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Yogita Ingle 6 years, 7 months ago
a3 – b3 = (a – b) (a2 + ab + b2 )
64m3 - 343n3 = (4m)3 - (7n)3
= (4m - 7n) [(4m)2 + 4m (7n) + (7n)2
= (4m - 7n) [16m2 + 28mn + 49n2]
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Gaurav Seth 6 years, 7 months ago
Let , us assume that √n is rational .
so, √n = a/b where a and b are integers and b is not equation to zero .
Let, a/b are co- prime
taking square both side ,we get
=> n = a^2/b^2
=> nb^2 = a^2 ......(1)
so, n divide a^2
it means n also divide a
for some integer c
a = nc
now squaring both side
a^2 = n^2c^2
=> nb^2 = n^2c^2 [ from (1) ]
=> b^2 = nc^2
so , n divide b^2
it means b also divide b
so, a and b have n as a prime factor
but this contradict the fact that a and b are co- prime .
therefore , our assumption is wrong .
hence, √n is irrational
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Gaurav Seth 6 years, 7 months ago
OBJECTIVE
To construct a square root spiral.
Materials Required
- Adhesive
- Geometry box
- Marker
- A piece of plywood
Prerequisite Knowledge
- Concept of number line.
- Concept of irrational numbers.
- Pythagoras theorem.
Theory
- A number line is a imaginary line whose each point represents a real number.
- The numbers which cannot be expressed in the form p/q where q ≠ 0 and both p and q are integers, are called irrational numbers, e.g. √3, π, etc.
- According to Pythagoras theorem, in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of other two sides containing right angle. ΔABC is a right angled triangle having right angle at B. (see Fig. 1.1)
Therefore, AC² = AB² +BC²
where, AC = hypotenuse, AB = perpendicular and BC = base
Procedure
- Take a piece of plywood having the dimensions 30 cm x 30 cm.
- Draw a line segment PQ of length 1 unit by taking 2 cm as 1 unit, (see Fig. 1.2)
- Construct a line QX perpendicular to the line segment PQ, by using compasses or a set square, (see Fig. 1.3)
- From Q, draw an arc of 1 unit, which cut QX at C(say). (see Fig. 1.4)
- Join PC.
- Taking PC as base, draw a perpendicular CY to PC, by using compasses or a set square.
- From C, draw an arc of 1 unit, which cut CY at D (say).
- Join PD. (see Fig. 1.5)
- Taking PD as base, draw a perpendicular DZ to PD, by using compasses or a set square.
- From D, draw an arc of 1 unit, which cut DZ at E (say).
- Join PE. (see Fig. 1.5)
Keep repeating the above process for sufficient number of times. Then, the figure so obtained is called a ‘square root spiral’.
Demonstration
- In the Fig. 1.5, ΔPQC is a right angled triangle.
So, from Pythagoras theorem,
we have PC² = PQ² + QC²
[∴ (Hypotenuse)² = (Perpendicular)² + (Base)²]
= 1² +1² =2
=> PC = √2
Again, ΔPCD is also a right angled triangle.
So, from Pythagoras theorem,
PD² =PC² +CD²
= (√2)² +(1)² =2+1 = 3
=> PD = √3 - Similarly, we will have
PE= √4
=> PF=√5
=> PG = √6 and so on.
Observations
On actual measurement, we get
PC = …….. ,
PD = …….. ,
PE = …….. ,
PF = …….. ,
PG = …….. ,
√2 = PC = …. (approx.)
√3 = PD = …. (approx.)
√4 = PE = …. (approx.)
√5 = PF = …. (approx.)
Result
A square root spiral has been constructed.
Posted by Neha Kushwaha 6 years, 7 months ago
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Gaurav Seth 6 years, 7 months ago
T+1/t=8
(t+1/t)³=t³+1/t³+3t*1/t(t+1/t)
8³=t³+1 /t³+3(8)
8³-24=question
512-24=488
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