# 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.

M manish painkra

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
$\Rightarrow$CD = EF
$\Rightarrow$ AB + CD  =  AB + EF ...........(i)

SInce $\Delta$BEC and  $\Delta$AFD 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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