Mathsblog Question (by Hari Govindan)

The picture illustrates a regular hexagon with the side length equal to √3. Quadrilaterals XABC and QPXR are squares . What is the area of the shaded triangle CPS?

The side of the hexagon is square root of 3. So the width (GD) of the hexagon is  $$\sqrt 3  \times \sqrt 3  = 3$$

From the right angled triangle ADX, we get,  $$\cos 30 = \frac{{AD}}{{AX}}$$ This implies  $$AX = \frac{{AD}}{{\cos 30}} = \frac{{\frac{{\sqrt 3 }}{2}}}{{\frac{{\sqrt 3 }}{2}}} = 1$$ Again, $$\sin 30 = \frac{{DX}}{{AX}} \Rightarrow DX = \frac{1}{2}$$ Since the side of the equilateral triangle XCP is 1 its altitude $$EX = \frac{{\sqrt 3 }}{2}$$ The altitude of the triangle CPS is SF, which is the same as GE (see figure)

But  $$GE = GD - (DX + EX) = 3 - \left( {\frac{1}{2} - \frac{{\sqrt 3 }}{2}} \right) = \frac{{\left( {5 - \sqrt 3 } \right)}}{2}$$ Finally, area of the triangle CPS is $$\frac{1}{2} \times {\rm{base }} \times {\rm{height}} = \frac{1}{2} \times 1 \times \frac{{5 - \sqrt 3 }}{2} = \frac{{5 - \sqrt 3 }}{4}$$