Find the lengths of the hypotenuses of all the right triangles in the figure which is referred to as the square root spiral.
Step-by-step solution
Idea: Each triangle is right-angled, with one leg equal to the previous hypotenuse and the other leg equal to 1. By the Baudhāyana–Pythagoras theorem, if the previous hypotenuse is √k, the next is √(k + 12) = √(k + 1).
- First triangle: legs 1 and 1. Hypotenuse = √(12 + 12) = √2.1 mark
- Second triangle: legs √2 (previous hypotenuse) and 1. Hypotenuse = √((√2)2 + 12) = √3. In general a triangle with legs √k and 1 has hypotenuse √(k + 1).1 mark
- The spiral has 10 triangles (11 sides marked 1), so the hypotenuses are √2, √3, √4 = 2, √5, √6, √7, √8 = 2√2, √9 = 3, √10, √11.1 mark
Check: Last triangle: (√10)2 + 12 = 11 ✓. Two of the lengths are whole numbers: √4 = 2 and √9 = 3 ✓.
Answer to write in the exam
1st: hyp2 = 12 + 12 = 2 ⇒ hyp = √2
2nd: hyp2 = (√2)2 + 12 = 3 ⇒ hyp = √3
Legs √k and 1 ⇒ hyp = √(k + 1)
∴ Hypotenuses: √2, √3, √4 (= 2), √5, √6, √7, √8, √9 (= 3), √10, √11
Common mistakes that cost marks
- Adding lengths instead of squares: writing the second hypotenuse as √2 + 1.
- Miscounting the triangles: there are 11 sides marked 1 but only 10 triangles (the first triangle uses two of them).
- Calling all the hypotenuses irrational; √4 = 2 and √9 = 3 are rational.
How this can come in the exam
In a square root spiral, the 15th right triangle has hypotenuse
- √15
- √16 = 4
- √14
- 15
Show answer
(B) √16 = 4
The nth triangle has hypotenuse √(n + 1); for n = 15 that is √16 = 4.
An art student draws a square root spiral on a wall, using 10 cm for each unit. She starts with a right triangle with both legs 10 cm, and keeps adding a 10 cm side at right angles to the last hypotenuse.
(i) Find the first hypotenuse. (ii) Find the 3rd hypotenuse. (iii) After how many triangles is a hypotenuse exactly 30 cm? (iv) Is the 5th hypotenuse a rational length in cm?
Show answer
(i) 10√2 cm (1 mark). (ii) 10√4 = 20 cm (1 mark). (iii) 10√(n + 1) = 30 ⇒ n + 1 = 9 ⇒ 8 triangles (1 mark). (iv) 10√6 cm: no, it is irrational (1 mark).Try one yourself
How many triangles must the square root spiral have before a hypotenuse of length 5 appears?
Show answer
√(n + 1) = 5 ⇒ n = 24 triangles.
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