The formula for a triangle is $A = \frac{1}{2} a b \sin C$ where a and b are the lengths of two sides that form angle C.
Let x denote the length PQ. Triangle PQR is an equilateral triangle so all the lengths are equal and all the angles are 60°. Find the area by substituting a = x, b = x and C = 60. Simplify the expression,

$\begin{array}{rcl} Area of P Q R & = & \frac{1}{2} \times x \times x \times \sin (60) \\ = & \frac{1}{2} \times x \times x \times \frac{\sqrt{3}}{2} \\ = & \frac{\sqrt{3}}{4} x^{2} \end{array}$

[1]

The length PQ = 1.5x. A regular hexagon is made up of 6 equilateral triangles. Find the area of the hexagon by finding substituting a = 1.5x, b = 1.5x and C = 60 into the formula and multiplying by 6.

$\begin{array}{rcl} Area of A B C D E F & = & \frac{1}{2} \times 1.5 x \times 1.5 x \times \sin (60) \times 6 \\ = & \frac{1}{2} \times 1.5 x \times 1.5 x \times \frac{\sqrt{3}}{2} \times 6 \\ = & \frac{27 \sqrt{3}}{8} x^{2} \end{array}$

[1]

Form an equation using the fact that the difference between the area of the triangle and the hexagon is equal to $72 \sqrt{3}$ cm².

$\begin{array}{rcl} \frac{27 \sqrt{3}}{8} x^{2} - \frac{\sqrt{3}}{4} x^{2} & = & 72 \sqrt{3} \end{array}$

[1]

Divide both sides √3 and multiply both sides by 8. Then solve the equation to find x (which is a positive value so you can ignore the negative square root).

$\begin{array}{rcl} 27 x^{2} - 2 x^{2} & = & 576 \\ 25 x^{2} & = & 576 \\ x^{2} & = & \frac{576}{25} \\ x & = & \sqrt{\frac{576}{25}} \\ x & = & \frac{24}{5} \end{array}$

The length of PQ is 4.8 cm [1]

Area & Perimeter (CIE IGCSE Maths: Extended)

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