Question

Perimeter of the each triangle,
when the mid poms of sides of an
equilateral triangle whose side is
24 cm are joined to form another
range, whose midpoints in turn
are joined to form still another
angle and the process continues
de
24​

Answer 1

Given :- An equilateral triangle with sides 24 cm. 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 ∆ = 24 cm

so,

→ Perimeter of Larger Equilateral ∆ = 3 * side = 3 * 24 = 72 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 ∆ = 24/2 = 12 cm .

then,

→ Perimeter of 2nd equilateral ∆ = 3 * 12 = 36 cm.

similarly,

→ Perimeter of 3rd equilateral ∆ = 3 * (12/2) = 18 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 = 72 + 36 + 18 + __________ ∞ .

as we can see that,

This is a geometric progression having common ratio = 36 / 72 = (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 = 72 / (1 - 1/2) = 72 / (1/2) = 72 * 2 = 144 cm. (Ans.)

now, we know that,

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

so,

→ Area of First ∆ = (√3/4) * (24)² = 24 * 6 * √3 = 144√3 cm².

→ Area of second ∆ = (√3/4) * 12 * 12 = 36√3 cm².

→ Area of third ∆ = (√3/4) * 6 * 6 = 9√3 cm².

therefore,

→ sum of the Area of all equilateral ∆'s = 144√3 + 36√3 + 9√3 + __________ ∞ .

here,

First term = 144√3 = a

common ratio = 36√3 ÷ 144√3 = (1/4) = r

hence,

→ sum of the Area of all equilateral ∆'s = a / (1 - r) = 144√3/(1 - 1/4) = 144√3 / (3/4) = (144√3 * 4)/3 = 192√3 cm² (Ans.)