# 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)    $ar(BYXD)=ar(ABMN)$

M manish painkra

Since, ar (||gm BYXD) = 2 .ar ($\Delta$MBC) ..........(i) [already proved in 2nd part]
and, ar (sq. ABMN) = 2. ar ($\Delta$MBC)............(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

$\small ar(BYXD)=ar(ABMN)$

Exams
Articles
Questions