Question

Prove that of all parallelograms of which the sides are given, the parallelogram which is rectangle has the greatest area

Answer 1

$\textbf{\huge{\underline{ Hello Mate! }}}$

Hi, here we need to know that

Area of ||gm = base × height or b × h

Area of □ = length × breadth or a × b

If you concentrate on 1st figure i.e. of ||gm then h is perpendicular and we are getting a small triangle where "a" is hypotenuse.

Since hypotensue is longest side in triangle,

a > h

Multiplying b both sides

a × b > h × b

Or,

Area of □ > Area of ||gm

$\textsf{\red{ Hence proved }}$

$\textbf{ Have great future ahead! }$

Answer 2

here is your answer OK


Let ABCD be a parallelogram in which AB = x and AD = y.
Let h be the altitude corresponding to the base AB. Then,
ar(gm ABCD) = AB h
= xh

Since, the sides x and y are given.
Therefore, with same sides x and y we can construct infinitely many parallelogram with different heights.
Now, ar(gm ABCD) = xh
ar(gm ABCD) is maximum when h is maximum.
(since, x is given i.e x is constant)
But the maximum value which h can attain is AD = y and this is possible when AD is perpendicular to AB i.e the gm ABCD becomes a rectangle.
Thus, ar(gm ABCD) is greatest when ADAB i.e when gm ABCD is a rectangle