1. Two chords of length 20 cm and 24 cm are drawn perpendicular to each other in a circle of radius is 15 cm. What is the distance between the Points of intersection of these chords (in cm) from the center of the circle?
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By: anil on 05 May 2019 01.48 am
Given : Radius = 15 cm, AB = 24 cm and CD = 20 cm To find : FG = ? Solution : In $$ riangle$$ BOG, => $$(OG)^2=(OB)^2-(BG)^2$$ => $$(OG)^2=(15)^2-(12)^2$$
=> $$(OG)^2=225-144=81$$ ----------(i) Similarly, in $$ riangle$$ COF, => $$(OF)^2=(OC)^2-(CF)^2$$ => $$(OF)^2=(15)^2-(10)^2$$ => $$(OF)^2=225-100=125$$ ----------(ii) Again, in $$ riangle$$ GOF, => $$(FG)^2=(OF)^2+(OG)^2$$ => $$(FG)^2=125+81=206$$ => $$FG=sqrt{206}$$ cm => Ans - (C)
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=> $$(OG)^2=225-144=81$$ ----------(i) Similarly, in $$ riangle$$ COF, => $$(OF)^2=(OC)^2-(CF)^2$$ => $$(OF)^2=(15)^2-(10)^2$$ => $$(OF)^2=225-100=125$$ ----------(ii) Again, in $$ riangle$$ GOF, => $$(FG)^2=(OF)^2+(OG)^2$$ => $$(FG)^2=125+81=206$$ => $$FG=sqrt{206}$$ cm => Ans - (C)