Exercise 1.4 Class 9 Maths

#mathsclass9 #rajansir Exercise 1.4 Class 9 Maths by @rajansir07 exercise 1.4 class 9 Class 9 Chapter 1 – Number Systems Exercise 1.4 1. Classify the following numbers as rational or irrational: (i) 2 –√5 Solution: We know that √5 = 2.2360679… Here, 2.2360679…is non-terminating and non-recurring. Now, substituting the value of √5 in 2 –√5, we get, 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 the 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 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 ) 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 ) = 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. 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… 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 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, We get, OD2=BD2+OB2 ⟹ (10.3/2)2 = BD2+(8.3/2)2 ⟹ BD2 = (10.3/2)2-(8.3/2)2 ⟹ (BD)2 = (10.3/2)-(8.3/2)(10.3/2)+(8.3/2) ⟹ BD2 = 9.3 ⟹ BD = √9.3 Thus, the length of BD is √9.3 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)/√72-√62 [denominator is obtained by the property, (a+b)(a-b) = a2-b2] = (√7+√6)/(7-6) = (√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)/(√52-√22) [denominator is obtained by the property, (a+b)(a-b) = a2-b2] = (√5-√2)/(5-2) = (√5-√2)/3 (iv) 1/(√7-2) Solution: 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) [denominator is obtained by the property, (a+b)(a-b) = a2-b2] = (√7+2)/(7-4) = (√7+2)/3