ABCD is a rectangle and EFCD is a parallelogram, with the same base CD and have equal areas.Prove that the perimeter of the parallelogram EFCD > perimeter of the rectangle ABCD
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 \(\angle DAE = 90 deg\)
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.
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 \(\angle DAE = 90 deg\)
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.
