Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 2 New Numbers Extra Questions and Answers Notes to clear their doubts.
Kerala SCERT Class 9 Maths Chapter 2 Solutions New Numbers
New Numbers Class 9 Kerala Syllabus Questions and Answers
Kerala State Syllabus 9th Standard Maths Chapter 2 New Numbers Solutions Questions and Answers
Class 9 Maths Chapter 2 Kerala Syllabus – Lengths and Numbers
Intext Questions And Answers
Question 1.
Look at this picture.
The diagonal of the smaller square makes it into two isosceles triangles.
A large square is made up of four similar triangles. Then the area of the larger square is twice the area of the smaller square.
- The square of any fraction is not equal to 2.
- There are many lengths that cannot be expressed as either a natural number or a fraction.
Eg: The length of the diagonal of a square whose sides are all 1 cannot be expressed as a fraction.
Textual Questions And Answers
Question 1.
We have seen in class 8 that any odd number greater than 1 can be written as the difference of two perfect squares. Use this to draw squares of area 7 square centimetres and 11 square centimetres. What are the lengths of the sides of these squares?
Answer:
Any odd number can be written in the form of 2x + 1.
2x + 1 = (x + 1)² – x²
If we take 7 = 2x + 1, we get,
x = \(\frac{7-1}{2}\) = 3
Also,
7 = (x + 1)² – x²
= 4² – 3²
Length of one side of a square = √7 cm.
Similarly, 11 = 2x + 1
Then,
x = \(\frac{11-1}{2}\) = 2
11 = (x + 1)² – x²
= 6² – 5²
Length of one side of a square = √11 cm.
Question 2.
What is the area of the square in the picture? What is the length of its sides?
Answer:
Area of the square = (3.5)² – (1.5)²
= 12.25 – 2.25 = 10 cm²
Length of one side = √10 cm
Question 3.
Calculate the area of the square and the length of its sides in each of the picture below:
Answer:
Area of square = (√3)3 = 3 cm²
Area of square = (√4)2 = 4 cm²
Area of square = (√5)2 = 5 cm²
Question 4.
Find three fractions greater than √2 and less than √3 .
Answer:
√2 = 1.414…
√3 = 1.732…
1.414……… < 1.5 < 1.6 < 1.7 < 1.732…
1.414……… < \(\frac{15}{10}<\frac{16}{10}<\frac{17}{10}\) < 1.732…
√2 < \(\frac{3}{2}<\frac{8}{5}<\frac{17}{10}\) < √3
Class 9 Maths Kerala Syllabus Chapter 2 Solutions – Addition And Subtraction
Intext Questions And Answers
Decimal numbers are represented by both integers and fractions. To avoid confusion between integers and fractions, they are separated by a period. Decimal numbers can also be negative to represent debts or losses.”
For eg.
Question 1.
What is the Perimeter of a right triangle of perpendicular sides 1 metre each?
Answer:
Perimeter = 1 + 1 + √2
= 2 + √2
= 2 + 1.41421…
= 3.41421…..
If we decide to have accuracy only up to a centimetre, we can take the perimeter as 3.41 metres; instead if we need accuracy up to a millimetre, we take the perimeter as 3.414 metres.
Question 2.
Calculate the perimeter of the triangle.
Answer:
Perimeter = 1 + √2 + √3
≅ 1 + 1.414 + 1.732
≅ 1 + 3.146
≅ 4.146 m
Textual Questions And Answers
Question 1.
The hypotenuse of a right triangle is 1\(\frac{1}{2}\) metres and one of the other sides is \(\frac{1}{2}\) metre. Calculate its perimeter, up to a centimetre.
Answer:
Question 2.
The picture below shows an equilateral triangle cut into two triangles along a line through the middle.
(i) What ¡s the perimeter of one of these?
(ii) How much is it less than the perimeter of the whole triangle?
Answer:
(i) By using Pythagoras theorem,
Length of the the third side = \(\sqrt{2^2-1^2}\)
= \(\sqrt{4-1}\)
= √3 m
Perimeter = 2 + 1 + √3 = 3 + √3 m
(ii) Perimeter of the whole triangle = 2 + 2 + 2 = 6m.
Difference in the perimeter = 6 – ( 3 + √3 ) = 6 – 3 – √3 = 3 – √3 m.
Question 3.
We’ve seen how we can go on drawing right triangles like this:
(i) What are the lengths of the sides of the tenth triangle ¡n this pattern?
(ii) How much more is the perimeter of the tenth triangle than that of the ninth?
Answer:
(i) Sides of first triangle = 1, 1, √2
Sides of 2nd triangle = 1, √2, √3
Sides of 3rd triangle = 1, √3 , √4
Sides of 9 th triangle = 1, √9 , √10
Sides of 10th triangle = 1, √10 , √11
(ii) Perimeter of 9th triangle = 1 + √9 + √10
Perimeter of 10th triangle = 1 + √10 + √11
Difference in the perimeter = (1 + √10 + √11 – (1 + √9 + √10)
= 1 + √10 + √11 – 1 – √9 – √10
= √11 – √9
= √11 – 3m
Question 4.
What is the hypotenuse of a right triangle with perpendicular sides √ centimetres and √3 centimetre? How much more is the sum of the perpendicular sides than the hypotenuse?
Answer:
In a right triangle
Hypotenuse² = base² + altitude²
= (√2)² + (√3)²
= 2 + 3
= 5
∴ Hypotenuse = √5 cm
Sum of vertical sides = √2 + √3
So the sum of the perpendicular sides √2 + √3 – √5 more than the hypotenuse.
New Numbers Class 9 Extra Questions and Answers Kerala Syllabus
Question 1.
In the figure ∠B =90° , AB = BC.
a) AB – BC = 1 cm. Then find the length of AC.
b) Find the perimeter of triangle ABC.
c) Find the area of triangle ABC.
d) If a square with side AC is drawn, then what is its Area?
Answer:
(a) AC = \(\sqrt{1^2+1^2}\) = √2 cm
(b) Perimeter = 1 + 1 + √2 = 2 + √2 ≈ 2 + 1.414 ≈ 3.414 cm
(c) Area = \(\frac{1}{2}\) × 1 × 1 = \(\frac{1}{2}\) cm²
(d) Area = 4 × \(\frac{1}{2}\) =2 cm²
Question 2.
If √3 + 1 and √3 – 1 are sides of a square, then
(a) Find its perimeter.
(b) Find its area.
(c) Find the length of its diagonal.
Answer:
(a) Perimeter = 2(√3 + 1 + √3 – 1)
= 2 × 2√3
= 4√3 ≈ 4 × 1.732
(b) Area = (√3 + 1)( √3 – 1) = (√3)² – 1²
= 3 – 1
= 2
(c) Length of its diagonal = \(\sqrt{(\sqrt{3}+1)^2+(\sqrt{3}-1)^2}\)
= \(\sqrt{(\sqrt{3})^2+2 \sqrt{3}+1+(\sqrt{3})^2-2 \sqrt{3}+1}\)
= \(\sqrt{3+3+2}\)
= √8
Question 3.
In an equilateral triangle one side is √3 + 1. Find its perimeter?
Answer:
Perimeter = 3 × side
= 3(√3 + 1)
= 3√3 + 3
≈ 3 × 1.732 + 3
= 5.196 + 3
= 8.196 cm
Question 4.
The area of a square is 3 cm,
(a) Find one of its length.
(b) Find its perimeter.
(c) Find the length of its diagonal.
Answer:
(a) One side = √3 cm
(b) perimeter = 4√3 cm
(c) Length of its diagonal = (√3)² + (√3)²
= √6 cm
Question 5.
Draw a square of area 10 cm²
Answer:
From the section Line and root, When x > 1 ,
x = \(\left(\frac{x+1}{2}\right)^2-\left(\frac{x-1}{2}\right)^2\)
So we can write,
10 = \(\left(\frac{10+1}{2}\right)^2-\left(\frac{10-1}{2}\right)^2\)
10 = (5.5)2 – (4.5)2
Draw as per the figure.
Here, the square of one side of the square is 10. Therefore, its area = 10 cm²