Question

If parallelogram and rectangle are drawn on same base with equal area prove that perimeter of parallelogram is greater than that of rectangle

Answer 1

Hello Mate!

Given : ||gm ABCD adn rect. CDEF is rectangle on same base and have same area.

To prove : Perimeter of ||gm > Perimeter of rectangle

Proof : Since they have same area and same base so they lie between same parallels.

So, AB = CD, since ABCD is ||gm

Similarly, EF = CD because CDEF is rectangle.

So, AB = CD and EF = CD

Therefore, AB = EF __(i)

Adding CD to equation (i)

CD + AB = EF + CD __(ii)

Now, because hypotenuse is greater than any side in right ∆.

AD > ED and BC > FC

Adding both equation we get,

AD + BC > ED + FC __(iv)

Adding eq (iii) and (iv)

AB + CD + AD + BC > EF + CD + ED + FC

Perimeter of ||gm > Perimeter of rectangle

Hence proved.

Have great future ahead!

Answer 2

here is your answer OK ☺☺☺☺☺☺☺☺



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.