This question was uploaded on 26/10/21 on social media accounts.
Square and an equilateral triangle such that one side of the square is on one side of the triangle and also one vertex for both figures is the same.
Solution:
Green Length (By cosine rule):
\[\cos30 = \frac{(\sqrt3)^2+(1+\sqrt3)^2-x^2}{2(\sqrt3)(1+\sqrt3)}\]
\[\Rightarrow x^2 = 4-\sqrt3\]
\[\Rightarrow x = \sqrt{4-\sqrt3}\]
Puzzle related to Geometry, Length, Square, Triangle.
