Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 1.4 8th Std Maths Answers Solutions Chapter 1 Rational and Irrational Numbers.

## Practice Set 1.4 8th Std Maths Answers Chapter 1 Rational and Irrational Numbers

Question 1.

The number √2 is shown on a number line. Steps are given to show √3 on the number line using √2. Fill in the boxes properly and complete the activity.

The point Q on the number line shows the number ……….

A line perpendicular to the number line is drawn through the point Q. Point R is at unit distance from Q on the line.

Right angled ∆OQR is obtained by drawing seg OR.

l(OQ) = √2, l(QR) = 1

∴By Pythagoras theorem,

[l(OR)]² = [l(OQ)]² + [l(QR)]²

Draw an arc with centre O and radius OR. Mark the point of intersection of the line and the arc as C. The point C shows the number √3

Solution:

The point Q on the number line shows the number √2

A line perpendicular to the number line is drawn through the point Q. Point R is at unit distance from Q on the line.

Right angled ∆OQR is obtained by drawing seg OR.

l(OQ) = √2, l(QR) = 1

∴By Pythagoras theorem,

[l(OR)]² = [l(OQ)]² + [l(QR)]²

.. .[Taking square root of both sides]

Draw an arc with centre O and radius OR. Mark the point of intersection of the line and the arc as C. The point C shows the number √3.

Question 2.

Show the number √5 on the number line.

Solution:

Draw a number line and take a point Q at 2

such that l(OQ) = 2 units.

Draw a line QR perpendicular to the number line through the point Q such that l(QR) = 1 unit.

Draw seg OR.

∆OQR formed is a right angled triangle.

By Pythagoras theorem,

[l(OR)]² = [l(OQ)]² + [l(QR)]²

= 2² + 1²

= 4 + 1

= 5

∴l(OR) = √5 units

…[Taking square root of both sides]

Draw an arc with centre O and radius OR. Mark the point of intersection of the number line and arc as C. The point C shows the number √5.

Question 3.

Show the number √7 on the number line.

Solution:

Draw a number line and take a point Q at 2 such that l(OQ) = 2 units.

Draw a line QR perpendicular to the number line through the point Q such that l(QR) = 1 unit.

Draw seg OR.

∆OQR formed is a right angled triangle.

By Pythagoras theorem,

[l(OR)]² = [l(OQ)]² + [l(QR)]²

= 2² + 1²

= 4 + 1

= 5

∴ l(OR) = √5 units

… [Taking square root of both sides]

Draw an arc with centre O and radius OR.

Mark the point of intersection of the number line and arc as C. The point C shows the number √5.

Similarly, draw a line CD perpendicular to the number line through the point C such that l(CD) = 1 unit.

By Pythagoras theorem,

l(OD) = √6 units

The point E shows the number √6 .

Similarly, draw a line EP perpendicular to the number line through the point E such that l(EP) = 1 unit.

By Pythagoras theorem,

l(OP) = √7 units

The point F shows the number √7.