A) 14
B) 10
C) 8
D) 5
Correct Answer: B
Solution :
Let AB and CD be parallel chords of a circle (on the same side of the centre) of lengths \[6\,cm\] and \[8\,cm\] respectively. Let OLM be the perpendicular to the chords, with \[LM=1\text{ }cm\]. If r is the radius of the circle, then \[OL=\sqrt{{{r}^{2}}-{{4}^{2}}}\] \[OM=\sqrt{{{r}^{2}}-{{3}^{2}}}\] \[\therefore \] \[\sqrt{{{r}^{2}}-{{3}^{2}}}-\sqrt{{{r}^{2}}-{{4}^{2}}}=1\] ??(i) Also \[({{r}^{2}}-{{3}^{2}})-({{r}^{2}}-{{4}^{2}})=7\] .....(ii) Adding corresponding sides, we get \[2\sqrt{{{r}^{2}}-{{3}^{2}}}=8\] or \[{{r}^{2}}=25\] or \[r=5\] Therefore diameter is \[10\text{ }cm.\]You need to login to perform this action.
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