10
A piece of wire of length $66 cm$ 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 $x cm$ and $y cm$ and the sides of the triangle measure $x cm$ , 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, $A {cm}^{2}$ , can be expressed by the formula
$A = 33 x - \frac{1}{4} (6 - \sqrt{3}) x^{2}$
[3 marks]

Answer

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Answer

$A = 33 x - \frac{1}{4} (6 - \sqrt{3}) x^{2}$

Steps

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, $3 x + 2 y = 66$

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

Step 5 ：Find the area of the rectangle: $A_{r e c t a n g l e} = x \cdot y = x \cdot \frac{66 - 3 x}{2}$

Step 6 ：Find the total area enclosed by the wire: $A = A_{t r i a n g l e} + A_{r e c t a n g l e} = \frac{x^{2} \sqrt{3}}{4} + x \cdot \frac{66 - 3 x}{2}$

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

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

Answer

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