question_answer

A square ABCD is inscribed in a circle of radius a. Another circle is inscribed in ABCD and a square EFGH is inscribed in this circle. The side EF is equal to

A)  a               

B)                         \[a\sqrt{2}\]

C)  \[\frac{a}{\sqrt{2}}\]                    

D)         \[\frac{a}{2}\]

Correct Answer: A

Solution :

 Let side of the square \[ABCD=\text{ }x\text{ }cm,\]and side of the square \[EFGH=y\text{ }cms\] In right-angle triangle OAE,                 \[O{{A}^{2}}=O{{E}^{2}}+E{{A}^{2}}\] or            \[{{a}^{2}}={{\left( \frac{x}{2} \right)}^{2}}+{{\left( \frac{x}{2} \right)}^{2}}=\frac{{{x}^{2}}+{{x}^{2}}}{4}\] or            \[2{{a}^{2}}={{x}^{2}}\] In right-angle triangle EHG,                 \[E{{G}^{2}}=E{{H}^{2}}+H{{G}^{2}}\] or            \[{{x}^{2}}={{y}^{2}}+{{y}^{2}}\] or            \[2{{y}^{2}}={{x}^{2}}=2{{a}^{2}}\] \[\therefore \]  \[y=a=EF\]