# 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)    $\small ar(CYXE)=ar(ACFG)$

M manish painkra

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

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

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

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