Question

A square of side $1$ is inscribed in a right angled triangle so that two sides of the square are along the sides of the triangle and one vertex of the square is on the hypotenuse.If the hypotenuse is $2\sqrt{6}$, then the ratio of the other two sides is

• A. $2$
• B. $\sqrt{2}+1$
• C. $\sqrt{2}+\sqrt{3}$
• D. $2+\sqrt{3}$

Answer 1

The correct option is D $2+\sqrt{3}$

From the figure.
$a^2+b^2={\left(2\sqrt{6}\right)}^2$
$\Rightarrow a^2+b^2=24$
Triangles $BDF$ and $FEA$ are similar.
$\therefore \frac{a-1}{1}=\frac{1}{b-1}$
$\Rightarrow (b-1)(a-1)=1$
$\Rightarrow a+b=ab$
Squaring both sides, we get
${(a+b)}^2=a^2{b}^2$
$\Rightarrow a^2{b}^2-2ab-24=0$ where $a^2+b^2=24$
$\Rightarrow ab=6=a+b$
$a,b$ are the roots of the equation $x^2-6x+6=0$
On using the formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ we have
$x=\frac{6\pm \sqrt{6^2-4\times 1\times 6}}{2}$
or $x=\frac{6\pm 2\sqrt{3}}{2}$
Thus, the roots are $a=3+\sqrt{3},b=3-\sqrt{3}$
$\frac{a}{b}=\frac{3+\sqrt{3}}{3-\sqrt{3}}$
$=\frac{3+\sqrt{3}}{3-\sqrt{3}}\times \frac{3+\sqrt{3}}{3+\sqrt{3}}$
$=\frac{9+3+6\sqrt{3}}{6}=2+\sqrt{3}$ (on simplifying)