Question 32 of 58advanced✨ CreateLong Answer5 marks

Try to extend this method for constructing line segments of lengths 3\sqrt{3} and 5\sqrt{5} using a ruler and a compass. Generalise this method to construct a line segment of any length of the form n\sqrt{n}, where nn is a positive integer.

Correct Answer

To construct a line segment of length 3\sqrt{3}, we extend the method used for 2\sqrt{2}. Starting with OA=1OA = 1 unit, draw a perpendicular at A and mark B such that AB=1AB = 1 unit, giving OB=2OB = \sqrt{2}. At B, draw a perpendicular to OB and mark C such that BC=1BC = 1 unit. By the Baudhāyana–Pythagoras Theorem:

OC=OB2+BC2=(2)2+12=3 units.OC = \sqrt{OB^2 + BC^2} = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{3} \text{ units.}

For 5\sqrt{5}, we continue this spiral. From C, draw a perpendicular and mark D such that CD=1CD = 1 unit, giving OD=4=2OD = \sqrt{4} = 2 units. Draw a perpendicular at D and mark E such that DE=1DE = 1 unit. Then:

OE=OD2+DE2=22+12=5 units.OE = \sqrt{OD^2 + DE^2} = \sqrt{2^2 + 1^2} = \sqrt{5} \text{ units.}

This method can be generalised to construct n\sqrt{n} for any positive integer nn. We form a spiral of right-angled triangles where each successive triangle has a base equal to the previous hypotenuse (n−1\sqrt{n-1}) and a perpendicular of 1 unit.

By the Baudhāyana–Pythagoras Theorem, the hypotenuse of the nthn^{th} triangle is:

(n−1)2+12=n units.\sqrt{(\sqrt{n-1})^2 + 1^2} = \sqrt{n} \text{ units.}

Using a compass with the origin as the centre and this hypotenuse as the radius, we draw an arc intersecting the number line to locate n\sqrt{n}.

Exercise: Think and Reflect | Q: 1 | (Page 16)
For More Understanding

Explanation

The textbook demonstrates constructing 2\sqrt{2} by creating a right-angled triangle with base 1 and perpendicular 1, then swinging an arc to the number line. The question asks students to extend this spiral method.

The student answer correctly iterates this process: adding a unit perpendicular to the previous hypotenuse to generate the next square root (3\sqrt{3}, 4\sqrt{4}, 5\sqrt{5}). It then successfully generalises this iterative process for n\sqrt{n} by applying the Pythagoras theorem algebraically to show that a right triangle with base n−1\sqrt{n-1} and perpendicular 1 yields a hypotenuse of n\sqrt{n}.

Solution Steps

  1. Step 1: Construct 2\sqrt{2} by drawing a 1-unit perpendicular at A (where OA=1OA = 1) to get point B, making OB=2OB = \sqrt{2}.

  2. Step 2: For 3\sqrt{3}, draw a 1-unit perpendicular at B to get point C, making OC=2+1=3OC = \sqrt{2 + 1} = \sqrt{3}.

  3. Step 3: For 5\sqrt{5}, continue the spiral by adding 1-unit perpendiculars at C (getting D, OD=2OD = 2) and at D (getting E, OE=5OE = \sqrt{5}).

  4. Step 4: Generalise by stating that adding a 1-unit perpendicular to a hypotenuse of n−1\sqrt{n-1} yields a new hypotenuse of n\sqrt{n}.

  5. Step 5: Use a compass with the origin as the centre and the final hypotenuse as the radius to draw an arc intersecting the number line at n\sqrt{n}.

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