Question

A square of side 1 cm is drawn inside an equilateral triangle. Find the minimum length of side of the triangle.

Answer 1

The area of triangle increases when we go from top to bottom of the triangle. Thus, the square with the greatest area will be placed at the bottom portion of the triangle.
AB || DC (opposite sides of a square)
$\therefore$ AB || QR (DC lies on QR)
Since AB || QR, A is the mid-point of PQ and B is the mid-point of PR.
Let the side of the equilateral triangle be 'x'
$\therefore AP=AR=\frac{x}{2}$
$BP=BR=\frac{x}{2}$
Now, since the triangle is equilateral and square is placed inside it QD = CR
Let QD = CR = y
In $\Delta AQD$
$A{D}^2+Q{D}^2=A{Q}^2$
$1^2+y^2={\left(\frac{x}{2}\right)}^2\Rightarrow y^2=\frac{x^2}{y}-1\Rightarrow y=\frac{\sqrt{x^2-4}}{2}$
$QR=QD+DC+CR=xx=y+1+yx=2y+1$
$x=2\times \frac{\sqrt{x^2-4}}{2}+1$
$x=\sqrt{x^2-4}+1$
$x-1=\sqrt{x^2-4}$
$(x-1{)}^2=x^2-4x^2-2x+1=x^2-4$
$-2x+1=-41+4=2x5=2xx=\frac{5}{2}=2.5cm$