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Sagot :
The areas of the square are obtained from the square of the lengths,
therefore, each subsequent area is four times the previous area.
The results are;
- [tex](i) \ \dfrac{T_3}{T_2} = \mathbf{ \dfrac{T_2}{T_1} = 4}[/tex]
- (ii) The common ratio is 4
- (ii) x is 2 cm.
- (iii) The area of the largest square is 1,048,576 cm²
Reasons:
The given parameters are;
Total length of the wire = 8184 cm.
Number of squares formed = 10 squares
The length of a square = 2 × The length of the previous square
Length of the first square = x
i) Given that the length of the first square is x, the next square is 2·x, the
following square is 4·x, which gives;
Area of the first square, T₁ = x²
Area of the second square, T₂ = (2·x)² = 4·x²
Area of the third square, T₃ = (4·x)² = 16·x²
[tex]\dfrac{T_3}{T_2} = \mathbf{ \dfrac{16 \cdot x^2}{4 \cdot x^2}} = 4[/tex]
[tex]\dfrac{T_2}{T_1} = \mathbf{ \dfrac{4 \cdot x^2}{ x^2}} = 4[/tex]
Therefore;
[tex]\dfrac{T_3}{T_2} = \dfrac{T_2}{T_1} = 4[/tex]
The common ratio = 4
ii) Given that the side lengths of the squares are a constant multiple of two and the previous square, we have;
The side lengths of the squares form an arithmetic progression having a first term of x, and a common ratio, r = 2
The sum of the first ten terms of the geometric progression is 8184
[tex]S_n = \dfrac{a \cdot (r^n - 1)}{(r - 1)}[/tex]
Therefore;
The perimeter of the first square, a = 4·x
- [tex]S_{10} = \dfrac{4 \times x \cdot (2^{10} - 1)}{(2 - 1)} = 8184[/tex]
- [tex]x = \dfrac{8184}{4 \times (2^{10} - 1)} = 2[/tex]
The length of the first square x = 2 cm.
iii) The length of the largest square, is therefore; a₁₀ = x·r¹⁰⁻¹
Which gives;
a₁₀ = 2 × 2⁹ = 1,024
The area of the largest square is therefore;
T₁₀ = 1,024² = 1,048,576
The area of the largest square is therefore, T₁₀ = 1,048,576 cm²
Learn more here:
https://brainly.com/question/14570161
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