In the given figure, a ||gm ABCD and a rectangle ABEF are of equal area.Then,

(a) perimeter of ABCD=perimeter of ABEF

(b) perimeter of ABCD<perimeter of ABEF

(c) perimeter of ABCD>perimeter of ABEF

(d) perimeter of ABCD= $\frac{1}{2}$ (perimeter of ABEF)

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Solution

ANSWER:
(c) perimeter of ABCD > perimeter of ABEF
Parallelogram ABCD and rectangle ABEF lie on the same base AB, i.e., one side is common in both the
figures.
In ||gm ABCD, we have:
AD is the hypotenuse of right angled triangle ADF.
So, AD > AF
∴ Perimeter of ABCD > perimeter of ABEF

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In the given figure, a ||gm ABCD and a rectangle ABEF are of equal area.Then, (a) perimeter of ABCD=perimeter of ABEF (b) perimeter of ABCD<perimeter of ABEF (c) perimeter of ABCD>perimeter of ABEF (d) perimeter of ABCD=12(perimeter of ABEF)

ANSWER: (c) perimeter of ABCD > perimeter of ABEF Parallelogram ABCD and rectangle ABEF lie on the same base AB, i.e., one side is common in both the figures. In ||gm ABCD, we have: AD is the hypotenuse of right angled triangle ADF. So, AD > AF ∴ Perimeter of ABCD > perimeter of ABEF

In the given figure, a ||gm ABCD and a rectangle ABEF are of equal area.Then,

(a) perimeter of ABCD=perimeter of ABEF

(b) perimeter of ABCD<perimeter of ABEF

(c) perimeter of ABCD>perimeter of ABEF

(d) perimeter of ABCD= $\frac{1}{2}$ (perimeter of ABEF)

ANSWER:
(c) perimeter of ABCD > perimeter of ABEF
Parallelogram ABCD and rectangle ABEF lie on the same base AB, i.e., one side is common in both the
figures.
In ||gm ABCD, we have:
AD is the hypotenuse of right angled triangle ADF.
So, AD > AF
∴ Perimeter of ABCD > perimeter of ABEF