പഞ്ഞം മാറുമോ എന്നവർ ചോദിക്കുന്നുണ്ട്. ഉന്നതവിദ്യാഭ്യാസം നേടാനൊന്നും സുകുമാരന് കഴിഞ്ഞില്ലെങ്കിലും കവിതയിലൂടെ കുതറിപ്പറക്കാനും പുതിയ ആകാശങ്ങൾ സൃഷ്ടിക്കാനും ഈ കവി കഠിനമായി പ്രയത്നിച്ചു. അങ്ങനെ അദ്ദേഹം നൂറുകണക്കിന് കവിതകൾ എഴുതി കൂട്ടി. ആദിവാസി ജീവിത ത്തിന്റെ വിവിധ ബിംബങ്ങളാണ് ഈ കവിതകളിൽ കാണുന്നത്. പ്ലസ് ടുവരെ മലയാളം പഠി ക്കാൻ സാധിച്ചതുകൊണ്ട് ഗോത്രഭാഷയായ റാവുളയോടൊപ്പം മലയാളത്തിലും സുകുമാരൻ എഴുതാറുണ്ട്. നമ്മുടെ പാഠ്യഭാഗമായ ‘ഏട്ടക്കൂരി’ റാവുള ഭാഷയിലും മലയാളഭാഷയിലും ചേർത്തി ട്ടുള്ളത് ശ്രദ്ധിക്കുമല്ലോ.

പാഠസംഗ്രഹം

കേളു അഥവാ ഏട്ടക്കൂരി ഗോത്രഭാഷയിലെഴുതിയ കവിതയാണ്. പ്രത്യേകിച്ച് ആദിമധ്യാന്ത പൊരുത്ത മുള്ളതും തുടർച്ചയായ ആശയങ്ങളുള്ളതുമായ കവിതയൊന്നുമല്ല ഇത്. ആദിവാസിയുടെ ആത്മാവിഷ്കാരം മാത്രമാണത്. ആനയോട് ഈ കവിക്ക് ഏറെ ആഭിമുഖ്യമുണ്ട്. ആനയെക്കുറിച്ച് വേറെയും കവിതകൾ ഇദ്ദേഹം എഴുതിയിട്ടുണ്ട്.

ഏട്ടക്കൂരി എന്ന കവിത ആരംഭിക്കുന്നതും ആനയുടെ പ്രസവം സൂചിപ്പിച്ചുകൊണ്ടാണ്. ചോലക്കാട്ടിൽ കാണ പ്പെടുന്ന നീർക്കാക്ക (കുളക്കോഴി) മുങ്ങാംകുഴിയിടുമ്പോൾ മീനുകൾ രക്ഷതേടി കരയോടടുത്തു നിരന്ന് നിൽക്കു ന്നു. മരത്തിൽ പടർന്നിരിക്കുന്ന മരവാഴയുടെ പൂക്കൾ അവ തലയിലണിയാൻ ആഗ്രഹിക്കുന്ന സ്ത്രീകളെ അന്വേഷിക്കുകയാണ്. കല്ലുന്തി മീനുകൾ വെള്ളത്തിൽ മുങ്ങാങ്കുഴിയിടുന്നു. നരസിപ്പുഴയിലെ കല്ലുകൾ വെള്ള ത്തിന്റെ നിരന്തരമായ ഒഴുക്കിൽപ്പെട്ടു നരച്ചിരിക്കുന്നു. പൊൻകുഴിപ്പുഴയുടെ തീരങ്ങൾ സ്വർണഖനികളായി. കൈതപ്പൂക്കൾ കാറ്റിലാടി. മുയലുകൾ കാട്ടിലെ പുല്ലിനിടയിൽ കുരുങ്ങിക്കിടന്നു. പുഴയിലൂടെ കുറുകെ കട ക്കുന്ന വാളമീനിനും, കരയ്ക്ക് കുറുകെ ചാടുന്ന ഏട്ടക്കൂരിക്കും ഏറെ ഭംഗിയുണ്ട്.

മലകൾ നശിച്ചുകൊണ്ടിരിക്കുന്നു. ഇടിഞ്ഞുമറിഞ്ഞ കല്ലുകളെല്ലാം മണ്ണൊലിച്ച് വെളുത്തനിറത്തിലുള്ള രക്തം ഛർദ്ദിക്കുന്നു. ഞാൻ ഓടിച്ചാടി നടന്ന കല്ലുകളെല്ലാം ഇപ്പോൾ അങ്ങാടിയിലെത്തി. ഞാൻ കുട്ടിക്കാലത്ത് ഇരുന്ന മരത്തിലെല്ലാം ഇന്ന് കാൽ വഴുതും. ഏഴുപുഴകൾ ചാടിക്കടന്നിട്ടും ഏട്ടക്കൂരിക്ക് പഴയ പടി തിരിച്ചുപോരേണ്ടി വന്നു. ഊത്തക്കാലത്ത് (പ്രജനനകാലത്ത്) പോലും മീനുകൾക്ക് രക്ഷയില്ലാത്ത അവസ്ഥയാണ്. ആശയങ്ങൾക്ക് തുടർച്ചയോ ഏകാതാനതയോ ഇല്ലെങ്കിലും സുകുമാരൻ അവതരിപ്പിക്കുന്നത് അവരുടെ വർഗം അനുഭവിക്കുന്ന സമകാലിക പ്രശ്നങ്ങളാണ്.

കഠിന പദങ്ങളും അർഥവും

പെറ്റുബൂന്ത = പെറ്റു വീണ
ഗുണ്ട് = കുഴി
ഗുണ്ടില = കുഴിയിലെ
കല്ലുഗേരി = കല്ലുകേറി (കല്ലുന്തി)
ബെളുത്ത = വെളുത്ത
ഏവുമല = ഏഴുമല
രച്ചെ = രക്ഷ
കാണി = ഇല്ല

Practicing with SSLC Malayalam Adisthana Padavali Class 10 Notes Pdf Unit 2 Chapter 3 കെത്തളു Kethalu Notes Questions and Answers improves language skills.

Kethalu Class 10 Notes Question Answer

Class 10 Malayalam Kethalu Notes Question Answer

Class 10 Malayalam Adisthana Padavali Unit 2 Chapter 3 Kethalu Notes Question Answer

പാഠപുസ്തകത്തിലെ ചോദ്യങ്ങളും അവയുടെ ഉത്തരങ്ങളും
Question 1.
കവിതയിൽ ആവിഷ്കരിച്ചിരിക്കുന്ന ദൃശ്യങ്ങൾ എന്തെല്ലാം? അവയുടെ സവിശേഷതകൾ അവതരിപ്പിക്കുക.
കവിതയിലെ ഗോത്രഭാഷാപദങ്ങൾക്ക് തുല്യമായ പ്രാദേശികപദങ്ങൾ കണ്ടെത്തി പട്ടികപ്പെടുത്തുക.
Answer:
ആദിവാസികളും അവരുടെ ജീവിത പരിസരങ്ങളുമായി ബന്ധപ്പെട്ട ചിത്രങ്ങളുമാണ് കവിതയിൽ അവതരി പ്പിക്കുന്നത്. കാട്ടാനയുടെ പ്രസവം, വനത്തിലെ ചോലകളിലെ ജലത്തിൽ മുങ്ങാംകുഴിയിടുന്ന നീർക്കാ ക്കൾ, നീർക്കാക്കയിൽ നിന്നും രക്ഷനേടാൻ നെട്ടോട്ടമോടുന്ന മീനുകൾ, കാട്ടിലെ മരവാഴകൾ (ഓർക്കി ഡുകൾ), മരവാഴയിലെ പൂക്കൾ ചൂടാൻ പെണ്ണുങ്ങളില്ലാത്തതിന്റെ വേദന, പുഴയൊഴുകിയൊഴുകി വെളുത്തു തേഞ്ഞുപോയ കല്ലുകൾ, കാട്ടിലെ കുന്നുകളിലെ സ്വർണസാന്നിധ്യം, ഒളിഞ്ഞുനോക്കുന്ന കൈതപ്പൂക്കൾ, പുല്ലുകൾക്കിടയിൽ കുരുങ്ങുന്ന മുയലുകൾ, പുഴയിലൂടെ വിലങ്ങനെ സഞ്ചരിക്കുന്ന വാളമീനുകൾ, ചാടി കളിക്കുന്ന ഏട്ടക്കുരീമീനുകൾ, വെളുത്ത് നരച്ചുപോയ മലകൾ, കച്ചവടച്ചരക്കായി മാറുന്ന കാട്ടുകല്ലു കൾ, മുറിച്ചു കടത്തപ്പെടുന്ന മരങ്ങൾ ഇതൊക്കെയാണ് കവിതയിലെ ദൃശ്യങ്ങൾ. ആദിവാസി പ്രകൃതിയു ടെയും ആദിവാസി ചൂഷണത്തിന്റെയും അസ്സൽക്കാഴ്ചയാണ് ഈ ചിത്രങ്ങളിലുള്ളത്.

Question 2.
കവിതയിലെ ഗോത്രഭാഷാ പദങ്ങൾക്ക് തുല്യമായ പ്രാദേശിക ഭാഷാപദങ്ങൾ കണ്ടെത്തി പട്ടികപ്പെടു
ത്തുക.
Answer:

ഗോത്രഭാഷാപദം	പ്രാദേശികപദം
കൊത്തളു	ഏട്ടക്കൂരി മത്സ്യം
കാണി	ഇല്ല
ബേണു	വേണം
കാടുമ്മുവൽ	കാട്ടുമുയൽ
ബാളെ	വാള
ചാന്തു	ചന്തം
നാൻ	ഞാൻ
ഏവു	ഏഴ്
പുവെ	പുഴയെ

Question 3.
“കുത്തിവീണ കല്ലുകളെല്ലാം
വെളുത്ത ചോരകക്കുന്നു”
കവിയുടെ വേദനയിൽ സമത്വബോധത്തിന്റെ വിശാലതലങ്ങൾ കണ്ടെത്താനാവുമോ? ചർച്ച ചെയ്യുക.
Answer:
സമൂഹത്തിന്റെ മുഖ്യധാരയിൽ നിന്നും പുറത്താക്കപ്പെടുകയോ അരികുവൽക്കരിക്കപ്പെടുകയോ ചെയ്ത ഒരു വിഭാഗം ജനങ്ങളാണ് ആദിവാസികൾ. അവരുടെ മണ്ണും, ഭൂമിയും ജലവുമെല്ലാം ചൂഷണത്തിന് വിധേ യമാവുന്ന കാഴ്ചയാണ് ഇന്ന് നാം കാണുന്നത്. ആദിവാസികളുടെ മലകളിൽ കണ്ണിടുന്ന പരിഷ്കൃത സമൂഹം സ്വാർഥതാൽപ്പര്യങ്ങൾക്കായി മലകളും മറ്റും ഇടിച്ചു നിരത്തുന്നു. ഏത് കറുപ്പിനെയും വെളുപ്പി ക്കാൻ കരുത്തുള്ള അധികാരി വർഗവും സമ്പന്നവർഗവും ആദിവാസികളുടെ സംസ്കാരത്തിന്റെ അന്ത കൻമാരായി നിലനിൽക്കുന്നതിന്റെ വേദനയാണിവിടെ കവി പങ്കു വയ്ക്കുന്നത്. എത്ര കിട്ടിയാലും, എന്തു കിട്ടിയാലും മതിവരാത്ത പരിഷ്കൃത സമൂഹത്തിന്റെ സമത്വബോധം വെറും പാഴ്വാക്കാണ്.

Question 4.
“ഏഴു മലകൾ ചാടിയിട്ടും
ഏഴു പുഴകൾ കടന്നിട്ടും
ഏട്ടക്കൂരി തിരിച്ചിറങ്ങി.
ഊത്തക്കാലത്തും രക്ഷയില്ല.”
പാർശ്വവൽക്കരിക്കപ്പെട്ട ഒരു ജനതയുടെ അതിജീവനത്തിന്റെ പ്രതീകാത്മകമായ ചിത്രീകരണമാണോ കാവ്യഭാഗത്തുള്ളത്? നിരീക്ഷണങ്ങൾ അവതരിപ്പിക്കുക.
Answer:
ഏട്ടക്കൂരി എന്ന സങ്കൽപ്പം കവിതയിൽ ആദിവാസിയെ പ്രതിനിധാനം ചെയ്യുന്ന ഒരു ബിംബമാണ്. കാലാ കാലങ്ങളായി മുഖ്യധാരയിലേക്ക് വരാൻ കിണഞ്ഞുപരിശ്രമിക്കുന്നവരാണ് ആദിവാസികൾ. ആരോഗ്യം, വിദ്യാഭ്യാസം, തൊഴിൽ മുതലായ മേഖലകളിൽ പുരോഗമനത്തിന്റെ പടവുകൾ കേറാൻ അവർ കഴിവതും ശ്രമിക്കുന്നു. ഭക്ഷണം, വസ്ത്രം, ഭാഷ മുതലായ കാര്യത്തിൽ നാട്ടിലെ ജനങ്ങളുടേതുമായി ഒത്തുപോ കാൻ ആദിവാസികൾക്ക് ആഗ്രഹമുണ്ട്. പക്ഷേ എത്ര പരിഷ്കരണപ്രസ്ഥാനങ്ങളും, പുനരധിവാസമ ങ്ങളും, കടന്നുപോയിട്ടും, ആദിവാസികൾക്ക് പഴയനിലയിൽ നിന്നും ഗണ്യമായ പുരോഗതിയൊന്നും ഇന്നും ഉണ്ടായിട്ടില്ല എന്നതാണ് യാഥാർത്ഥ്യം. അതിജീവനത്തിനായി അതികഠിനമായി അധ്വാനിക്കുന്നവ രാണ് ഇന്നത്തെ ഗോത്രവർഗ്ഗക്കാർ. പക്ഷേ ചെമ്മീൻ ചാടിയാൽ ചട്ടിയോളം എന്ന അവസ്ഥയാണ് അവ രുടേത്. ഇത്തരമൊരു ദയനീയാവസ്ഥയ്ക്ക് മാറ്റം വരണമെങ്കിൽ പുതിയ ക്ഷേമപദ്ധതികൾ യാഥാർത്ഥ്യ മാക്കാൻ കഴിയണം.

Question 5.
“കന്നാരംപുഴയിലെ കല്ലുകൾ ഉരസി ഉരസി
നരസിക്കു നരവീണു”
വരികളിലെ കാവ്യഭംഗിക്ക് കാരണമാകുന്ന ഘടകങ്ങൾ എന്തെല്ലാം? വിവരിക്കുക
Answer:
ഈ വരികൾ ഏട്ടക്കൂരി എന്ന കവിതയിലേതാണ്. ഈ വരികൾക്ക് ആശയഭംഗിയും രൂപഭംഗിയുമുണ്ട്. പുഴയൊഴുകുമ്പോൾ കല്ലുകൾ മിനുസപ്പെടുകയും, അവയുടെ നിറം മാറുകയും ചെയ്യുക സ്വാഭാവികമാ ണ്. പുഴയുടെ അനേക കാലമായുള്ള ഒഴുക്കിന്റെ നൈരന്ത്യം ഇവിടെ വ്യക്തമാകുന്നു. കല്ലിനെപ്പോലും വെളുപ്പിക്കുന്ന ഒഴുക്കിന്റെ ശക്തിയും ഇവിടെ ധ്വനിക്കുന്നു. കൂടാതെ “ക ‘ന’ മുതലായ അക്ഷരങ്ങ ളുടെ ആ വർത്തനവും ഈ വരികളുടെ ഭംഗി വർധിപ്പിക്കുന്നുണ്ട്.

Question 6.
കേരളത്തിൽ അനവധി ഗോത്രഭാഷകളുള്ളതിൽ ഒന്നിലാണ് നാം പരിചയപ്പെട്ട ‘കെത്ത്’ എന്ന കവിത രചിച്ചിട്ടുള്ളത്. ഇതുപോലെ മറ്റു ഗോത്രഭാഷകളിൽ നിന്ന് ലഭ്യമാകുന്ന രചനകളും വാമൊഴിപ്പാട്ടുകളും ശേഖരിച്ച് പാരമ്പര്യവാദ്യങ്ങളും സംഗീതോപകരണങ്ങളും പ്രയോജനപ്പെടുത്തി പാട്ടരങ്ങ് സംഘടിപ്പിക്കുക. ശേഖരിച്ചവ പതിപ്പാക്കുകയും വേണം.
Answer:
ഉത്തരസൂചന :
കേരളത്തിൽ അനേകം ഗോത്രഭാഷകളുണ്ട്. അവയിൽ പ്രധാനപ്പെട്ട ഒന്നാണ് വയനാട്ടിലെ ആദിവാസികൾ സംസാരിക്കുന്ന റാവുള ഭാഷ. ഈ ഭാഷയിലാണ് കെത്തളു എന്ന കവിത എഴുതിയിരിക്കുന്നത്. കേരള ത്തിലെ പ്രധാന ഗോത്രവർഗ്ഗക്കാർ ചോലനായ്ക്കൻമാർ, കാട്ടുനായ്ക്കർ, കാടർ കുറുമ്പർ, കൊറഗർ, കുറി ച്യർ, മുതുവർ മുതലായവയാണ്. മുതുവർഗോത്രഭാഷ, ഇരുളഗോത്രഭാഷ, ‘റാവുള ഭാഷ മുതലായവ യാണ് പ്രധാന ഗോത്രഭാഷകൾ. അശോകൻ മറയൂർ, സുകുമാരൻ ചാലി ഗദ്ധ, ധന്യവേങ്ങച്ചേരി, ലിജിന കടുമേനി, ബിന്ദു ഇരുളം, ശാന്തിപനയ്ക്കൽ മുതലായവരാണ് പ്രധാന ഗോത്രഭാഷാകവികൾ: ഇവരുടെ കവിതകൾ പലതും യൂട്യൂബിൽ ലഭ്യമാണ്. പാട്ടരങ്ങ് സംഘടിപ്പിക്കുവാനായി ഈ കവിതകൾ ശേഖരി ക്കാവുന്നതാണ്.

Question 7.
പരിസ്ഥിതി പ്രമേയമായ കവിതകൾ കോർത്തിണക്കി ഓഡിയോ ആൽബം നിർമ്മിക്കുക. സ്കൂളിലെ റെക്കോർഡിങ്, ഐ.ടി. സംവിധാനങ്ങൾ ഉപയോഗപ്പെടുത്തി ക്ലാസിൽ അവതരിപ്പിക്കുക.
Answer:
പരിസ്ഥിതി പ്രമേയപരമായ കവിതകൾ അതാതു പുസ്തകങ്ങളിൽ നിന്നോ, യൂട്യൂബിൽ നിന്നോ ശേഖരിക്കാം. ഓരോ കവിതയും റിക്കോർഡ് ചെയ്യുമ്പോൾ കവിയുടെ പേര്, കവിതയുടെ ഉള്ളടക്കം, രചനയുടെ വർഷം പാടിയ വ്യക്തിയുടെ പേര് മുതലായവിവരങ്ങളും സൂചിപ്പിക്കേണ്ടതാണ്.

കെത്തളു Extra Questions and Answers

അധികചോദ്യങ്ങളും ഉത്തരങ്ങളും
Question 1.
മരബാവു എന്ന ഗോത്രഭാഷാ പദം ഏത് അർഥ ത്തിലാണ് പ്രയോഗിക്കാറുള്ളത്
(a) മരങ്ങൾ
(b) മരത്തണൽ
(c) മരച്ചീനി
(d) മരവാഴ
Answer:
(d) മരവാഴ

Question 2.
‘ഴ’ എന്ന അക്ഷരത്തിന് പകരം ഗോത്രഭാഷയിൽ കടന്നു വരുന്ന അക്ഷരം ഏത്?
(a) യ
(b) ര
(c) വ
(d) ല
Answer:
(c) വ

Question 3.
‘ഉരച്ചി’ എന്ന ഗോത്രഭാഷാപദത്തിന് സമാനമായ മലയാളപദം ഏത്?
(a) ഉരസി
(b) ഉരച്ചു
(c) ഉരൽ
(d) ഉറങ്ങി
Ans.
(a) ഉരസി

Question 4.
“പു പെട്ടി നീന്തി ബാളെമ്മു
കരെ ബെട്ടിചാടിന്റെ കെളുക്കുമ്മു
എന്ന ചാന്തു” ഗോത്ര കവിതയുടെ ആസ്വാദ്യ തയ്ക്ക് ഉദാഹരണമാണ് ഈ വരികൾ. നിങ്ങളുടെ നിരീക്ഷണം കുറിക്കുക.
Answer:
വയനാട്ടിലെ ഗോത്ര ഭാഷകളിലൊന്നായ റാവുള യിൽ ആണ് ഈ കവിത രചിച്ചിരിക്കുന്നത്. ഗോത കവിതകളുടെ വിഷയം ഗോത്ര ജീവിതവുമായി ദൈനം ദിനം ബന്ധപ്പെടുന്ന വിഷയം തന്നെ യാവും. കാട്ടാനയുടെ പ്രസവമോ, മീൻപിടിത്ത ത്തിന്റെ വിശേഷങ്ങളോ, ആഹാരസമ്പാദന ത്തിന്റെ പ്രശ്നങ്ങളോ ഒക്കെയാവാം കാവ്യ വിഷയം. എന്തു വിഷയമായാലും ഗോത്രകവിത യിലും കാവ്യാത്മകമായ സൗന്ദര്യം പ്രകടമാണ്. കവിത സാർവജനീനമായതുകൊണ്ടുതന്നെ ഭാഷ കാര്യമായി വശമില്ലെങ്കിലും നമ്മൾ ആ കവിത ആസ്വദിച്ചു പോകുന്നു. ഗോത്ര ഭാഷയ്ക്ക് മല യാളത്തോടും തമിഴിനോടും കന്നഡത്തിനോടു മൊക്കെ പുലമുണ്ട്.
‘പുവെ ബെട്ടി നീന്തിണ ബാള’ എന്നതിന്
‘പുഴയ്ക്ക് കുറുകെ നീന്തുന്ന വാള’

എന്നാണർഥം. ‘കരെ ബെട്ടി ചാടിന്റെ കെത്തളു’ എന്നാൽ കരയിലേക്ക് ചാടുന്ന ഏട്ടക്കൂരി എന്ന അർഥം. ആറ്റിലെ വാള മീനുകൾ ആറ്റുവെള്ളത്തി ലൂടെ നീന്തി മറിയുന്നതിന്റെ ഭംഗി നമ്മെ ആസ്വദി പ്പിക്കാൻ ഈ വരികൾക്ക് കഴിയുന്നു. മീനുകളുടെ പ്രജനനകാലത്ത് ചില മത്സ്യങ്ങൾ കരയിൽ നിന്നും കരയിലേക്ക് ചാടിയുയർന്ന് വളരെ സുരക്ഷിത മായ ഒരിടത്തു കെട്ടിനിൽക്കുന്ന വെള്ളത്തിൽ മുട്ട യിടാൻ ശ്രമിക്കും. അക്കാര്യമാണ് ഇവിടെ സൂചി പ്പിക്കുന്നത്. മീനുകളുടെ ചാട്ടം ഏറ്റവും നല്ല ദൃശ്യാനുഭവമാണല്ലോ.