Question

llgm ABCD and rect.ABEF are on the same base AB and have equal areas. Show that the perimeter of the llgm is greater than that of the rectangle.

please solv

Answer 1

Hi there!

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For the solution of your question, see the above picture.

Now, let's see the related topics :-

1️⃣ Parallelogram : In geometry, parallelogram is a simple quadrilateral with two pairs of parallel sides.

2️⃣ Rectangle : In geometry, a rectangle is a quadrilateral with four right angles.

3️⃣ Angle : In geometry , an angle is the figure formed by two rays.

4️⃣ Line : A line is defined as a line of points that extends infinitely in two directions.

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Thanks for the question :).

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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.

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