Q: 8    In Fig. \small 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment \small AX\perp DE  meets BC at Y. Show that:

(vi)    \small ar(CYXE)=ar(ACFG)

    

Answers (1)


Since ar( ||gm CYXE) = 2. ar(\DeltaACE)  {in part (v)}...................(i)
Also, \DeltaFCB and quadrilateral ACFG lie on the same base FC and between the same parallels FC and BG.
Therefore, ar (quad. ACFG) = 2.ar(\DeltaFCB) ................(ii)

From eq (i) and eq (ii), we get

\small ar(CYXE)=ar(ACFG)
Hence proved

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