(a) Side Length of Square:

\[ A = s^2 \quad P = 4s \quad (A1) \]

\[ A = P \implies s^2 = 4s \quad (M1) \]

\[ s(s – 4) = 0 \implies s = 4 \quad (s > 0) \quad (A1) \]

Note: Award A1M1A0 if both \( s = 4 \) and \( s = 0 \) are stated.

[3 marks]

(b) Area of Isosceles Triangle:

\[ A_T = \frac{1}{2} x^2 \sin \frac{2\pi}{n} \quad (A1) \]

Note: Award A1 for a correct alternative form in terms of \( x \) and \( n \).

[1 mark]

(c) Derive \( y \):

Method 1:

Using \( \sin \theta = \frac{\text{opp}}{\text{hyp}} \quad (M1) \):

\[ \sin \frac{\pi}{n} = \frac{\frac{y}{2}}{x} \implies \frac{y}{2} = x \sin \frac{\pi}{n} \quad (A1) \]

\[ y = 2x \sin \frac{\pi}{n} \]

Method 2:

Using Pythagoras: \[ \left( \frac{y}{2} \right)^2 + h^2 = x^2 \], where \[ h = x \cos \frac{\pi}{n} \quad (M1) \]

\[ \left( \frac{y}{2} \right)^2 = x^2 – x^2 \cos^2 \frac{\pi}{n} = x^2 \sin^2 \frac{\pi}{n} \quad (A1) \]

\[ y = 2x \sin \frac{\pi}{n} \]

Method 3:

Using cosine rule: \[ y^2 = x^2 + x^2 – 2x^2 \cos \frac{2\pi}{n} = 2x^2 \left( 1 – \cos \frac{2\pi}{n} \right) = 4x^2 \sin^2 \frac{\pi}{n} \quad (M1)(A1) \]

\[ y = 2x \sin \frac{\pi}{n} \]

Method 4:

Using sine rule: \[ \frac{y}{\sin \frac{2\pi}{n}} = \frac{x}{\sin \left( \frac{\pi}{2} – \frac{\pi}{n} \right)} \quad (M1) \]

\[ y \cos \frac{\pi}{n} = x \sin \frac{2\pi}{n} = 2x \sin \frac{\pi}{n} \cos \frac{\pi}{n} \quad (A1) \]

\[ y = 2x \sin \frac{\pi}{n} \]

[2 marks]

(d) Show \( A = P = 4n \tan \frac{\pi}{n} \):

\[ A = n A_T = \frac{1}{2} n x^2 \sin \frac{2\pi}{n} \], \[ P = n y = n \cdot 2x \sin \frac{\pi}{n} \quad (M1) \]

\[ A = P \implies \frac{1}{2} n x^2 \sin \frac{2\pi}{n} = 2n x \sin \frac{\pi}{n} \quad (A1) \]

Use: \[ \sin \frac{2\pi}{n} = 2 \sin \frac{\pi}{n} \cos \frac{\pi}{n} \quad (M1) \]

\[ \frac{1}{2} n x^2 \cdot 2 \sin \frac{\pi}{n} \cos \frac{\pi}{n} = 2n x \sin \frac{\pi}{n} \]

\[ n x^2 \sin \frac{\pi}{n} \cos \frac{\pi}{n} = 2n x \sin \frac{\pi}{n} \quad (A1) \]

Factorize: \[ x \sin \frac{\pi}{n} \left( x \cos \frac{\pi}{n} – 2 \right) = 0 \quad (M1) \]

Since \( x \sin \frac{\pi}{n} \neq 0 \), \[ x = \frac{2}{\cos \frac{\pi}{n}} \quad (A1) \]

Substitute into \( P \): \[ P = n \cdot 2 \cdot \frac{2}{\cos \frac{\pi}{n}} \cdot \sin \frac{\pi}{n} = 4n \frac{\sin \frac{\pi}{n}}{\cos \frac{\pi}{n}} = 4n \tan \frac{\pi}{n} \quad (M1)(A1) \]

\[ A = P = 4n \tan \frac{\pi}{n} \]

[7 marks]

(e)(i) Limit:

Use Maclaurin series: \[ \tan \frac{\pi}{n} = \frac{\pi}{n} + \frac{\left( \frac{\pi}{n} \right)^3}{3} + \frac{2 \left( \frac{\pi}{n} \right)^5}{15} + \dots \quad (M1) \]

\[ 4n \tan \frac{\pi}{n} = 4n \left( \frac{\pi}{n} + \frac{\pi^3}{3n^3} + \frac{2\pi^5}{15n^5} + \dots \right) = 4\pi + \frac{4\pi^3}{3n^2} + \frac{8\pi^5}{15n^4} + \dots \quad (A1) \]

\[ \lim_{n \to \infty} \left( 4n \tan \frac{\pi}{n} \right) = 4\pi \quad (A1) \]

Note: Award M1A1A0 if \( \lim_{n \to \infty} \) is not stated.

[3 marks]

(e)(ii) Geometric Interpretation:

As \( n \to \infty \), the polygon becomes a circle with radius 2, area \( 4\pi \), and perimeter \( 4\pi \quad (R1) \).

Note: Award R1 for stating the polygon becomes a circle with area or perimeter \( 4\pi \), or \( A = P = 4\pi \).

[1 mark]

(f) Derive \( a \):

\[ A = \frac{1}{2} ab \], \[ P = a + b + \sqrt{a^2 + b^2} \quad (A1)(A1) \]

\[ A = P \implies \frac{1}{2} ab = a + b + \sqrt{a^2 + b^2} \quad (M1) \]

\[ \sqrt{a^2 + b^2} = \frac{1}{2} ab – (a + b) \quad (M1) \]

Square both sides: \[ a^2 + b^2 = \left( \frac{1}{2} ab – a – b \right)^2 \quad (M1) \]

\[ a^2 + b^2 = \frac{1}{4} a^2 b^2 – a^2 b – ab^2 + a^2 + 2ab + b^2 \quad (A1) \]

Simplify: \[ \frac{1}{4} a^2 b^2 – a^2 b – ab^2 + 2ab = 0 \quad (A1) \]

\[ a \left( \frac{1}{4} ab – a – b + 2 \right) = 0 \implies \frac{1}{4} ab – a – b + 2 = 0 \]

\[ ab – 4a = 4b – 8 \implies a = \frac{4b – 8}{b – 4} = \frac{8}{b – 4} + 4 \quad (M1)(A1) \]

[7 marks]

(g)(i) Right-Angled Triangles:

Substitute integer \( b \) into \( a = \frac{8}{b – 4} + 4 \), ensure \( a, b, \sqrt{a^2 + b^2} \in \mathbb{Z} \quad (M1) \):

For \( b = 5 \): \[ a = \frac{8}{5 – 4} + 4 = 12 \], hypotenuse \[ \sqrt{12^2 + 5^2} = 13 \]. Triangle: \( (5, 12, 13) \quad (A1) \).

For \( b = 6 \): \[ a = \frac{8}{6 – 4} + 4 = 8 \], hypotenuse \[ \sqrt{8^2 + 6^2} = 10 \]. Triangle: \( (6, 8, 10) \quad (A1) \).

[3 marks]

(g)(ii) Area and Perimeter:

For \( (5, 12, 13) \): \[ A = \frac{1}{2} \cdot 5 \cdot 12 = 30 \], \[ P = 5 + 12 + 13 = 30 \].

For \( (6, 8, 10) \): \[ A = \frac{1}{2} \cdot 6 \cdot 8 = 24 \], \[ P = 6 + 8 + 10 = 24 \quad (A1) \]

[1 mark]