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?

Given : Radius = 15 cm, AB = 24 cm and CD = 20 cm

To find : FG = ?

Solution : In $$\triangle$$ BOG,

=> $$(OG)^2=(OB)^2-(BG)^2$$

=> $$(OG)^2=(15)^2-(12)^2$$

=> $$(OG)^2=225-144=81$$ ----------(i)

Similarly, in $$\triangle$$ COF,

=> $$(OF)^2=(OC)^2-(CF)^2$$

=> $$(OF)^2=(15)^2-(10)^2$$

=> $$(OF)^2=225-100=125$$ ----------(ii)

Again, in $$\triangle$$ GOF,

=> $$(FG)^2=(OF)^2+(OG)^2$$

=> $$(FG)^2=125+81=206$$

=> $$FG=\sqrt{206}$$ cm

=> Ans - (C)