question_answer

AB and CD are two parallel chords of a circle of lengths 10 cm and 4 cm, respectively. If the chords are on the same side of the centre and the  distance between them is 3 cm, then the diameter of the circle is [SSC CGL Tier II, 2015]

A) \[\sqrt{29}\,cm\]

B) \[2\sqrt{21}\,cm\]

C) \[\sqrt{21}\,cm\]

D) \[2\sqrt{29}\,cm\]

Correct Answer: A

Solution :

[a] According to the question,                         OB = OD = radius of sphere In \[\Delta OMB,\] \[{{(OB)}^{2}}={{(OM)}^{2}}+{{(MB)}^{2}}\] \[\Rightarrow \]   \[{{r}^{2}}={{x}^{2}}+{{5}^{2}}\]                        (Let OM = x) \[\therefore \]      \[{{r}^{2}}={{x}^{2}}+25\]                         (i) In \[\Delta {\mathrm O}\Nu D,\] \[{{(OD)}^{2}}={{(ON)}^{2}}+{{(ND)}^{2}}\] \[\Rightarrow \]   \[{{r}^{2}}={{(x+3)}^{2}}+{{2}^{2}}\] \[\Rightarrow \]   \[{{r}^{2}}={{x}^{2}}+9+6x+4\] \[\Rightarrow \]   \[{{x}^{2}}+25={{x}^{2}}+6x+13\]            [From Eq.(i)] \[\Rightarrow \]   \[6x=25-13=12\] \[\therefore \]      \[x=2\,cm\] Then, \[r=\sqrt{4+25}\]\[\Rightarrow \]\[r=\sqrt{29}\] \[\therefore \]      \[2r=2\sqrt{29}\]