As the midpoints of the triangles are joined successively to get another term and this is being a repeatedly infinite number of terms.

So we will be having an infinite number of side length for an infinite number of triangles.

Let ΔABC represents the equilateral triangle with side 18 cm.

D,E and F are the midpoints of side AB,BC and AC respectively

And thus ΔDEF represents another equilateral triangle.

We can find the length of DE using midpoint theorem of triangles.

If the midpoint of the 2 sides of a triangle are joined,it is parallel to the third side and is equal to 1/2 of it.

∴ DE = 1/2 × 18 = 9 cm

Similarly triangle inside DEF will have side = 9/2, and so on for other triangles.

We need to find sum of perimeters of all the triangles.

Sum of Perimeter of all the triangles = P(say)

∴ P = 3×18 + 3×9 + 3×(9/2) + 3×(9/4) + …∞

⇒ P = 54 + 27 (1 + 1/2 + 1/4 +…∞ )

⇒ P = 54 + 27S

Where S = (1 + 1/2 + 1/4 +…∞ )

We observe that above progression possess a common ratio. So it is a geometric progression.

Common ratio = 1/2 and first term (a) = 1

Sum of infinite GP = ,where a is the first term and k is the common ratio.

Note: We can only use the above formula if |k|<1

∴ we can use the formula for sum of infinite GP.

⇒ S =

∴ P = 54 + 27×2 = 54+54 = 108

∴ Sum of the perimeters of all the triangles is 108 cm

We need to find sum of Area of all the triangles.

Sum of Perimeter of all the triangles = A(say)

As the area of an equilateral triangle is given by - ,where l represents the length of side of triangle.

∴ A =

⇒ A =

⇒ A =

⇒ P = 81√3 (1 + 1/4 + 1/16 +…∞ )

⇒ P = 81√3 S’

Where S’ = (1 + 1/4 + 1/16 +…∞ )

We observe that the above progression possess a common ratio. So it is a geometric progression.

Common ratio = 1/4 and first term (a) = 1

Sum of infinite GP = ,where a is the first term and k is the common ratio.

Note: We can only use the above formula if |k|<1

∴ we can use the formula for the sum of infinite GP.

⇒ S’ =

∴ A = = 108√3

∴ Sum of the Area of all the triangles is 108√3 cm²