Q.10 (a) (iii) PF = (10-x) cm , show that PS= (20-x)cm PQ = x cm Q.10 (b)    The diagram shows the vertices of a square KLMN touching the sides of the same hexagon ABCDEF , With KN parallel to FE . USE your result from part (a) (ii) and part (ii) to find the length of a side of the square .

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10(a)(iii)


$$\begin{gathered} Solution : Perpendiculars from F and E is drawn on the line PS meeting PS at M and N respectively. \\ So, MN will be 10cm as FEMN forms a rec\tan gle \\ let PM=y=NS \left(say\right) \\ Also, \\ \angle APQ+\angle QPS+\angle FPS={180}^O (Angles on a stright line) \\ \Rightarrow {30}^o+{90}^o+\angle FPS={180}^o \\ \Rightarrow \angle FPS={180}^o-{120}^o={60}^o \\ Now, in △FPM \\ \cos {60}^o=\frac{PM}{PF} \\ \Rightarrow \frac{1}{2}=\frac{y}{10-x} \\ \Rightarrow y=\frac{10-x}{2} \\ and PS=y+10+y=2y+10 \\ \Rightarrow PS=(\cancel{2}\times \frac{10-x}{\cancel{2}})+10 \\ \Rightarrow PS=10-x+10=20-x \\ Hence, PS=20-x \end{gathered}$$

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