Question

Find the perimeter of the rectangle whose length is 40 cm and a diagonal 41 cm

Answer 1

The diagonal of the rectangle divides it into two right-angled triangles, so we can use the pythagorean theorem to solve this problem. 
41cm is the diagonal and 40cm is the length.
hyp² = length² + breadth²
41²   = 40² + breadth²
1681 = 1600 + breadth²
1681 - 1600 = breadth²
breadth² = 81
breadth = √81
= 9cm
Perimeter = 2( l + b )
               = 2 * (40 + 9)
               = 2 * 49
               = 98cm
The perimeter of the rectangle is 98cm.

Answer 2

let ABCD be rectangle.
Then AC will be diagonal. And angle B = 90.
Length(BC) = 40 cm
diagonal(AB) = 41 cm
By using pythagoras theorem, we have:-
$AB^{2} + BC^{2} = AC^{2}$
$x^{2} + 40^{2} = 41^{2}$
$AB^{2} = 41^{2} - 40^{2}$
$AB^{2} = 1681 - 1600 = 81$
$AB = \sqrt{81} = 9 cm$
Now, Perimeter of rectangle = 2(L+B) = 2(BC+ AB)
                                         = 2(9 + 40) = 2(49)
                                         = 98 cm