Try to extend this method for constructing line segments of lengths √3 and √5 using a ruler and a compass. Generalise this method to construct a line segment of any length of the form √n, where n is a positive integer.
- 1. Try to extend this method for constructing line segments of lengths √3 and √5 using a ruler and a compass.
- 2. Generalise this method to construct a line segment of any length of the form √n, where n is a positive integer.
Step-by-step solution
Idea: A right triangle with legs √a and 1 has hypotenuse √(a + 1). Starting from length 1 and adding a unit perpendicular each time climbs through √2, √3, √4, √5, … (the square root spiral).
1. Try to extend this method for constructing line segments of lengths √3 and √5 using a ruler and a compass.
- √3: mark P at √2 on the number line (as before). At P draw a perpendicular and mark Q with PQ = 1. Then OQ2 = OP2 + PQ2 = 2 + 1 = 3, so OQ = √3. An arc with centre O and radius OQ cuts the line at √3.1 mark
- √5: mark A at 2 on the number line. At A draw a perpendicular and mark B with AB = 1. Then OB2 = 22 + 12 = 5, so OB = √5. An arc with centre O and radius OB cuts the line at √5 (see the diagram). (Or continue from √3: √3 and 1 give 2, then 2 and 1 give √5.)1 mark
2. Generalise this method to construct a line segment of any length of the form √n, where n is a positive integer.
- Suppose a segment of length √(n − 1) has been made, with O at one end and R at the other. At R draw a perpendicular and mark S with RS = 1.½ mark
- Then OS2 = (√(n − 1))2 + 12 = n − 1 + 1 = n, so OS = √n.½ mark
- Start with length 1 and repeat: 1 → √2 → √3 → √4 = 2 → √5 → … So every √n can be reached in n − 1 steps; an arc centred at O carries each one onto the number line. (Shortcut: if n = a2 + b2, use legs a and b directly.)1 mark
Check: √3 ≈ 1.732 and √5 ≈ 2.236: on the diagram the arcs meet the line between 1 and 2, and between 2 and 3 ✓.
Answer to write in the exam
1.
OP = √2 on the number line; PQ ⊥ OP, PQ = 1
OQ2 = 2 + 1 = 3 ⇒ OQ = √3; arc (centre O, radius OQ) marks √3
OA = 2; AB ⊥ OA, AB = 1
OB2 = 4 + 1 = 5 ⇒ OB = √5; arc (centre O, radius OB) marks √5
2.
OR = √(n − 1); RS ⊥ OR, RS = 1
OS2 = (n − 1) + 1 = n
∴ OS = √n; repeating from OR = 1 gives √2, √3, √4, …, √n.
Common mistakes that cost marks
- Drawing the new perpendicular at 1 or 2 on the line instead of at the point √2 for √3: legs 1 and 1 give √2 again, not √3.
- Thinking √5 needs legs 2 and 3 (that gives √13). Choose legs whose squares add to 5.
- Forgetting to transfer the hypotenuse to the number line with a compass centred at O.
How this can come in the exam
To construct √6 by the step method, the perpendicular of length 1 is drawn at the point
- √5
- √6
- 5
- 6
Show answer
(A) √5
Legs √5 and 1 give hypotenuse √(5 + 1) = √6.
Find two whole numbers a, b so that a right triangle with legs a, b has hypotenuse √17, and explain how to mark √17 on the number line.
Show answer
42 + 12 = 17, so a = 4, b = 1 (1 mark). Take OA = 4 on the line, AB = 1 perpendicular at A, and draw an arc centre O, radius OB to cut the line at √17 (1 mark).Try one yourself
Which legs would you use to construct √8 in a single step?
Show answer
Legs 2 and 2: 4 + 4 = 8.
More questions like this
- We know what it means to add 2, 100, or even a lakh terms. What does it mean to add an infinite number of terms?
- It terminates: The division eventually leaves a remainder of 0. The decimal stops. 38 = 0.375. (Can you tell for which rational numbers the decimal will be terminating?)
- It repeats: The division never reaches a remainder of 0, but the sequence of digits begins to loop infinitely. 511 = 0.454545 … = 0.45.
- Try to find the decimal expansions of 103 and 1112. What do you observe about the repetition of the digits after the decimal point?
- Why do some rational numbers have repeating decimal representations? Imagine calculating 17 using long division. You are dividing by 7. What are the possible remainders at each step? They can only be 1, 2, 3, 4, 5, or 6 (why not 0?).