Mathematics Part I Solutions Solutions for Class 9 Math 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 Math 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 Math 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 Math 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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