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 Meritnationâ€™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.

#### 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.

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.

#### Question 2:

Compute the perimeter of the quadrilateral with measures as given below, correct to centimetres:

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.

#### 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?

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.

#### 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?

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.

#### Question 1:

For each of the products below, find out whether the answer is rational or irrational.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

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.

#### Question 2:

Find the larger one of each pair below, doing the computations in head.

(i)

(ii)

(iii)

(iv)

(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.

#### Question 3:

Which is the largest of ?

As 32 > 27 > 24,>>

> >

Thus, the largest number among the given numbers is .

#### Question 1:

Compute the length of the diagonal of a square of side 10 metres, correct to centimetres.

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.

#### Question 2:

Without using calculating devices, compute the following correct to two decimals.

(i)

(ii)

(iii)

(i)

Thus, is approximately equal to 11.28.

(ii)

Thus, is approximately equal to 29.61.

(iii)

Thus, is approximately equal to 1.73.

#### Question 3:

What is the total length of the line shown below?

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.

#### Question 4:

A, B, C are three points such that. Do they lie on a straight line?

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.

#### Question 1:

Using correct to three decimals.

≈ 1.732

= 0.866

is approximately equal to 0.866.

#### Question 2:

Using correct three decimals.

≈ 3.162

= 0.316

is approximately equal to 0.316.

#### Question 3:

Using correct to three decimals.

≈ 2.449

is approximately equal to 2. 041.

#### Question 4:

Compute up to two decimals.