A square is inscribed in a quarter-circle in such a manner that two of its adjacent vertices lie on the two radii at an equal distance from the centre, while the other two vertices lie on the circular arc. If the square has sides of length $$x$$. then the radius of the circle is:
A. $$\frac{{16x}}{{\pi + 4}}$$
B. $$\frac{{2x}}{{\sqrt x }}$$
C. $$\frac{{\sqrt 5 x}}{{\sqrt 2 }}$$
D. $$\sqrt 2 x$$
Answer: Option C
Solution(By Examveda Team)
Let ABCD is a square of x unit side
Then ∠AOD = 90°
Then OD = $$\frac{{\text{x}}}{{\sqrt 2 }}$$
Diagonal of square ABCD = $$\sqrt 2 $$ x
Line MB || OD
i.e OD = MB = $$\frac{{\text{x}}}{{\sqrt 2 }}$$
⇒ Then MBOD will be a rectangle become MB || OD, MB = OD = $$\frac{{\text{x}}}{{\sqrt 2 }}$$
BD || MO, MO = BD = $$\sqrt 2 $$ x
$${\text{R}} = \sqrt {{{\left( {\frac{{\text{x}}}{{\sqrt 2 }}} \right)}^2} + {{\left( {\sqrt 2 {\text{x}}} \right)}^2}} = \frac{{\sqrt 5 {\text{x}}}}{{\sqrt 2 }}{\text{ Ans}}{\text{.}}$$
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