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Question 64
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
Solution
Let, the distance between the centre and the vertices on the radii be 'a'.
So, $$x^2=a^2+a^2$$
a = $$\frac{x}{\sqrt{2}}$$
Now , the diagonal of the square , radius of the circle and the distance 'a' forms a right angle triangle with radius as hypotenuse.
so, radius = $$\sqrt{a^2+x^2}=\sqrt{(\frac{x}{\sqrt{2}})^2+x^2}$$
= $$\frac{\sqrt{5}x}{\sqrt{2}}$$
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