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 .

Dear Student,


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 rectanglelet PM=y=NS (say)Also,APQ+QPS+FPS=180O (Angles on a stright line)30o+90o+FPS=180oFPS=180o-120o=60oNow, in FPMcos 60o=PMPF12=y10-xy=10-x2and PS=y+10+y=2y+10PS=(2×10-x2)+10PS=10-x+10=20-xHence, PS=20-x

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Can you explain part a(ii)
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