Question

A square DEAF is constructed inside a ${30}^{\circ }-{60}^{\circ }-{90}^{\circ }$ triangle ABC with the hypotenuse BC = 4 units, 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. $2-\sqrt{3}$
• D. 1.5 units

Answer 1

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

In the given question,

BC = 4

So, using trigonometric relation,

AC = ABcos60 = 2

applying Pythagoras theorem,

${BC}^2={AB}^2+{AC}^2$

So, we get,

${AB}^2=\sqrt{4^2-2^2}{AB}^2=\sqrt{12}AB=2\sqrt{3}$

Now, Area of triangle ABC

= $\frac{1}{2}\times AB\times AC$

=$\frac{1}{2}\times 2\times 2\sqrt{3}=2\sqrt{3}$

Now, let the side of the square is $x$

Area of square AEDF = $x^2$

From figure,

$BF=2\sqrt{3}-xEC=2-x$

Area of ABC = Area of BFD + Area of AEDF + Area of CDE

$2\sqrt{3}=[\frac{1}{2}\times x\times (2\sqrt{3}-x)\times x]+x^2+[\frac{1}{2}\times x\times (2-x)]onsolvingtheaboveequationweget,x=3-\sqrt{3}$

So, correct answer is option A.