This question was uploaded on 16/09/21 on social media accounts.
A square is inscribed in a regular hexagon such that one vertex is on any vertex of the hexagon and the other two vertices are on the sides of the hexagon making an equal angle with the sides of the hexagon. The fourth vertex is inside the hexagon. Find are of the square.
Solution:
Method 1:
\[\frac{Diagonal}{2} = 2cos(30)=\sqrt3\]
\[\Rightarrow Diagonal =2\sqrt3\]
Now,\[Area = \frac{(Diagonal)^2}{2} = \frac{12}{2} = 6\]
Method 2:
Use sine rule:
\[\frac{2}{sin(45)}=\frac{x}{sin(120)}\]
\[\Rightarrow\frac{2}{\frac{1}{\sqrt2}}=\frac{x}{\frac{\sqrt3}{2}}\]
\[\Rightarrow x=\sqrt6\]
\[\Rightarrow Area=x^2=6\]
Puzzle related to Geometry, Square, Hexagon, Area
