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Maths CBSE Live product
Class 9 (2026-27)
The World of Numbers
Introduction to irrational numbers
8.
Constructing a square root spiral
Question:
3
m.
Produce
a square root spiral for the number
11
on the number line.
Arrange the procedure:
Step - 1:
Using Pythagoras theorem determine the measure of the hypotenuse \(OB = √2\).
Join \(BO\) to get a right-angled triangle \(OAB\).
Proceeding in the same way, we can locate \(√11\) for any positive integer \(n\), after \(√10\).
Mark \(0\) as \(O\) and \(1\) as \(A\). At \(A\), draw a perpendicular line segment \(AB\) of length \(1\) unit.
Using \(OB\) as the base, construct a perpendicular unit segment at \(B\) to form right triangle OBC, whose hypotenuse is \(√3\).
Draw a number line with \(0\) at the centre, positive integers on the right, and negative integers on the left.
Step - 2:
Mark \(0\) as \(O\) and \(1\) as \(A\). At \(A\), draw a perpendicular line segment \(AB\) of length \(1\) unit.
Using Pythagoras theorem determine the measure of the hypotenuse \(OB = √2\).
Using \(OB\) as the base, construct a perpendicular unit segment at \(B\) to form right triangle OBC, whose hypotenuse is \(√3\).
Proceeding in the same way, we can locate \(√11\) for any positive integer \(n\), after \(√10\).
Join \(BO\) to get a right-angled triangle \(OAB\).
Draw a number line with \(0\) at the centre, positive integers on the right, and negative integers on the left.
Step - 3:
Join \(BO\) to get a right-angled triangle \(OAB\).
Draw a number line with \(0\) at the centre, positive integers on the right, and negative integers on the left.
Mark \(0\) as \(O\) and \(1\) as \(A\). At \(A\), draw a perpendicular line segment \(AB\) of length \(1\) unit.
Using Pythagoras theorem determine the measure of the hypotenuse \(OB = √2\).
Proceeding in the same way, we can locate \(√11\) for any positive integer \(n\), after \(√10\).
Using \(OB\) as the base, construct a perpendicular unit segment at \(B\) to form right triangle OBC, whose hypotenuse is \(√3\).
Step - 4:
Proceeding in the same way, we can locate \(√11\) for any positive integer \(n\), after \(√10\).
Draw a number line with \(0\) at the centre, positive integers on the right, and negative integers on the left.
Join \(BO\) to get a right-angled triangle \(OAB\).
Mark \(0\) as \(O\) and \(1\) as \(A\). At \(A\), draw a perpendicular line segment \(AB\) of length \(1\) unit.
Using Pythagoras theorem determine the measure of the hypotenuse \(OB = √2\).
Using \(OB\) as the base, construct a perpendicular unit segment at \(B\) to form right triangle OBC, whose hypotenuse is \(√3\).
Step - 5:
Join \(BO\) to get a right-angled triangle \(OAB\).
Mark \(0\) as \(O\) and \(1\) as \(A\). At \(A\), draw a perpendicular line segment \(AB\) of length \(1\) unit.
Draw a number line with \(0\) at the centre, positive integers on the right, and negative integers on the left.
Using \(OB\) as the base, construct a perpendicular unit segment at \(B\) to form right triangle OBC, whose hypotenuse is \(√3\).
Proceeding in the same way, we can locate \(√11\) for any positive integer \(n\), after \(√10\).
Using Pythagoras theorem determine the measure of the hypotenuse \(OB = √2\).
Step - 6:
Mark \(0\) as \(O\) and \(1\) as \(A\). At \(A\), draw a perpendicular line segment \(AB\) of length \(1\) unit.
Using \(OB\) as the base, construct a perpendicular unit segment at \(B\) to form right triangle OBC, whose hypotenuse is \(√3\).
Join \(BO\) to get a right-angled triangle \(OAB\).
Proceeding in the same way, we can locate \(√11\) for any positive integer \(n\), after \(√10\).
Draw a number line with \(0\) at the centre, positive integers on the right, and negative integers on the left.
Using Pythagoras theorem determine the measure of the hypotenuse \(OB = √2\).
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