question_answer

PQRS is a trapezium in which \[\mathbf{PS}\parallel \mathbf{QR}\] and \[PQ=SR=12\]m. then the distance of PS from QR is:

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 \[\Delta RSF\], \[sin45{}^\circ =\frac{SF}{SR}\] \[\Rightarrow \]\[\frac{1}{\sqrt{2}}=\frac{SF}{SR}\] \[\Rightarrow \] \[SF=\frac{12}{\sqrt{2}}=6\sqrt{2}\]