Mathematics Part I Solutions Solutions for Class 9 Maths Chapter 4 Irrational Numbers are provided here with simple step-by-step explanations. These solutions for Irrational Numbers are extremely popular among Class 9 students for Maths Irrational Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part I Solutions Book of Class 9 Maths Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part I Solutions Solutions. All Mathematics Part I Solutions Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.
Page No 60:
Question 1:
The hypotenuse of a right angled triangle is 3.5 metres long and another of its sides is 2.5 metres long.
Calculate its perimeter correct to centimetres.
Answer:
Hypotenuse of the given right-angled triangle = 3.5 m
Length of the second side = 2.5 m
Let the length of the third side be a m.
By Pythagoras theorem, we have:
We know that perimeter of a triangle is equal to the sum of lengths of all its sides.
∴ Perimeter of the given right-angled triangle ≈ (3.5 + 2.5 + 2.45) m = 8.45 m
Thus, the perimeter of the given right-angled triangle is approximately equal to 8.45 m.
Page No 60:
Question 2:
Compute the perimeter of the quadrilateral with measures as given below, correct to centimetres:
Answer:
In ΔADC, using Pythagoras theorem, we have:
In ΔABC, ∠BAC =∠BCA = 60°
Using angle sum property:
∠ABC + ∠BCA + ∠BAC = 180°
⇒ ∠ABC + 60° + 60° = 180°
⇒ ∠ABC = 180° − 120° = 60°
Therefore, ΔABC is an equilateral triangle.
⇒ AB = BC = CA ≈ 1.41 m
We know that perimeter of a quadrilateral is equal to the sum of lengths of all its sides.
∴ Perimeter of quadrilateral ABCD ≈ (1 + 1 + 1.41 + 1.41) m = 4.82 m
Thus, the perimeter of the given quadrilateral is approximately equal to 4.82 m.
Page No 60:
Question 3:
The sides of a square are 4 centimetres long. The midpoints of the sides are joined to form another square as shown below
What is the perimeter of the smaller square?
Answer:
Given: ABCD and PQRS are squares.
P, Q, R and S are the mid points of the sides AD, AB, BC and CD of the square ABCD.
AB = BC = CD = DA = 4 cm
P and Q are the midpoints of AD and AB respectively.
Now, in ΔAPQ, ∠PAQ = 90° (Angle of a square)
So, by using Pythagoras theorem, we have:
Now, PQRS formed inside the square ABCD is also a square.
∴ PQ = QR = RS = SP ≈ 2.83 cm
We know that perimeter of a square is equal to four times the length of its side.
∴ Perimeter of square PQRS ≈ 4 × 2.83 cm = 11.32 cm
Thus, the perimeter of square PQRS is approximately equal to 11.32 cm.
Page No 61:
Question 1:
The figure below shows a square, each of whose sides is 3 centimetres long. Each side is divided into three equal parts and these points are joined to form an octagon:
What is the perimeter of the octagon?
Answer:
Given: PQRS is a square.
PQ = QR = RS = SP = 3 cm
Each side of the square PQRS is divided into three equal parts.
Similarly, CD = FE = GH = 1 cm
Now, in ΔAHP, ∠APH = 90° (Angle of a square)
So, by using Pythagoras theorem, we have:
Similarly, BC = DE = FG ≈ 1.41 cm
We know that perimeter of an octagon is equal to the sum of lengths of all its sides.
∴ Perimeter of the octagon ABCDEFGH ≈ (1 + 1.41 + 1 + 1.41 + 1 + 1.41 + 1 + 1.41) cm
= 9.64 cm
Thus, the perimeter of the octagon is approximately equal to 9.64 cm.
Page No 64:
Question 1:
For each of the products below, find out whether the answer is rational or irrational.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Answer:
We know that for any two positive numbers x and y,
(i)
We know that is an irrational number.
∴is an irrational number.
(ii)
We know that 2 is a rational number.
∴is a rational number.
(iii)
We know that 6 is a rational number.
∴is a rational number.
(iv)
We know that is an irrational number.
∴is an irrational number.
(v)
We know that 0.6 is a rational number.
∴is a rational number.
(vi)
We know that 5 is a rational number.
∴is a rational number.
Page No 64:
Question 2:
Find the larger one of each pair below, doing the computations in head.
(i)
(ii)
(iii)
(iv)
Answer:
(i)
As 18 > 12, > .
⇒>
Thus, is larger than.
(ii)
As 45 > 44, > .
⇒>
Thus, is larger than.
(iii)
As 0.5 > 0.2 , > .
⇒>
Thus, is larger than.
(iv)
As > , > .
⇒>
Thus, is larger than.
Page No 64:
Question 3:
Which is the largest of ?
Answer:
As 32 > 27 > 24,>>
⇒> >
Thus, the largest number among the given numbers is .
Page No 65:
Question 1:
Compute the length of the diagonal of a square of side 10 metres, correct to centimetres.
Answer:
Length of the side of the square = 10 m
Let the length of the diagonal be a m.
We know that each angle in a square measures 90°.
By applying Pythagoras theorem in the right-angled triangle formed by joining the diagonal, we have:
Thus, the length of the diagonal is approximately equal to 14.1 m.
Page No 65:
Question 2:
Without using calculating devices, compute the following correct to two decimals.
(i)
(ii)
(iii)
Answer:
(i)
Thus, is approximately equal to 11.28.
(ii)
Thus, is approximately equal to 29.61.
(iii)
Thus, is approximately equal to 1.73.
Page No 65:
Question 3:
What is the total length of the line shown below?
Answer:
Length of the first part of the line segment = cm
Length of the second part of the line segment = cm
Total length = cm + cm
Thus, the length of the given line segment is approximately equal to 5.64 cm.
Page No 65:
Question 4:
A, B, C are three points such that. Do they lie on a straight line?
Answer:
Length of line segment AB = cm
Length of line segment BC = cm
Length of line segment CA = cm
For three points, A, B and C to lie on a straight line, the sum of the lengths of line segments AB and BC should be equal to the length of line segment CA.
AB + BC
CA =
⇒ AB + BC = CA
Thus, the given points lie on a straight line.
Page No 68:
Question 1:
Using correct to three decimals.
Answer:
≈ 1.732
⇒
= 0.866
∴ is approximately equal to 0.866.
Page No 68:
Question 2:
Using correct three decimals.
Answer:
≈ 3.162
⇒
= 0.316
∴ is approximately equal to 0.316.
Page No 68:
Question 3:
Using correct to three decimals.
Answer:
≈ 2.449
∴ is approximately equal to 2. 041.
Page No 68:
Question 4:
Compute up to two decimals.
Answer:
≈ 1.73
⇒
= 2.89
∴ is approximately equal to 2.89.
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