Question

one side of an equilateral triangle is 18 cm. the mid point of its sides are joined to form another triangle whose mid point , in turn , are joined to form another triangle. the process is continued indefinitely. find the sum of the perimeters of all triangles and areas of all triangle?

Answer 1

side length of first triangle = b
perimeter = 3b
side length of second triangle = b/2 , as it mid point
so perimeter = 3b/2
similarly, side length of third triangle = b/4
perimeter = 3b/4
and so on
sum of all perimeter will be 
3b + 3b/2 + 3b/4 + 3b/8 +......to ∞
it is a geometrical progression series.
sum of ∞ terms = a/(1 -r)
a= 3b = 54
r = 1/2
so sum = 54/(1 - 1/2) = 108 cm
now area,
area of first triangle = b²/4 ×(√3)
for second triangle = b²/16 ×(√3)
for third triangle = b²/64 ×(√3)
and so on
sum =  b²/4 ×(√3)+ b²/16 ×(√3) +  b²/64 ×(√3) + .... to ∞
this is also a GP with a = b²/4 ×(√3) and r =1/4
so sum = a/(1 -r) 
            = b²/4 ×(√3)/(1 -1/4) = 108√3 cm²

Answer 2

Given :- An equilateral triangle with sides 18cm. The midpoints of its sides are joined to form another triangle whose midpoints ,in turn ,are joined to form another triangle .The process is continued indefinitely.

To Find :-

• The sum of perimeters of all the triangles will be ?
• The area of all triangles will be . ?

Solution :-

→ Side of an equilateral ∆ = 18 cm

so,

→ Perimeter of Larger Equilateral ∆ = 3 * side = 3 * 18 = 54 cm.

now, we know that,

• Line segment joining mid point of two sides of a triangle is parallel to third side and half of it.

it has been said that, the mid-points of its sides are joined to form another triangle whose mid-points are joined to form another triangle. This process continues indefinitely.

so,

→ Side of 2nd equilateral ∆ = Half of Larger ∆ = 18/2 = 9 cm .

then,

→ Perimeter of 2nd equilateral ∆ = 3 * 9 = 27 cm.

similarly,

→ Perimeter of 3rd equilateral ∆ = 3 * (9/2) = (27/2) cm.

we can conclude that,

• Perimeter of is reducing in the form of divide by 2.

therefore,

→ sum of the perimeter of all equilateral ∆'s = 54 + 27 + (27/2) + (27/4) + __________ ∞ .

as we can see that,

• This is a geometric progression having common ratio = 27 / 54 = (1/2)

now, we know that,

• Sum of an infinite terms of GP = first term / (1 - common ratio .

hence,

→ sum of the perimeter of all equilateral ∆'s = 54 / (1 - 1/2) = 54 / (1/2) = 54 * 2 = 108 cm. (Ans.)

now, we know that,

• Area of Equaliteral ∆ = (√3/4) * (side)²

so,

→ Area of First ∆ = (√3/4) * (18)² = 9 * 9 * √3 = 81√3 cm².

→ Area of second ∆ = (√3/4) * 9 * 9 = (81√3/4) cm².

→ Area of third ∆ = (√3/4) * (9/2)² = (81√3/16) cm².

therefore,

→ sum of the Area of all equilateral ∆'s = 81√3 + (81√3/4) + (81√3/16) + __________ ∞ .

here,

• First term = 81√3 = a
• common ratio = (81√3/4) ÷ 81√3 = (1/4) = r

hence,

→ sum of the Area of all equilateral ∆'s = a / (1 - r) = 81√3/(1 - 1/4) = 81√3 / (3/4) = (81√3 * 4)/3 = 108√3 cm² (Ans.)