# one side of an equailateral triangle is 18cm. the mid points of its sides are joined to form another traingle whose midpoints 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 the triangles

### Asked by gurpreet kaur | 16th Jun, 2013, 08:17: PM

Expert Answer:

### Using the mid point theorem, In a given triangle, the line joining the mid points of 2 sides of a triangle is parallel to the third side and equal to half of it.
Since, here the triangle given is equilateral triangle, so the lengths of all 3 line segments formed by joining the mid points of all sides of the triangle would be equal and equal to half of the side of the parent triangle.
So, the sides of each consecutive triangle would be given by a, a/2, a/4, a/8 etc ...if a is the length of side of the first triangle.
Here a = 18cm
Perimeter of an equilateral triangle = 3s where s is the length of side
and area = sqrt(3)/4 * s^2
Hence sum of perimeter of all triangles formed = 3a + 3(a/2) + 3(a/4) + ....
Sum of perimeter = 3(a+a/2+a/4+a/8 + ....)
Now this is an infinite GP with first term = a and common ratio = 1/2. Sum of infinite terms of a GP = a/(1-r)
Hence, sum of perimeter = 3 *a/(1-1/2) = 6a = 6*18 = 108 cm
Now, sum of area of all triangles = root(3)/4 (a^2 + a^2/4 + a^2/16 + a^2/64+ ...)
Now this is an infinite GP with first term = a^2 and common ratio = 1/4. Sum of infinite terms of a GP = a/(1-r)
Hence, sum of area of all triangles = root(3)/4 *a^2/(1-1/4) = root(3)/4 * a^2*4/3 = a^2/root(3) = 364/roo(3) cm^2

Now this is an infinite GP with first term = a^2 and common ratio = 1/4. Sum of infinite terms of a GP = a/(1-r)

Hence, sum of area of all triangles = root(3)/4 *a^2/(1-1/4) = root(3)/4 * a^2*4/3 = a^2/root(3) = 364/roo(3) cm^2

### Answered by | 16th Jun, 2013, 09:06: PM

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