# Number System

## Exercise 1.5 Part 2

Question 4: Represent 9.3 on number line.

Answer:

Step-1 – Draw a horizontal line.
Step 2 – Cut a line segment CA equal to 9.3 unit
Step 3 – Cut AB equal to 1 unit.
Step 4 - Find the centre point of line CB. Assume centre point comes at O.
Step 5 – Draw a semicircle taking OB as radius.
Step 6 – Draw a perpendicular at point A. This perpendicular cut the semicircle at point D.
Step 7 – AD will become equal to
Step 8 – Now measure AD using compass and cut equal to AD on horizontal line from A, which will come at E.
Here AD = AE =

Proof: In the above figure
CA = 9.3 unit
AB = 1 unit
Thus, diameter of circle = 9.3 + 1 = 10.3 unit
Therefore, radius OB = OC = OD of the circle = 10.3/2 unit
Now, OA = OB – AB

Or, OA=(10.3)/(2)-1

Or, OA=(10.3-2)/2

Or, OA=8.3/2

Since AD is perpendicular to CA, thus, ODA is a right angle triangle.

So, OD^2=OA^2+AD^2

Or, (10.3/2)^2=(8.3/2)^2+AD^2

Or, AD^2=(10.3/2)^2-(8.3/2)^2

Or, AD^2=106.09/4-68.09/4

Or, AD^2=37.2/4=9.3

Or, AD=sqrt9.3

Question 5: Rationalise the denominators of the following:

(i) 1/sqrt7

Answer: Given, 1/sqrt7

=(1/sqrt7)xx(sqrt7/sqrt7)

=(1xxsqrt7)/(sqrt7xxsqrt7)=(sqrt7)/7

(ii)1/(sqrt7-sqrt6)

Answer: Given, 1/(sqrt7-sqrt6)

=(1/(sqrt7-sqrt6))xx((sqrt7+sqrt6)/(sqrt7+sqrt6))

(sqrt7+sqrt6)/((sqrt7-sqrt6)(sqrt7+sqrt6))

Since, a^2-b^2=(a+b)(a-b)

So, we have, (sqrt7+sqrt6)/((sqrt7)^2-(sqrt6)^2)

=(sqrt7+sqrt6)/(7-6)

(sqrt7+sqrt6)/1=sqrt7+sqrt6

(iii) 1/(sqrt5+sqrt2)

Answer: Given, 1/(sqrt5+sqrt2)

=(1/(sqrt5+sqrt2))xx((sqrt5-sqrt2)/(sqrt5-sqrt2))

(sqrt5-sqrt2)/((sqrt5+sqrt2)(sqrt5-sqrt2))

=(sqrt5-sqrt2)/((sqrt5)^2-(sqrt2)^2)

=(sqrt5-sqrt2)/(5-2)=(sqrt5-sqrt2)/3

(iv) 1/(sqrt7-2)

Answer: Given, 1/(sqrt7-2)

=(1)/(sqrt7-2)xx(sqrt7+2)/(sqrt7+2)

=(sqrt7+2)/((sqrt7)^2-2^2)

=(sqrt7+2)/(7-4)=(sqrt7+2)/3

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