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:

(v)    \small ar(CYXE)=2ar(FCB)

        

Answers (1)


Since ||gm CYXE and \DeltaACE lie on the same base CE and between the same parallels CE and AX.
Therefore, 2. ar(\DeltaACE) = ar (||gm CYXE) 
B
ut, \DeltaFCB  \cong  \DeltaACE (in iv part)
Since the congruent triangle has equal areas. So,
\small ar(CYXE)=2ar(FCB)
Hence proved

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