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Answers
Answer:
question 1 :-
a). √5 is an irrational number because it can't be represented in the form of rational numbers. We know that the sum or difference of a rational and an irrational number is an irrational number. Hence, 2−√5 is irrational.
1. Classify the following numbers as rational or irrational:(i) 2 –√5
Solution:
We know that, √5 = 2.2360679---
So, 2-√5 = 2-2.2360679--- = -0.2360679
Since the number, – 0.2360679---, is non-terminating non-recurring, 2 –√5 is an irrational number.
(ii) (3 +√23)- √23
Solution:
(3 +√23) –√23 = 3+√23–√23
= 3
= 3/1
Since the number 3/1 is in p/q form, (3 +√23)- √23 is rational.
(iii) 2√7/7√7
Solution:
2√7/7√7 = ( 2/7)× (√7/√7)
We know that (√7/√7) = 1
Hence, ( 2/7)× (√7/√7) = (2/7)×1 = 2/7
Since the number, 2/7 is in p/q form, 2√7/7√7 is rational.
(iv) 1/√2
Solution:
Multiplying and dividing numerator and denominator by √2 we get,
(1/√2) ×(√2/√2)= √2/2 ( since √2×√2 = 2)
We know that, √2 = 1.4142…
√2/2 = 1.4142/2 = 0.7071..
Since the number , 0.7071..is non-terminating non-recurring, 1/√2 is an irrational number.
(v) 2
Solution:
We know that, the value of = 3.1415
Hence, 2 = 2×3.1415.. = 6.2830…
Since the number, 6.2830---, is non-terminating non-recurring, 2 is an irrational number.
2. Simplify each of the following expressions:(i) (3+√3)(2+√2)
Solution:
(3+√3)(2+√2 )
(3×2)+(3×√2)+(√3×2)+(√3×√2)
= 6+3√2+2√3+√6
(ii) (3+√3)(3-√3 )
Solution:
(3+√3)(3-√3 )
= 32-(√3)2
= 9-3
= 6
(iii) (√5+√2)2
Solution:
(√5+√2)2
= √52+(2×√5×√2)+ √22
= 5+2×√10+2 = 7+2√10
(iv) (√5-√2)(√5+√2)
Solution:
(√5-√2)(√5+√2)
= (√52-√22)
= 5-2
= 3
3. Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter, (say d). That is, π =c/d. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?
Solution:
There is no contradiction. When we measure a value with a scale, we only obtain an approximate value. Therefore, we may not realize whether c or d is irrational. The value of π is almost equal to 22/7 or 3.14---
4. Represent (√9.3) on the number line.
Solution:(Attachment 03)
Step 1: Draw a 9.3 units long line segment, AB. Extend AB to C such that BC=1 unit.
Step 2: Now, AC = 10.3 units. Let the centre of AC be O.
Step 3: Draw a semi-circle of radius OC with centre O.
Step 4: Draw a BD perpendicular to AC at point B intersecting the semicircle at D. Join OD.
Step 5: OBD, obtained, is a right angled triangle.
Here, OD 10.3/2 (radius of semi-circle)
OC = 10.3/2
BC = 1
OB = OC – BC
⟹ (10.3/2)-1 = 8.3/2
Using Pythagoras theorem,
The length of BD is √9.3.
Step 6: Taking BD as radius and B as centre draw an arc which touches the line segment. The point where it touches the line segment is at a distance of √9.3 from O.
5. Rationalize the denominators of the following:(Attachment)
