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A parallelogram and a rectangle are on the same base between the same parallels. Prove that perimeter of the parallelogram is greater than that of the



Construct the diagram and you will find two right angled triangles. The hypotenuse of this triangle is the side of the parallelogram, and the altitude is the side of the rectangle. It's a given fact that the hypotenuse is always greater than the altitude, so therefore sum of two sides of the parallelogram are greater than sum of two sides of the rectangle. The other two sides of the rectangle and parallelogram are equal, so they're sums are the same.
Hence, because of two larger sides, parallelogram has greater perimeter than rectangle. 
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I would have provided diagram, but it would have been deleted then
Given ;A parallelogram  and a  rectangle ABEF with the same base AB and equal areas.
To Prove ;Perimeter of parallelogram ABCD> perimeter of rectangle ABEF.
proof; Since opposite side of a parallelogram and a rectangle are equal.
therefore AB=DC                    [ABCD  IS A PARALLELOGRAM}
and   AB=EF                            [ IS ABEF  IS A RECTANGLE ]
       DC=EF                 ..............(1)
;AB+DC=AB+EF        ...............(11)
since of all the segment that can be drawn to a give line from a point not lying on it   the perpendicular segment is the shortest.
;BE<BC and AF<AD
BC>BE and  AD>AF
BC+AD>BE+AF                          .....................(111)
Adding (11) and (111), we get


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