4. Six congruent rhombuses, each of area $5 \mathrm{~cm}^{2}$, form a star. The tips of the star are joined to draw a regular hexagon, as shown. What is the area of the hexagon?
A $36 \mathrm{~cm}^{2}$
B $40 \mathrm{~cm}^{2}$
C $45 \mathrm{~cm}^{2}$
D $48 \mathrm{~cm}^{2}$
E $60 \mathrm{~cm}^{2}$
Solution
C

The six angles in the centre of the hexagon are each $360^{\circ} \div 6=60^{\circ}$. Therefore the obtuse angles in the rhombuses are each $180^{\circ}-60^{\circ}=120^{\circ}$. The obtuse angles in the unshaded triangles are each $360^{\circ}-120^{\circ}-120^{\circ}=120^{\circ}$. Each unshaded triangle is isosceles with two short sides equal to the side-length of the rhombuses. If we split each rhombus in half along the long diagonal we create isosceles triangles with an angle of $120^{\circ}$. These are congruent to the unshaded triangles since the two short sides have the same length and the angle between each side is $120^{\circ}$. Hence the area of each unshaded triangle is half that of a rhombus, namely $2.5 \mathrm{~cm}^{2}$. The total area is then $(6 \times 5+6 \times 2.5) \mathrm{cm}^{2}=45 \mathrm{~cm}^{2}$.