Construct the square root of spiral of three.
Answers
Answer:
Step-by-step explanation:
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) NCERT Class 9 Maths Lab Manual - Construct a Square Root Spiral 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)
NCERT Class 9 Maths Lab Manual - Construct a Square Root Spiral 2
Construct a line QX perpendicular to the line segment PQ, by using compasses or a set square, (see Fig. 1.3)NCERT Class 9 Maths Lab Manual - Construct a Square Root Spiral 3
From Q, draw an arc of 1 unit, which cut QX at C(say). (see Fig. 1.4)
NCERT Class 9 Maths Lab Manual - Construct a Square Root Spiral 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)
NCERT Class 9 Maths Lab Manual - Construct a Square Root Spiral 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.
Answer:
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. ...
Construct a line QX perpendicular to the line segment PQ, by using compasses or a set square, (see Fig. ...
From Q, draw an arc of 1 unit, which cut QX at C(say). ( ...
Join PC.
Step-by-step explanation:
hope it helps
make it brainlist
follow me
thank you
