A) \[\frac{2r}{(\sqrt{2}+1)}\]
B) \[\frac{r}{\sqrt{2}}\]
C) \[(\sqrt{2}-1)r\]
D) \[\sqrt{2}r\]
Correct Answer: C
Solution :
[c] Each side of square\[ABCD=2r\] Diagonal, \[BD=\sqrt{2}(2r)=2\sqrt{2}r\] \[BO=\frac{1}{4}\times BD=\sqrt{2}r\] \[\Rightarrow \] \[BP+PO=\sqrt{2}r\] \[\Rightarrow \] \[r+PO=\sqrt{2}r\] \[\Rightarrow \] \[PO=(\sqrt{2}-1)r\] |
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