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.

Answers (1)



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

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