Question

7.
Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.​

Answer 1

Breadth = 9cm

Given ,

• Length of rectangle = 40cm
• Diagonal of rectangle = 41 cm

We need to find ,

• Breadth of rectangle = ?

As we know that ,

• Diagonal = √l² + b²

By simplifying ,

• d² = l² + b²
• d² - l² = b²
• Breadth = √d² - l²

Now , substituting the known values ,

=> Breadth = √41² - 40²

=> Breadth = √1681 - 1600

=> Breadth = √81

=> Breadth = √9²

=> Breadth = 9 cm

Answer 2

Given :-

The length of the side of the rectangle is l = 40cm 

and

The length of the diagonal is d = 41cm

let breadth be bcm

Therefore, 

$l {}^{2} + b {}^{2} = d {}^{2}$

$⟹40^2+b^2=41^2$

$⟹b^2=81$

$⟹b=9cm$

We know that ,

The perimeter of the rectangle is 

$2(l+b)=2(40+9)=98cm$