Question

A polygon is regular when all angles are equal and all sides are equal (otherwise
it is "irregular"). Below given figure is 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. Answer
the questions given below:

i The sum of perimeters of all the triangles will be:
a)100 cm b) 110cm c) 118cm d)108cm

Answer 1

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 :-

a)100 cm b) 110cm c) 118cm d)108cm

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.) (Option D) .

Answer 2

Answer:

d) 108

Step-by-step explanation:

given: the triangles are equilateral,

           side of ABC=18cm

perimeter of ABC = 18*3 = 54cm^2 ___ i

side of DEF=18/2 = 9cm

perimeter of DEF = 9*3 = 27cm^2 ____ ii

side of GHI=9/2cm

perimeter of GHI = 9/2 *3 = 27/2 _____ iii

From i , ii , iii,

common ratio r = 1/2

They are in infinite G.P

G.P = 54,27...  , where a = 54 and r = 1/2

the sum of terms in an infinite GP = $Sn_{} =\frac{a}{1-r}$

                                                                 = $\frac{54}{1-\frac{1}{2} } = \frac{54}{\frac{2-1}{2} } = \frac{54*2}{1}$

                                                                 = 108

Hope this helps :)