Question

A square is inscribed in an equilateral triangle such that one of the sides of the square completely lies on one of the sides of the equilateral triangle. Find the ratio of the perimeters of the square and equilateral triangle.

• A. \(\frac{4}{3+2\sqrt3}\)
• B. \(\frac{2\sqrt3}{3\sqrt 3+6}\)
• C. \(\frac{\sqrt3}{3\sqrt 3+6}\)
• D. \(\frac{5}{3+2\sqrt 3}\)
• E. None of these

Answer 1

The Correct Option is A

Let the side of the square be \(‘a’\) cm.

\(PQ||BC\), so \(∠APQ = ∠AQP = 60°\)
So, \(△ \;APQ\) is an equilateral triangle.
Thus, \(AP = AQ = PQ = a\)
AS PS⊥BC
So, in \(△\;PSB\),

\(\frac{a}{PB} = \sin 60\)

\(⇒ \frac{a}{PB} = \frac{\sqrt3}{2}\)

\(⇒ Pb = \frac{2a}{\sqrt3} = QC\)

Thus, the side of the equilateral triangle = \(a + \frac{2a}{√3}\)

So, the perimeter of the equilateral triangle = \(3(a + \frac{2a}{√3})\)

So, the required ratio = \(4a : 3(a + \frac{2a}{√3} ) = 4 : 3(\frac{√3+2}{√3}) = 4√3 : 3(√3+2) = \frac{4√3}{3√3+6 }= \frac{4}{3+2√3}\)

Hence, option A is the correct answer.