The midpoints of the sides of a series of a square of side 1 are joined to form a new sqaure. This procedure is repeated for each new square.
a) find the sum of the areas all the squares
b) find the sum of the perimeters of all the aquares
the area of the first square is 1
clearly each successive square is 1/2 of the previous one.
check: side of second square:
(1/2)^2 + (1/2)^2 = x^
x^2 = 1/2
x = 1/√2
so the area = (1/√2)^2 = 1/2
so we have an infinite series
1 + 1/2 + 1/4 + ...
a=1, r=1/2
S∞ = 1/(1-r) = 1/(1-1/2) = 2
hint for b)
side of first square = 1
side of second square = 1/√2
....
take it from there.
yeah that is what I was asking...thanx.!
so the perimeter is four and the series stay the same?
To solve this problem, we need to break it down into smaller steps:
a) Finding the sum of the areas of all the squares:
- First, let's look at the pattern. When we join the midpoints of the sides of a square, we are effectively dividing each side into two equal parts, creating smaller squares.
- The area of each smaller square is (1/2)^2 = 1/4 of the original square. This is because the side length of the new square is half of the side length of the original square.
- So, the area of each subsequent square is 1/4 of the previous square.
- We can represent the areas of the squares as a geometric sequence:
1, (1/4)^2, (1/4)^4, (1/4)^6, ...
The first term is 1, and the common ratio is (1/4)^2 = 1/16.
- The sum of all the areas of the squares can be found using the formula for the sum of an infinite geometric series:
sum = a / (1 - r), where a is the first term and r is the common ratio.
- Plugging in the values, we get:
sum = 1 / (1 - 1/16) = 1 / (15/16) = 16/15.
- Therefore, the sum of the areas of all the squares is 16/15.
b) Finding the sum of the perimeters of all the squares:
- The perimeter of each square is equal to 4 times the length of its side.
- As we saw in part a), the side length of each subsequent square is half of the previous square.
- So, the perimeter of each subsequent square is also half of the previous square's perimeter.
- We can represent the perimeters of the squares as a geometric sequence:
4, 4(1/2), 4(1/2)^2, 4(1/2)^3, ...
The first term is 4, and the common ratio is 1/2.
- The sum of all the perimeters of the squares can be found using the formula for the sum of an infinite geometric series:
sum = a / (1 - r), where a is the first term and r is the common ratio.
- Plugging in the values, we get:
sum = 4 / (1 - 1/2) = 4 / (1/2) = 8.
- Therefore, the sum of the perimeters of all the squares is 8.
That's ok but not good to understand .....
whats 2+2
Huh?
the perimeter of the first square =4
the perimeter of the second square = 4*(1/√2) = 4/√2
so a=4, r = 1/√2
sum = 4/(1 - 1/√2)
=
you finish it , ok?
Absolutely right and easy to understand
.
Good
Good