A) \[10\sqrt{2}\]m
B) \[4\sqrt{2}\]m
C) \[5\sqrt{2}\]m
D) \[6\sqrt{2}\]m
Correct Answer: D
Solution :
(d): \[PE\bot QR;SF\bot QR\] \[\therefore \]\[\angle SRQ={{45}^{{}^\circ }}\] In \[\Delta RSF\], \[sin45{}^\circ =\frac{SF}{SR}\] \[\Rightarrow \]\[\frac{1}{\sqrt{2}}=\frac{SF}{SR}\] \[\Rightarrow \] \[SF=\frac{12}{\sqrt{2}}=6\sqrt{2}\]You need to login to perform this action.
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