Question

find the perimeter of the rectangle where length is 40 cm and a diagonal is 41 cm​

Answer 1

$\huge\bold{\gray{\sf{Answer:}}}$

Explanation:

Given :

Length of Rectangle =40 cm

Diagonal =41 cm

To Find :

Perimeter of the rectangle.?

Formula used here :-

$\bold{Perimeter \:of \:rectangle=2(l+b)}$

$\bold{\red{\boxed{Diagonal = \sqrt{ {l}^{2} + {b}^{2} }}}}$

Now ,come to the question:-

$= > p = 2(l + b)$

$= > p = 2l + 2 \sqrt{ {d}^{2} - {l}^{2} }$

$= > p = 2 \times 40 + 2 \times \sqrt{ {(41)}^{2} - {(40)}^{2} }$

$= > p = 80 + 2 \times \sqrt{1681 - 1600}$

$= > p = 80 + 2 \times \sqrt{81}$

$= > p = 80 + 2 \times 9 = 98$

Therefore, Perimeter of rectangle is 98cm

Answer 2

Answer:

98cm

Step-by-step explanation:

given that length of the side is of the rectangle is l = 40cm

length of the diagonal is d = 41 cm

let breath be a bcm

therefore l'2 + b`2 = d'2

=> 40'2+b'2 =41'2

now if we add both we got 81

b'2 = 81

b= 9 cm

we know that perimeter of the rectangle = 2 ( l+b)

therefore

2(40+9)

= 2×49

= 98cm