Question
Prove that of all parallelograms of which the sides are given, the parallelogram which is a rectangle, has the greatest area.
Solution
Given: A parallelogram ABCD whose sides are given.
To Prove: The area of the parallelogram ABCD is the greatest when it is a rectangle.
Construction: Draw DE ⊥ AB.
Proof: ar(|| gm ABCD)
= Base x Corresponding altitude = AB x DE
When parallelogram ABCD is a rectangle, then its area = AB x DA
| ∵ Then ∠DAB = 90° as such DA will be the altitude
In right triangle DEA,
∠DEA > ∠DAE
DA > DE
∵ Side opposite to greater angle of a triangle is longer ∴ ar(rectangle ABCD) > ar(|| gm ABCD)
i.e., the area of the parallelogram ABCD is the greatest when it is a rectangle.
