Question 3590: a side of a square is 12cm. the midpoints of its sides are joined to form an inscribed square, and this process is continued as shown in the diagram. find the sumof the perimeters of the squares if this process is continued withoutend. a side of an equilateral triangle is 10cm. the midpoints of its sides are joined t fom an inscribed equilateral triangle, and this process is continued without end
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! Let the side of the 1st square be s1, consider the diagram below
s1/2
A --------- B
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C
the side of the 2nd square s2 = BC = sqrt(2(1/2s1)^2) = s1 sqrt(1/2)
Let the side of the nth square of this process be sn.
We have sn = sn-1 sqrt(1/2) for n>=2
and so sn = s1 [sqrt(1/2)]^(n-1) for all n>=1
oo oo
Hence, the summation of the total sides E sn = E s1 [sqrt(1/2)]^(n-1)
n=1 n=1
= s1/ (1- sqrt(1/2)) = s1/(1 - sqrt(2)/2) = 2s1/(1-sqrt(1/2))
= 2s1(1+sqrt(1/2))/(1-1/2) = 4s1 (1+sqrt(1/2)) = 4s1 (1+sqrt(2)/2)
= 2s1 (2+sqrt(2)).
The perimeter of a square = 4 * side,
Since the sum of all perimeters = 8 s1 (2+sqrt(2))
= 96 (2+sqrt(2)) cm, when s1 = 12 cm.
The 2nd one about triangle is similar but easier.
Let s(1)=10 be the side of the first equilateral triangle.
and s(n) be the side of the first equilateral triangle for n>=2.
Since s(2) = 1/2 s(n)= 1/2 s(n-1) for n>=2 (why?)
oo oo
Hence, the summation of the total sides E s(n) = E s(1) [1/2)]^(n-1)
n=1 n=1
= s(1)/(1- 1/2) = 2 s(1) = 20 cm.
The perimeter of a square = 3 * side,
Since the sum of all perimeters of the equilateral triangles = 60 cm
Kenny
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