Q 31. If a parallelogram and a square (or a rectangle) stand on the same base and between the same parallels, then the perimeter of the square (or the rectangle) is less than that of the gm.
Lets prove it using rectangle:
given: ABCD is a rectangle and EFCD is a parallelogram, with the same base CD and have equal areas.
TPT: perimeter of the parallelogram EFCD > perimeter of the rectangle
proof:
since parallelogram and rectangle have equal areas with same base CD, therefore it will be between same set of parallel lines.
CD=EF............(1) [opposite sides of the parallelogram]
CD=AB...........(2) [opposite sides of the rectangle]
from (1) and (2), EF=AB...........(3)
in the triangle DAE,
since ∠DAE=90 deg
ED>AD [since length of the hypotenuse is greater than other sides]..........(4)
CF>BC [since CF=ED and BC=AD]...............(5)
perimeter of parallelogram EFCD
=EF+FC+CD+DE
=AB+FC+CD+DE [using (3)]
>AB+BC+CD+AD [using (5)]
which is the perimeter of the rectangle ABCD
therefore perimeter of parallelogram with the same base and with equal areas is greater than the perimeter of the rectangle.