Question

find the diagonal of a rectangle whose length is 40 cm and breath 11 cm​

Answer 1

Step-by-step explanation:

Pythagoras theorem,

(Diagonal)^2=(Length)^2+(Width)^2

Now as given in the question,

Diagonal = 41 cm.

Length = 40 cm.

So, Putting these value we get,

(41)^2=(40)^2+(Width)^2

(Width)^2=(41)^2-(40)^2

(Width)^2=1681-1600

(Width)^2=81

Width=9cm

Hence the width of the rectangle is 9 cm.

So

The perimeter of the rectangle = 2 ( Length + Width )

= 2 ( 40 cm + 9 cm )

= 2 x 49 cm

= 98 cm

Hence the perimeter of the rectangle is 98 cm.

Answer 2

Step-by-step explanation:

We know that diagonal of rectangle

$diagonal \: of \: recangle \: = \sqrt{l ^{2} + {b}^{2} }$

Length = 40 cm

Breadth = 11 cm

$\sqrt{ {40}^{2} + {11}^{2} } \\ \sqrt{1600 + 121 } \\ \sqrt{1721}$

$41.5cm \:$