Consider a square ABCD of side 60 cm. lt 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?

Given : Side of square = CD = 60 cm

=> AB = CD = 60 cm , => Radius of circles centered at A and D have equal radius of 60 cm

To find : OP = $$r = ?$$

Solution : $$AO = 60 - r$$ and $$AQ = OP = r$$

In $$\triangle$$ AOQ

=> $$(OQ)^2 = (AO)^2 - (AQ)^2$$

=> $$(OQ)^2 = (60 - r)^2 - (r)^2$$

=> $$(OQ)^2 = 3600 - 120r + r^2 - r^2$$

=> $$(OQ)^2 = 3600 - 120r$$

Now, $$OD = 60 + r$$ and $$QD = 60 - r$$

In $$\triangle$$ DOQ

=> $$(OD)^2 = (QD)^2 + (OQ)^2$$

=> $$(60 + r)^2 = (60 - r)^2 + (3600 - 120r)$$

=> $$(3600 + 120r + r^2)$$ = $$(3600 - 120r + r^2) + (3600 - 120r)$$

=> $$120r + 120r + 120r = 3600$$

=> $$r = \frac{3600}{360} = 10 cm$$