We have seen how we can draw a series of right triangles as in the picture.

i) What are the lengths of the sides of the tenth triangle?

ii) How much more is the perimeter of the tenth triangle than the perimeter of the ninth triangle?

iii) How do we write in algebra, the difference in perimeter of the n^th triangle and that of the triangle just before it?

We know that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Consider the 1^st triangle.

⇒ Hypotenuse of 1^st triangle, x² = 1² + 1²

= 2

∴ Hypotenuse, x = √2 m

Consider 2^nd triangle.

⇒ Hypotenuse of 2^nd triangle, y² = (√2)² + 1²

= 2 + 1

= 3

∴ Hypotenuse, y = √3 m

Consider 3^rd triangle.

⇒ Hypotenuse of 3^rd triangle, z² = (√3)² + 1²

= 3 + 1

= 4

∴ Hypotenuse, y = √4 m

And so on.

(i) The lengths of the 10^th triangle will be √10 m, 1 m and hypotenuse.

Let hypotenuse be a.

⇒ a² = (√10)² + 1²

= 10 + 1

= 11

⇒ a = √11 m

∴ The lengths of the sides of the 10^th triangle are √10 m, 1 m and √11 m.

(ii) We know that perimeter of a polygon is the sum of all its sides.

The perimeter of 10^th triangle = √10 + 1 + √11

[√10 ≈ 3.16; √11 ≈ 3.31]

∴ Perimeter = 3.16 + 1 + 3.31

= 7.47 m

The perimeter of 9^th triangle = √9 + 1 + √10

= 3 + 1 + 3.16

= 7.16 m

Perimeter of 10^th triangle – Perimeter of 9^th triangle = 7.47 – 7.16 = 0.31

∴ The perimeter of the 10^th triangle is 0.31 m more than the perimeter of 9^th triangle.

(iii) Perimeter of nth triangle = √n + 1 + √(n+1)

Perimeter of (n-1)th triangle = √(n – 1) + 1 + √(n – 1 + 1)

= √(n - 1) + 1 + √n

Difference between perimeters of nth triangle and (n – 1)th triangle = √n + 1 + √(n+1) – [√(n - 1) + 1 + √n]

= √n + 1 + √(n+1) – √(n - 1) - 1 - √n

= √(n + 1) - √(n – 1)