1.Classify the following numbers as rational or irrational:
(i) 2 –√5
Solution:
√5 = 2.2360679…
Here, 2.2360679…is non-terminating and non-recurring. So, it is an irrational number.
Now,
2-√5 = 2-2.2360679… = -0.2360679
Since the number, – 0.2360679…, is non-terminating non-recurring.
Hence, 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, which is a rational number.
Hence, (3 +√23)- √23 is rational number.
(iii) 2√7/7√7
Solution:
According to question,
2√7/7√7 = (2/7) × (√7/√7)
So, (√7/√7) = 1
Now,
(2/7) × (√7/√7) = (2/7)×1 = 2/7
Since, the number, 2/7 is in p/q form, which is a rational number.
Hence, 2√7/7√7 is rational number.
(iv) 1/√2
Solution:
We have to multiply and divide the numerator and denominator by √2,
(1/√2) × (√2/√2) = √2/2 (since √2×√2 = 2)
√2 = 1.4142…
Then, √2/2 = 1.4142/2 = 0.7071..
Since the number 0.7071.. is non-terminating non-recurring which is an irrational number.
Hence, 1/√2 is an irrational number.
(v) 2π
Solution:
The value of π = 3.1415
Hence, 2 = 2 × 3.1415.. = 6.2830…
Since the number, 6.2830…, is non-terminating non-recurring, which is an irrational number.
Hence, 2π 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)
[a2 – b2 = (a + b) (a – b)]
= 32-(√3)2
= 9-3
= 6
(iii) (√5+√2)2
Solution:
(a + b)2 = a2 + b2 +2ab)
= (√5+√2)2
= (√5)2+ (2×√5×√2) + (√2)2
= 5+2×√10+2
= 7+2√10
(iv) (√5-√2) (√5+√2)
Solution:
(a + b) (a – b) = (a2 – b2)
= (√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:
When we measure a value with a scale or any device, we only obtain an approximate value. We never obtain an exact value. Therefore, we may not realize whether c or d is irrational. The value of π is almost equal to 22/7 or 3.142857… .
Let c and d both are irrational.
c/d is irrational and hence π is irrational.
Hence, there is no contradiction in saying that it is irrational.
4. Represent (√9.3) on the number line.
Solution:
Step 1: Draw a line segment 9.3 unit long, AB. Extend AB to C such that BC=1 unit.
Step 2: Let the centre of AC be O. Now, AC = 10.3 units.
Step 3: With center o draw a semi-circle of radius OC.
Step 4: At B point draw a BD perpendicular to AC which intersecting the semicircle at D. Now, Join OD.
Step 5: We obtained a right angled triangle OBD.
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
Now, using Pythagoras theorem,
OD2 = BD2 + OB2
⟹ (10.3/2)2 = BD2+(8.3/2)2
⟹ BD2 = (10.3/2)2 – (8.3/2)2
⟹ BD2 = 9.3
⟹ BD = √9.3
Hence, the length of BD is √9.3.
Step 6: Draw an arc taking BD as radius and B as centre which touches the line segment. The point where it touches the line segment is at a distance of √9.3 from O which is shown in the figure.

5. Rationalize the denominators of the following:
(i) 1/√7
Solution:
We have to multiply and divide 1/√7 by √7
= (1×√7)/(√7×√7)
= √7/7
(ii) 1/(√7-√6)
Solution:
We have to multiply and divide 1/(√7-√6) by (√7+√6)
= [1/(√7-√6)]×(√7+√6)/(√7+√6)
= (√7+√6)/(√7-√6)(√7+√6)
= (√7+√6)/√72-√62
[Denominator is obtained by the property, a2 – b2 = (a + b) (a – b)]
= (√7+√6)/(7-6)
= (√7+√6)/1
= √7+√6
(iii) 1/(√5+√2)
Solution:
We have to multiply and divide 1/(√5+√2) by (√5-√2)
[1/(√5+√2)]×(√5-√2)/(√5-√2)
= (√5-√2)/(√5+√2)(√5-√2)
= (√5-√2)/(√52-√22)
= (√5-√2)/(5-2)
= (√5-√2)/3
(iv) 1/(√7-2)
Solution:
We have to multiply and divide 1/(√7-2) by (√7+2)
= 1/(√7-2)×(√7+2)/(√7+2)
= (√7+2)/(√7-2)(√7+2)
= (√7+2)/(√72-22)
= (√7+2)/(7-4)
= (√7+2)/3
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