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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:

(ii)    \small ar(BYXD)=2ar(MBC)


Answers (1)


SInce ||gm BYXD and \DeltaABD are on the same base BD and between same parallels BD and AX
Therefore, ar(\DeltaABD) = 1/2. ar(||gm BYXD)..........(i)

But, \DeltaABD \cong \DeltaMBC (proved in 1st part)
Since congruent triangles have equal areas.
Therefore, By using equation (i) we get
 \small ar(BYXD)=2ar(MBC)
Hence proved

Posted by

manish painkra

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