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Shubham Sharma 5 years, 6 months ago
fistly we divided the 1/3=0.33333333333333
now find the vale of underroot 2 =1.41421
now add the both value = 1.747
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Sia ? 5 years, 6 months ago
Eucid's second axiom: If equals are added to equals, the wholes are equal.
Algebraically: If x = y, and if a = b, then x + a = y + b
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Gaurav Seth 5 years, 6 months ago
Since adding a number (a) to itself for some number of times (n) is multiplication (a + a + a … = n * a),
root2 + root2 = 2 * root2 = 2 root2
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Gaurav Seth 5 years, 6 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.
Application
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