Question

One side of an equilateral triangle is 18 cm. The mid-point of its sides are joined to form another triangle whose mid-points, in turn, are joined to form still another triangle. The process is continued indefinitely. Find the sum of the (i) perimeters of all the triangles. (ii) areas of all triangles.

Answer 1

Let `Delta ABC` be the given triangle having each side 18 cm. Let D, E, F be the midpoints of BC, CA, AB respectively to form `Delta DEF`. Let G, H, I be the midpoints of DE, EF and FD respectively to form `Delta GHI`.
We continue this process indefinitely.
The sides of these triangle are 18 cm, 9 cm, `9/2` cm, ..., and so on.
(i) Sum of the perimeters of all triangles so formed
`=3 {18+9+9/2+9/4+...oo} cm`
`={3xxa/((1-r))} cm`, where `a=18` and `r=9/18=1/2`
`={3xx18/((1-1/2))} cm=(3xx36)cm=108 cm`.
(ii) Sum of the areas of all the triangles so formed
`=sqrt(3)/4.{(18)^(2)+(9)^(2)+(9/2)^(2)+(9/4)^(2)+...oo} cm^(2)`
`=sqrt(3)/4.{324+81+81/4+81/16+...oo} cm^(2)`
`=sqrt(3)/4. a/((1-r)) cm^(2)`, where `a=324` and `r=81/324=1/4`
`={sqrt(3)/4xx324/((1-1/4))} cm^(2)=108 sqrt(3) cm^(2)`.