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:

(iv)    ar(BYXD)=ar(ABMN)

            

Answers (1)


Since, ar (||gm BYXD) = 2 .ar (\DeltaMBC) ..........(i) [already proved in 2nd part]
and, ar (sq. ABMN) = 2. ar (\DeltaMBC)............(ii)
[Since ABMN and AMBC are on the same base MB and between same parallels MB and NC]
From eq(i) and eq (ii), we get

\small ar(BYXD)=ar(ABMN)

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