1. 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?
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By: anil on 05 May 2019 02.37 am
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 $$ riangle$$ 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 $$ riangle$$ DOQ => $$(OD)^2 = (QD)^2 + (OQ)^2$$ => $$(60 + r)^2 = (60 - r)^2 + (3600 - 120r)$$ => $$(3600 + 120r + r^2)$$ = $$(3600 - 120r + r^2) + (3600 - 120r)$$ => $$120r + 120r + 120r = 3600$$ => $$r = frac{3600}{360} = 10 cm$$
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