Question

5. a) A wire has a total length of 8184cm. It was cut into various pieces to form a series of 10 squares. The а length of each subsequent square doubles of the previous square. If the length of the first square is x cm,
i) If the area of the squares follows geometric progression, show that T3/T2=T2/T3. Hence determine the common ratio. [2 marks]
ii) Find the value of x. [3 marks]
iii) Find the area of the largest square. [2 marks] ​

Answer 1

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;

• $(i) \ \dfrac{T_3}{T_2} = \mathbf{ \dfrac{T_2}{T_1} = 4}$

• (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

$\dfrac{T_3}{T_2} = \mathbf{ \dfrac{16 \cdot x^2}{4 \cdot x^2}} = 4$

$\dfrac{T_2}{T_1} = \mathbf{ \dfrac{4 \cdot x^2}{ x^2}} = 4$

Therefore;

$\dfrac{T_3}{T_2} = \dfrac{T_2}{T_1} = 4$

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

$S_n = \dfrac{a \cdot (r^n - 1)}{(r - 1)}$

Therefore;

The perimeter of the first square, a = 4·x

• $S_{10} = \dfrac{4 \times x \cdot (2^{10} - 1)}{(2 - 1)} = 8184$

• $x = \dfrac{8184}{4 \times (2^{10} - 1)} = 2$

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

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