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:

(vii)    \small ar(BCED)=ar(ABMN)+ar(ACFG)

        

Answers (1)


We have,
ar(quad. BCDE) = ar(quad, CYXE) + ar(quad. BYXD)

                           = ar(quad, CYXE) + ar (quad. ABMN)        [already proved in part (iii)]
Thus, \small ar(BCED)=ar(ABMN)+ar(ACFG)

Hence proved

                            

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