Question

Consider a square ABCD of side 60 cm. It contains arcs BD and AC drawn with centres at A and D respectively. A circle is drawn such that it is tangent to side AB, and the arcs BD and AC. What is the radius of the circle?

Accepted Answer

B. Let radius of circle be $r$. Center of circle $O$ is at distance $r$ from $AB$. $OD = 60 + r$. $QD = 60 - r$. In right $\Delta DOQ$, $(60 + r)^2 = (60 - r)^2 + OQ^2$. From $\Delta AOQ$, $AO = 60 - r, AQ = r \implies OQ^2 = (60 - r)^2 - r^2 = 3600 - 120r$. Thus $(60+r)^2 = (60-r)^2 + 3600 - 120r \implies 240r = 3600 - 120r \implies 360r = 3600 \implies r = 10$ cm.

Geometry: Inscribed Circle in Square Arcs

Explanation

Practise this under exam conditions