A) 10 cm
B) 15 cm
C) 12 cm
D) 14 cm
Correct Answer: B
Solution :
(b): \[PS\bot QR\] \[QS=SR=12\]cm. \[OR=OP=Circum-radius=r\,cm.\] \[PS=\sqrt{P{{Q}^{2}}-Q{{S}^{2}}}\] \[=\sqrt{{{\left( 12\sqrt{5} \right)}^{2}}-{{(12)}^{2}}}\] \[=\sqrt{144\times 5-144}\] \[=\sqrt{144(5-1)}=\sqrt{144\times 4}\] 24 cm. In \[\Delta ORS\], \[OS=\left( 24-r \right)\]cm. \[\therefore \]\[O{{R}^{2}}=O{{S}^{2}}+R{{S}^{2}}\] \[\Rightarrow \]\[{{r}^{2}}={{(24-r)}^{2}}+{{12}^{2}}\] \[\Rightarrow \] \[{{r}^{2}}=576-48r+{{r}^{2}}+144\] \[\Rightarrow \] \[r=\frac{720}{48}=15\]cmYou need to login to perform this action.
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