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 Meritnationâ€™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 Δ*ADC*, using Pythagoras theorem, we have:

In Δ*ABC*, ∠*BAC* =∠*BCA* = 60°

Using angle sum property:

∠*ABC* + ∠*BCA* + ∠*BAC* = 180°

⇒ ∠*ABC* + 60° + 60° = 180°

⇒ ∠*ABC* = 180° − 120° = 60°

Therefore, Δ*ABC* 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 Δ*APQ*, ∠*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 Δ*AHP*, ∠*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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