Question

A square ABCD is inscribed in a circle of radius $r$, Another circle is inscribed in ABCD and a square EFGH is inscribed in this circle, then the side EF is equal to ?

• A. $r$
• B. $\sqrt{2}r$
• C. $r/2$
• D. $r/\sqrt{2}$

Answer 1

The correct option is A $r$

In given question $a=r$ is taken in the figure.

Let O be the centre of both the circles.

OD = OA = r (Radii of the outer circle)

AC and BD are diameters of the outer circle which bisect each other at right angle. Also, AC and BD are diagonals of square ABCD.

Therefore, on applying Pythagoras theorem in right angled ΔAOD,

${AD}^2={OA}^2+{OD}^2$

${AD}^2=r^2+r^2$

${AD}^2=2r^2$

$AD=\sqrt{2}r$

$AD=AB=BC=CD=\sqrt{2}r$

Since, the circle inscribed in square ABCD will pass through the mid points of all the sides, EG and HF will be equal to the side of square ABCD.

$EG=HF=AB$.

Also, EG and HF will bisect each other at O; Also they would be perpendicular to each other.

Therefore $OE=OF=\frac{AB}{2}=\frac{\sqrt{2}r}{2}$

Therefore, on applying Pythagoras theorem in right angled ΔOEF,

${EF}^2={OE}^2+{OF}^2$

$EF=2{\left(\frac{r\sqrt{2}}{2}\right)}^2$

${EF}^2=r^2$

$EF=r$

So, option A is the answer.