AB and CD are two parallel chords of a circle such that AB = 10 cm and CD = 24 cm. If the chords are on the opposite sides of the centre and distances between them is 17 cm, then the radius of the circle is |
A) 12 cm
B) 13 cm
C) 10 cm
D) 11 cm
Correct Answer: B
Solution :
Let radius of the circle be r. |
Given, \[MN=17\,cm\] |
Let \[ON=x\] |
Then, \[OM=17-x\] |
In \[\Delta AOM,\]\[O{{A}^{2}}=O{{M}^{2}}+A{{M}^{2}}\] |
\[\Rightarrow \] \[{{r}^{2}}={{(17-x)}^{2}}+{{5}^{2}}\] |
\[=289+{{x}^{2}}-34x+25\] |
\[{{r}^{2}}=314+{{x}^{2}}-34x\] ... (i) |
In \[\Delta CON,\]\[O{{C}^{2}}=O{{N}^{2}}+C{{N}^{2}}\] |
\[{{r}^{2}}={{x}^{2}}+{{12}^{2}}={{x}^{2}}+144\] ... (ii) |
From Eqs. (i) and (ii), we get |
\[314+{{x}^{2}}-34x={{x}^{2}}+144\] |
\[\Rightarrow \] \[34x=170\]\[\Rightarrow \]\[x=5\] |
From Eq. (ii), |
\[{{r}^{2}}=25+144=169\] |
\[\Rightarrow \] \[r=13\,cm\] |
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