Let ABC be the triangle with AB = BC = AC = 20 cm
Let D, E and F be midpoints of AC, CB and AB respectively which are joined to form an equilateral triangle DEF
Now we have to find the length of side of ΔDEF
Consider ΔCDE
CD = CE = 10 cm … D and E are midpoints of AC and CB
Hence ΔCDE is isosceles
⇒ ∠CDE = ∠CED … base angles of isosceles triangle
But ∠DCE = 60° …∠ABC is equilateral
Hence ∠CDE = ∠CED = 60°
Hence ΔCDE is equilateral
Hence DE = 10 cm
Similarly, we can show that GH = 5 cm
Hence the series of sides of equilateral triangle will be 20, 10, 5 …
The series is GP with first term a = 20 and common ratio r = ½
To find the perimeter of 6th triangle inscribed we first have to find the side of 6th triangle that is the 6th term in the series
nth term in GP is given by tn = arn-1
⇒ t6 = (20) (1/2)6-1
⇒ t6 = 20/ 25
= 20/ (4 × 23)
⇒ t6 = 5/8
Hence the side of 6th equilateral triangle is 5/8 cm and hence its perimeter would be thrice its side length because it’s an equilateral triangle
Perimeter of 6th equilateral triangle inscribed is 3 × 5/8 = 15/8 cm