Question

construct square root spiral of √10​

Answer 1

Answer:-

Objective

To make a square root spiral by using paper folding.

Prerequisite Knowledge

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

eg., √2 = √(12 +12). By using this Concept, we will represent irrational numbers on a number line by paper folding.

Materials Required

Tracing paper, pencil, geometry box.

Procedure

To represent √2 on a number line.

Draw a line OX on the tracing paper. Mark point O on one end and mark points 0, 1,2, 3, … at equal distances of 1 unit by paper folding.

Fold the paper along the line that passes through the point marked ‘1’ and perpendicular to the line OX, i.e., fold the paper in such a way that point ‘O’ coincides with point ‘2’. Make a crease and unfold it. From the point marked ‘1’, draw a line of length 1 unit moving along the crease. Mark the point as M such that PM = 1 unit. Join OM, clearly OM= √2 units.

cbse-class-9-maths-lab-manual-square-root-spiral-1

Fold the paper along the line ( fold on point M in such a way that point O joined with any point lie on OX,) that passes through point M and perpendicular to OM at M. Make a crease and unfold it. From the point M, draw a line of 1 unit moving upward, along the crease. Mark the point as N such that MN = 1 unit. Join ON, where ON = √3.

Keep this process continuously to get √4, √5, √6, ……….

In this way, we get a square root spiral pattern by using paper folding.

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Answer 2

Concept:

The Square Root Spiral is a geometrical structure which is based on the three basic constants.

Given:

√10

Find:

Square root spiral of √10

Solution:

Let O will be the center of the square root spiral.

From O, we need to make a straight line, OA, of 1cm horizontally.

From A, we will draw a perpendicular line of 1 cm, AB.

Now, we will join OB to get √2.

Now, from B, draw a perpendicular line, BC, of 1 cm.

Join OC to get √3.

Repeat the same procedure till we obtain √10.

Therefore, we get the square root spiral of √10 as shown in the figure attached.

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