exercise 1.5 class 9th math
Answers
Answer:
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Step-by-step explanation:
Exercise 1.5
1. Classify the following numbers as rational or irrational:
(0)2-15 Solution:
We know that, vs 2.2360679...
Here, 2.2360679...is non-terminating and non-recurring. Now, substituting the value of 5 in 2 v5, we get.
2-52-2.2360679... -0.2360679
Since the number, 0.2360679..., is non-terminating non-recurring, 2 √5 is an irrational number.
(ii)(3+√23)-V23
Solution:
(3+23)-V233+V23-V23
=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) (17/17)
We know that (7/7) = 1
Hence, (2/7) (√7/V7) = (2/7)x1= 2/7
Since the number, 2/7 is in p/q form, 2v7/7/7 is rational.
(iv)1/2 Solution:
Multiplying and dividing numerator and denominator by v2 we get, (1/2) x(√2/2) 2/2 (since v2xv2 = 2)
We know that, v2= 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 x= 3.1415
Hence, 22x3.1415.6.2830...
Since the number, 6.2830..., is non-terminating non-recurring, 2x is an irrational number.
2. Simplify each of the following expressions:
(i) (3+√3)(2+2)
Solution:
(3+√3)(2+√2)
NCERT Solutions for Class 9 Maths Chapter 1- Number System
Opening the brackets, we get, (3x2)+(3x√2)+(√3x2)+(√3x√2)
6+3√2+2√3+V6
(ii) (3+√3)(3-√3) Solution:
(3+√3)(3-√3)=3²-(√3)² = 9-3 =6
(iii) (5+2) Solution:
(V5+√2)² = √5²+(2x√5√2)+ √2² = 5+2xv10+2= 7+2√10
(iv) (5-√2)(√5+√2) Solution:
(VS-√2)(√5+√2)(√5²-√2²) = 5-2 = 3
3. Recall, is defined as the ratio of the circumference c) of a circle to its diameter, (say d). That is, c/d. This seems to contradict the fact that 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
is almost equal to 22/7 or 3.142857...
4. Represent (19.3) on the number line.
Solution:
Step 1: Draw a 9.3 units long line segment, AB. Extend AB OC 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
OBOC BC
(10.3/2)- 18.3/2
Using Pythagoras theorem, We get,
OD=BD²+OB
(10.3/2)2 BD²+(8.3/2)2
BD2 (10.3/2)²-(8.3/2)2
(BD) (10.3/2)-(8.3/2) (10.3/2)+(8.3/2) BD²=9.3
BD 19.3
Thus, the length of BD is √9.3 units.
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 19.3 from O as shown in the figure.
Answer:
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