This question was uploaded on 28/10/21 on social media accounts.
A right-angled triangle is drawn inside a regular hexagon such that the right angle of the triangle is apart from a vertex by a distance same as the side of the hexagon.
Solution:
ABC is an equilateral triangle.
ABD and ACE are symmetrical triangles drawn.
Due to this, ADE is also an equilateral triangle.
ADB is right-angled triangle:
\[a^2+b^2=(x\sqrt3)^2\]
Apply cosine rule on the triangle CDE:
\[\cos30=\frac{a^2+b^2-x^2}{2ab}\]
\[\Rightarrow A_{Blue} = \frac{ab}{2}=\frac{x^2\sqrt3}{3}\]
\[A_{Hexagon} = \frac{3\sqrt3}{2}x^2\]
\[Fractio = \frac{A_{Blue}}{A_{Hexagon}} = \frac29\]
Puzzle related to Geometry, Area, Polygon, Ratio.
