Parallelogrom ABCD and rectangle ABEF have the same base AB and also have equal areas.Show that the perimeter of the parallelogram is greater than that of the rectangle.
consider the above diagram
opposite sides of rectangle and parallelogram are equal so
AB= EF and AB= CD so EF=CD
therefore AB+CD=AB+EF ...... (1)
Now since of all the segments that can be drawn to a given line from a point, the perpendicular segment is the shortest.so,
AF< AD and BE<BC
AD+BC > AF+ BE ...... (2)
Adding (1) and (2)
AB+BC+CD+AD > AB+BE+EF+AF
⇒ Perimeter of parallelogram ABCD > perimeter of rectangle ABEF