10
A piece of wire of length $66 \mathrm{~cm}$ is bent to form the five sides of a pentagon.
The pentagon consists of three sides of a rectangle and two sides of an equilateral triangle.
The sides of the rectangle measure $x \mathrm{~cm}$ and $y \mathrm{~cm}$ and the sides of the triangle measure $x \mathrm{~cm}$, as shown in the diagram below.
10 (a) (i) You are given that $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$
Explain why the area of the triangle is $\frac{\sqrt{3}}{4} x^{2}$
[1 mark]
10 (a) (ii) Show that the area enclosed by the wire, $A \mathrm{~cm}^{2}$, can be expressed by the formula
\[
A=33 x-\frac{1}{4}(6-\sqrt{3}) x^{2}
\]
[3 marks]

Step 1 ：Given that the sides of the rectangle are x cm and y cm, and the sides of the triangle are x cm.

Step 2 ：Using the formula for the area of an equilateral triangle, we find the area of the triangle: \(\frac{x^2 \sqrt{3}}{4}\)

Step 3 ：The perimeter of the pentagon is equal to the length of the wire, which is 66 cm. So, \(3x + 2y = 66\)

Step 4 ：Express y in terms of x: \(y = \frac{66 - 3x}{2}\)

Step 5 ：Find the area of the rectangle: \(A_{rectangle} = x \cdot y = x \cdot \frac{66 - 3x}{2}\)

Step 6 ：Find the total area enclosed by the wire: \(A = A_{triangle} + A_{rectangle} = \frac{x^2 \sqrt{3}}{4} + x \cdot \frac{66 - 3x}{2}\)

Step 7 ：Simplify the expression: \(A = 33x - \frac{1}{4}(6 - \sqrt{3})x^2\)

Step 8 ：\(\boxed{A = 33x - \frac{1}{4}(6 - \sqrt{3})x^2}\)