Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 5 Irrational Multiplication Extra Questions and Answers Notes to clear their doubts.
Kerala SCERT Class 9 Maths Chapter 5 Solutions Irrational Multiplication
Irrational Multiplication Class 9 Kerala Syllabus Questions and Answers
Kerala State Syllabus 9th Standard Maths Chapter 5 Irrational Multiplication Solutions Questions and Answers
Class 9 Maths Chapter 5 Kerala Syllabus – Multiplication
Textual Questions And Answers
Question 1.
We can make a rectangle using four equilateral triangles of the same size by cutting two of them along their heights and rearranging these pieces and the other two whole triangles:
If the sides of all equilateral triangles are 2 centimetres, what is the perimeter and area of the rectangle?
Answer:
If we consider the sides of the triangles, we will get it as follows:
So, the sides of the given rectangle are:
Now,
Perimeter of the rectangle = 2(length + breadth)
= 2(√3 + √3 + 2)
= 2(2√3 + 2)
= 4(√3 + 1)
= 4(1.732 + 1)
= 4 × 2.732
= 10.928 cm
Area of the rectangle = length × breadth
= 2√3 × 2
= 4√3 – 4 × 1.732
= 6.928 cm²
Question 2.
We can make a trapezium by cutting a square and an equilateral triangle with sides twice that of the square, and rearranging the pieces as below:
If the side of the square is 2 centimeters, what is the perimeter and area of the trapezium?
Answer:
Consider one part of the square. Its measures are:
Consider one part of the triangle. Its measures are:
Now, the measures of the trapezium are;
Perimeter of the trapezium = 2√2 + 2 + 2√3 + 2 + 2√2 + 2√3
= 4 + 4√2 + 4√3
= 4(1 + √2 + √3)
= 4(1 + 1.414+ 1.732)
= 4 × 4.146
= 16.584 cm
Area of the trapezium = \(\frac{1}{2}\) × sum of the lengths of the parallel sides x height
= \(\frac{1}{2}\) × (2 + 2√3 +2 + 2√3) × 2
= 4 + 4√3
= 4(1 + √3)
= 4(1 + 1.732)
= 4 × 2.732
= 10.928 cm²
Question 3.
The picture shows the figure formed by joining two squares
Calculate the length of the bottom side of this figure, correct to a centimetre.
Answer:
Side length of the big square = √27
= \(\sqrt{9 \times 3}\)
= √9 × √3
= 3 × √3
= 3 × 1.73
= 5.19 cm
Side length of the small square = √12
= \(\sqrt{4 \times 3}\)
= √4 × √3
= Side length of the big square + Side length of the small square
= 5.19 + 3.46
= 8.65 cm
Question 4.
The figure shows two squares with two corners joined:
Find the length of the slanted line.
Answer:
We know that the diagonal length of a square with side length a is a√2.
Diagonal length of the square with side 3 = 3√2 cm
Diagonal length of the square with side 2 = 2√2 cm
Length of the slanted line
= Diagonal length of the square with side 3 + Diagonal length of the square with side 2
= 3√2 + 2√2
= 5√2
≈ 5 x 1.414
≈ 7.07 cm
Question 5.
Calculate the length of the third side of the right triangle in the picture and also its perimeter.
Answer:
Side length of the big square = \(\sqrt{50}\)
= \(\sqrt{25 \times 2}\)
= \(\sqrt{25}\) × √2
= 5 × √2
≈ 5 × 1.414
≈ 7.07 cm
Side length of the small square = \(\sqrt{18}\)
= \(\sqrt{9 \times 2}\)
= √9 × √2
= 3 × √2
≈ 3 × 1.414
≈ 4.242 cm
Using Pythagoras theorem,
Third side of the right triangle
= \(\sqrt{(\text { side length of the big square })^2-(\text { side length of the small square })^2}\)
= \(\sqrt{(5 \sqrt{2})^2-(3 \sqrt{2})^2}\)
= \(\sqrt{(25 \times 2)-(9 \times 2)}\)
= \(=\sqrt{2(25-9)}\)
= \(=\sqrt{2 \times 16}\)
= √2 × √16
= √2 × 4
= 4√2
Perimeter of the right triangle = 5√2 + 3√2 + 4√2
= 12√2
≈ 12 × 1.414
≈ 16.968 cm
Question 6.
The product of some of the pairs of numbers below are natural numbers or fractions. Find those pairs.
(i) √3, √12
Answer:
√3 × √12 = √3 × \(\sqrt{3 \times 4}\)
= √3 × √3 × √4
= 3 × 2
= 6
Product is a natural number.
(ii) √3, √1.2
Answer:
\(\sqrt{3} \times \sqrt{1.2}=\sqrt{3 \times 1.2}\)
= \(\sqrt{3.6}\)
(iii) √5, √8
Answer:
\(\sqrt{5} \times \sqrt{8}=\sqrt{5 \times 8}\)
= \(\sqrt{40}\)
= \(\sqrt{4 \times 10}\)
= √4 × √10
= 2 × √10
= 2√10
(vi) √0.5, √8
Answer:
√0.5 × √8 = \(\sqrt{0.5 \times 8}\)
= √4
= 2
Product is a natural number.
(v) \(\sqrt{7 \frac{1}{2}}, \sqrt{3 \frac{1}{3}}\)
Answer:
\(\sqrt{7 \frac{1}{2}} \times \sqrt{3 \frac{1}{3}}=\sqrt{\frac{15}{2}} \times \sqrt{\frac{10}{3}}\)
= \(\sqrt{\frac{15}{2} \times \frac{10}{3}}\)
= \(\sqrt{5 \times 5}\)
= \(\sqrt{25}\)
= 5
Product is a fraction.
(vi) \(\sqrt{\frac{2}{5}}, \sqrt{\frac{1}{10}}\)
Answer:
\(\sqrt{\frac{2}{5}} \times \sqrt{\frac{1}{10}}=\sqrt{\frac{2}{5} \times \frac{1}{10}}=\sqrt{\frac{1}{25}}=\frac{1}{5}\)
Product is a natural number.
Class 9 Maths Kerala Syllabus Chapter 5 Solutions – (Division)
Textual Questions And Answers
Question 1.
Prove that (√2 + 1) (√-2 – 1) = 1. Using this:
i. Compute \(\frac{1}{\sqrt{2}-1}\) up to two decimal places.
ii. Compute \(\frac{1}{\sqrt{2}+1}\) up to two decimal places.
Answer:
(√2 + 1)(√2 – 1) = (√2)² – 1²
= 2 – 1
= 1
i. \(\frac{1}{\sqrt{2}-1}=\frac{(\sqrt{2}+1)(\sqrt{2}-1)}{\sqrt{2}-1}\)
= √2 + 1
= 1.41 + 1
= 2.41
ii. \(\frac{1}{\sqrt{2}+1}=\frac{(\sqrt{2}+1)(\sqrt{2}-1)}{\sqrt{2}+1}\)
= √2 – 1
= 1.41 – 1
= 0.41
Question 2.
Compute the lengths of the sides of the equilateral triangle shown below, correct to a millimetre.
Answer:
Consider the triangle given below.
Using Pythagoras theorem;
(2x)² – x² = 4²
4x² – x² = 16
3x² = 16
x² = \(\frac{16}{3}\)
x = \(\sqrt{\frac{16}{3}}\)
= \(\frac{\sqrt{16}}{\sqrt{3}}\)
= \(\frac{4}{1.732}\)
= 2.30 cm
Length of the sides = 2x
= 2 × 2.30
= 4.60 cm
Question 3.
All red triangles in the picture are equilateral and of the same size. What is the ratio of the sides of the outer and inner squares?
Answer:
Let ‘a’ be the side length of the equilaleral triangle
Side length of the outer square = a + a√3 = a( 1 + √3)
Side length of the inner square = a + a√3 – (a + a)
= a + a√3 – 2a
= a√3 – a
= a(√3 – 1)
Ratio of the sides of the outer and inner squares
= \(\frac{\text { Side length of the outer square }}{\text { Side length of the inner square }}\)
= \(\frac{\mathrm{a}(1+\sqrt{3})}{\mathrm{a}(\sqrt{3}-1)}\)
= \(\frac{\sqrt{3}+1}{\sqrt{3}-1}\)
= (√3 + 1) (√3 – 1)
Question 4.
Prove that \(\sqrt{2 \frac{2}{3}}=2 \sqrt{\frac{2}{3}}\) and \(\sqrt{3 \frac{3}{8}}=3 \sqrt{\frac{3}{8}}\) Find other numbers like this.
Answer:
The general form of such numbers can be defined as follows;
Question 5.
Among the pairs of numbers given below, find those for which the quotient of the first by the second is a natural number or a fraction.
(i) √72, √2
(ii) √27, √3
(iii) √125, √50
(iv) √10, √2
(v) √20, √5
(vi) √18, √8
Answer:
(i) \(\frac{\sqrt{72}}{\sqrt{2}}=\frac{\sqrt{36 \times 2}}{\sqrt{2}}=\frac{\sqrt{36} \times \sqrt{2}}{\sqrt{2}}\)
= \(\sqrt{36}\)
= 6
When we divide the first by the second the result is natural number.
(ii) \(\frac{\sqrt{27}}{\sqrt{3}}=\frac{\sqrt{9 \times 3}}{\sqrt{3}}=\frac{\sqrt{9} \times \sqrt{3}}{\sqrt{3}}\)
= √9
= 3
When we divide the first by the second the result is natural number.
(iii) \(\frac{\sqrt{125}}{\sqrt{50}}=\frac{\sqrt{25 \times 5}}{\sqrt{25 \times 2}}\)
= \(\frac{\sqrt{25} \times \sqrt{5}}{\sqrt{25} \times \sqrt{2}}\)
= \(\frac{\sqrt{5}}{\sqrt{2}}\)
(iv) \(\frac{\sqrt{10}}{\sqrt{2}}=\frac{\sqrt{2 \times 5}}{\sqrt{2}}\)
= \(\frac{\sqrt{2} \times \sqrt{5}}{\sqrt{2}}\)
= √5
(v) \(\frac{\sqrt{20}}{\sqrt{5}}=\frac{\sqrt{4 \times 5}}{\sqrt{5}}\)
= \(\frac{\sqrt{4} \times \sqrt{5}}{\sqrt{5}}\)
= √4
= 2
When we divide the first by the second the result is a Natural fraction.
(vi) \(\frac{\sqrt{18}}{\sqrt{8}}=\frac{\sqrt{9 \times 2}}{\sqrt{4 \times 2}}=\frac{\sqrt{9} \times \sqrt{2}}{\sqrt{4} \times \sqrt{2}}\)
= \(\frac{\sqrt{9}}{\sqrt{4}}=\frac{3}{2}\)
When we divide the first by the second the result is a fraction.
SCERT Class 9 Maths Chapter 5 Solutions – Areas of Triangles
Textual Questions And Answers
Question 1.
For each of the lengths below, calculate the area of the equilateral triangle with that as the lengths of the sides:
(i) 10 cm
(ii) 5 cm
(iii) √3 cm
Answer:
(i) Here, a = 10
Area = \(\frac{\sqrt{3}}{4}\) × a
= \(\frac{\sqrt{3}}{4}\) × 10²
= \(\frac{\sqrt{3}}{4}\) × 100
= √3 × 25
~ 1.732 × 25
~ 43.3 cm²
(ii) Here, a = 5
Area = \(\frac{\sqrt{3}}{4}\) × a²
= \(\frac{\sqrt{3}}{4}\) × 5²
= \(\frac{\sqrt{3}}{4}\) × 25
= √3 × 6.25
~ 6.25 × 1.732
~ 10.825 cm²
(iii) Here, a = √3
Area = \(\frac{\sqrt{3}}{4}\) × a²
= \(\frac{\sqrt{3}}{4}\) × (√3)²
= \(\frac{\sqrt{3}}{4}\) × 3
= √3 × 0.75
~ 1.732 × 0.75
~ 1.299 cm²
Question 2.
Calculate the area of the regular hexagon with lengths of the sides 6 centimetres.
Answer:
A regular hexagon can be divided into six equilateral triangles. In such cases; side length of the equilateral triangle = side length of the regular hexagon
So,
Area of the regular hexagon
= sum of the areas of these six equilateral triangles
= 6 × area of one equilateral triangle
= 6 × \(\frac{\sqrt{3}}{4}\) × 6²
= 6 × √3 × 9
= 54 × √3
= 54 × 1.732
= 93.528 cm²
Question 3.
Calculate the perimeter and area of the equilateral triangle with height 12 centimetres.
Answer:
We know that height of an equilateral triangle of side length a is \(\frac{\sqrt{3}}{4}\) × a.
Here, height = 12 cm
ie, \(\frac{\sqrt{3}}{4}\) × a = 12
a = \(=\frac{12 \times 2}{\sqrt{3}}\)
= \(\frac{3 \times 4 \times 2}{\sqrt{3}}\)
Perimeter of the equilateral triangle = 3 × 8√3
= 24√3
≈ 24 × 1.732
≈ 41.568 cm
Area of the equilateral triangle = \(\frac{\sqrt{3}}{4}\) × a²
= \(\frac{\sqrt{3}}{4}\) × (8√3)²
= \(\frac{\sqrt{3}}{4}\) × 64 × 3
= √3 × 16 × 3
≈ 1.732 × 48
≈ 83.136 cm²
Question 4.
Calculate the perimeter and area of the regular hexagon with the distance between parallel sides 6 centimetres.
Answer:
Here,
distance between parallel sides = 2 × height of an equilateral triangle
6 = 2 × \(\frac{\sqrt{3}}{2}\) × a
= √3 × a
a = \(\frac{6}{\sqrt{3}}\)
= \(\frac{3 \times 2}{\sqrt{3}}\)
= \(\frac{\sqrt{3} \times \sqrt{3} \times 2}{\sqrt{3}}\)
= √3 × 2
= 2√3 cm
Perimeter of the regular hexagon = 6 × 2√3
= 12√3
≈ 12 × 1.732
≈ 20.784 cm
Area of the regular hexagon = 6 × area of one equilateral triangle
= 6 × \(\frac{\sqrt{3}}{4}\) × (2√3)²
= 6 × \(\frac{\sqrt{3}}{4}\) × 4 × 3
= 18 × √3
≈ 18 × 1.732
≈ 31.176 cm²
Question 5.
Calculate the height and area of the triangle with sides 8 centimetres, 6 centimetres,6 centimetres.
Answer:
Consider the picture given below;
Using Pythagoras theorem, height
= \(\sqrt{6^2-4^2}\)
= \(\sqrt{36-16}\)
= \(\sqrt{20}\)
= \(\sqrt{4 \times 5}\)
= √4 × √5
= 2 × √5
= 2√5 cm
Area = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × 8 × 2√5
= 8√5 cm²
Question 6.
For each of the set of three lengths given below, calculate the area of the triangle with these as the lengths of sides:
(i) 4 cm, 5 cm, 7cm
(ii) 4cm, 13 cm, 15 cm
(iii) 5 cm, 12 cm, 13 cm
Answer:
If the lengths of the sides of a triangle are a, b, c and we take the semi perimeter
s = \(\frac{1}{2}\) (a + b + c), then the area of the triangle is \(\sqrt{s(\mathrm{~s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}\)
(i) Here, a = 4 cm
b = 5 cm
c = 7 cm
(ii) Here, a = 4 cm
b = 13 cm
c = 15 cm
(iii) Here, a = 5 cm
b = 12 cm
c = 13 cm
Irrational Multiplication Class 9 Extra Questions and Answers Kerala Syllabus
Question 1.
Multiply 3√8 and 7√8
Answer:
3√8 × 7√8 = 3 × √8 × 7 × √8
= 3 × 7 × √8 × √8
= 21 × 8
= 168
Question 2.
Write in ascending order.
(i) 3√5 and 4√3
(ii) 2√5, 5√2 and 3√7
Answer:
(i) \(\sqrt{53}=\sqrt{3 \times 3 \times 5}=\sqrt{45}\)
4√3 = \(\sqrt{4 \times 4 \times 3}=\sqrt{48}\)
48 > 45 ⇒ \(\sqrt{48}>\sqrt{45}\)
⇒ 4√3 > 3√5
When arrange in ascending order; 3√5, 4√3
(ii) 2√5 = √2 × 2 × 5 = √20
5√2 = √5 × 5 × 2 = √50
3√7 = √3 × 3 × 7 = √63
20 < 50 < 63 ⇒ √20 <√50 < √63
⇒ 2√5 < 5√2 < 3√7
When arrange in ascending order; 2√5, 5√2, 3√7
Question 3.
x = \(\sqrt{0.5}\), y = \(\sqrt{32}\), z = \(\sqrt{128}\)
(a) Find xy, yz and xz.
Answer:
(b) Find xy + yz + xz
Answer:
xy + yz + xz = 4 + 64 + 8
= 76
(c) Prove that y = 8x
Answer:
8x = 8 x \(\sqrt{0.5}\)
= \(\sqrt{64 \times 0.5}\)
= \(\sqrt{32}\)
= y
Question 4.
√8 can be written as \(\sqrt{4 \times 2}\) = 2√2
a) Write √18, √32 as the product of an integer and √2
Answer:
\(\sqrt{18}=\sqrt{9 \times 2}\) = 3√2
\(\sqrt{32}=\sqrt{16 \times 2}\) = 4√2
b) Write the simplified form of √2 + √8 + √18 – √32
Answer:
\(\sqrt{2}+\sqrt{8}+\sqrt{18}-\sqrt{32}\)
= √2 + 2√2 + 3√2 – 4√2
= 6√2 – 4√2
= 2√2
Question 5.
Find the area of equilateral triangle if perimeter is 18 cm.
Answer:
Perimeter =18
3 × side = 18
side = \(\frac{18}{3}\)
= 6 cm
Area = \(\frac{\sqrt{3}}{4}\) × (side)²
= \(\frac{\sqrt{3}}{4}\) × 6²
= √3 × 9
= 9√3 cm²