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Q: 1     Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.

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We have ||gm ABCD and a rectangle ABEF both lie on the same base AB such that, ar(||gm ABCD) = ar(ABEF)
for rectangle,  AB = EF 
and for ||gm AB =  CD 
\RightarrowCD = EF 
\Rightarrow AB + CD  =  AB + EF ...........(i)

SInce \DeltaBEC and  \DeltaAFD are right angled triangle
Therefore, AD > AF and BC > BE
\Rightarrow (BC + AD ) > (AF + BE)...........(ii)

Adding equation (i) and (ii), we get

(AB + CD)+(BC + AD) > (AB + BE) + (AF + BE)
\Rightarrow (AB + BC + CD + DA) > (AB + BE + EF + FA)
Hence proved, perimeter of ||gm ABCD is greater than perimeter of rectangle ABEF

Posted by

manish painkra

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