Number System

Exercise 1.5 Part 2

Question 4: Represent 9.3 on number line.

Answer:

class nine math number system 1.5_A

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 class nine math number system 1.5_17
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 = class nine math number system 1.5_17


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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