Question

A square DEAF is constructed inside a ${30}^{\circ }-{60}^{\circ }-{90}^{\circ }$ triangle ABC with the hypotenuse BC = 4 D on side BC, E on side AC and F on side AB. The length of the side of the square is

• A. $3-\sqrt{3}$
• B. $3-\sqrt{2}$
• C. $2$
• D. $1.5$

Answer 1

The correct option is A $3-\sqrt{3}$

In $△DEC,tan{60}^0=\frac{x}{EC}\Rightarrow \sqrt{3}=\frac{x}{EC}\Rightarrow EC=x/\sqrt{3}$

& $sin{60}^0=\frac{x}{DC}\Rightarrow \frac{\sqrt{3}}{2}=\frac{x}{DC}\Rightarrow DC=\frac{2x}{\sqrt{3}}$

In $△BFD,cos{60}^0=\frac{x}{BD}\Rightarrow \frac{1}{2}=\frac{x}{BD}\Rightarrow BD=2x$

& $tan{60}^0=\frac{BF}{x}\Rightarrow \sqrt{3}=\frac{BF}{x}\Rightarrow BF=x\sqrt{3}$

Now, $AB=AF+BF=x+\sqrt{3}x=x(\sqrt{3}+1)$

$AC=AE+EC=x+\frac{x}{\sqrt{3}}=\frac{(\sqrt{3}+1)x}{\sqrt{3}}$

$BC=BD+DC=2x+\frac{2x}{\sqrt{3}}=\frac{2(\sqrt{3}+1)x}{\sqrt{3}}$

$BC=4$

$\frac{2(\sqrt{3}+1)x}{\sqrt{3}}=4\Rightarrow x=\frac{2\sqrt{3}}{\sqrt{3}+1}=\frac{2\sqrt{3}(\sqrt{3}-1)}{{\left(\sqrt{3}\right)}^2-{\left(1\right)}^2}=\frac{2(3-\sqrt{3})}{2}=3-\sqrt{3}$