Q.1. Classify the following numbers as rational or irrational:
(i) 2 –√5
Solution:
We know that, √5 = 2.2360679…
Here, 2.2360679…is nonterminating and nonrecurring.
Now, substituting the value of √5 in 2 –√5, we get,
2√5 = 22.2360679… = 0.2360679
Since the number, – 0.2360679…, is nonterminating nonrecurring, 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…
Then, √2/2 = 1.4142/2 = 0.7071..
Since the number , 0.7071..is nonterminating nonrecurring, 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 nonterminating nonrecurring, 2 is an irrational number.
Q.2. Simplify each of the following expressions:
(i) (3+√3)(2+√2)
Solution:
(3+√3)(2+√2 )
Opening the brackets, we get, (3×2)+(3×√2)+(√3×2)+(√3×√2)
= 6+3√2+2√3+√6
(ii) (3+√3)(3√3 )
Solution:
(3+√3)(3√3 ) = 3^{2}(√3)^{2} = 93
= 6
(iii) (√5+√2)^{2}
Solution:
(√5+√2)^{2 }= √5^{2}+(2×√5×√2)+ √2^{2}
= 5+2×√10+2 = 7+2√10
(iv) (√5√2)(√5+√2)
Solution:
(√5√2)(√5+√2) = (√5^{2}√2^{2}) = 52 = 3
Q.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. 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…
Q.4. Represent (√9.3) on the number line.
Solution:
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 semicircle 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 semicircle), OC = 10.3/2 , BC = 1
OB = OC – BC
⟹ (10.3/2)1 = 8.3/2
Using Pythagoras theorem,
We get,
OD^{2 }= BD^{2 }+ OB^{2}
⟹ (10.3/2)^{2} = BD^{2}+(8.3/2)^{2}
⟹ BD^{2 = }(10.3/2)^{2}(8.3/2)^{2}
⟹ (BD)^{2 }= (10.3/2)(8.3/2)(10.3/2)+(8.3/2)
⟹ BD^{2} = 9.3
⟹ BD = √9.3
Thus, 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 as shown in the figure.
Q.5. Rationalize the denominators of the following:
(i) 1/√7
Solution:
Multiply and divide 1/√7 by √7
(1×√7)/(√7×√7) = √7/7
(ii) 1/(√7√6)
Solution:
Multiply and divide 1/(√7√6) by (√7+√6)
[1/(√7√6)]×(√7+√6)/(√7+√6) = (√7+√6)/(√7√6)(√7+√6)
= (√7+√6)/√7^{2}√6^{2 }[denominator is obtained by the property, (a+b)(ab) = a^{2}b^{2}]
= (√7+√6)/(76)
= (√7+√6)/1
= √7+√6
(iii) 1/(√5+√2)
Solution:
Multiply and divide 1/(√5+√2) by (√5√2)
[1/(√5+√2)]×(√5√2)/(√5√2) = (√5√2)/(√5+√2)(√5√2)
= (√5√2)/(√5^{2}√2^{2}) [denominator is obtained by the property, (a+b)(ab) = a^{2}b^{2}]
= (√5√2)/(52)
= (√5√2)/3
(iv) 1/(√72)
Solution:
Multiply and divide 1/(√72) by (√7+2)
1/(√7  2)×(√7 + 2)/(√7 + 2) = (√7 + 2)/(√7  2)(√7 + 2)
= (√7 + 2)/(√7^{2 } 2^{2}) [denominator is obtained by the property, (a + b)(a  b) = a^{2}b^{2}]
= (√7 + 2)/(7  4)
= (√7 + 2)/3
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Exercise 1.6 NCERT Solutions: Number System Doc  1 pages 
Ex 6.3 NCERT Solutions Lines and Angles Doc  3 pages 
Ex 7.5 NCERT Solutions  Triangles Doc  2 pages 
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Exercise 1.6 NCERT Solutions: Number System Doc  1 pages 
Ex 6.3 NCERT Solutions Lines and Angles Doc  3 pages 
Ex 7.5 NCERT Solutions  Triangles Doc  2 pages 

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