Question

find the perimeter of rectangle whose length is 40m and diagonal is 41m also find area​

Answer 1

Answer:

The Perimeter of the Rectangle is 90 cm and Area is 360 cm²

Step-by-step explanation:

Given :

Length of the Rectangle = 40m

Diagonal of the Rectangle = 41m

To find :

Its Perimeter and Area

Solution :

First find the breadth.

Let the Breadth be as x

★ $\textbf{\small{\underline{By Pythagoras theorem - }}}$

• Base = 40 cm
• Height = x
• Hypotenuse = 41 cm

$\boxed{\sf{{Hypotenuse}^{2} = Base^{2} + Height^{2} }}$

$\sf{\implies} \: {Hypotenuse}^{2} = Base^{2} + Height^{2} \\ \sf{\implies} \: {41}^{2} = {40}^{2} + {x}^{2} \\ \sf{\implies} \:1681 = 1600 + {x}^{2} \\ \sf{\implies} \: {x}^{2} = 1681 - 1600 \\ \sf{\implies} \: {x}^{2} = 81 \\ \sf{\implies} \:x = \sqrt{81} \\ \sf{\implies} \:x = 9$

Breadth = 9 cm

$\rule{300}{1.5}$

✵ Perimeter of the Rectangle -

★ $\boxed{\sf{2(Length+Breadth)}}$

$\sf{\implies} \: 2(40 + 9) \\ \sf{\implies} \:80 + 18 \\ \sf{\implies} \:98$

Perimeter of the Rectangle = 90 cm

$\rule{300}{1.5}$

✵ Area of the Rectangle -

★ $\boxed{\sf{Length \times Breadth}}$

$\sf{\implies} \:40 \times 9 \\ \sf{\implies} \:360$

Area of the Rectangle = 360 cm²

$\therefore$ The Perimeter of the Rectangle is 90 cm and Area is 360 cm²

Answer 2

Answer :-

Perimeter of the rectangle is 98 m

Area of the rectangle is 360 m²

Explanation :-

Length of the recatangle (l) = 40 m

Diagonal of the rectangle (d) = 41 m

Breadth of the rectangle = b

Consider the right triangle formed by length, breadth and diagonal.

By pythgoras theorem

d² = l² + b²

41² = 40² + b²

1681 = 1600 + b²

1681 - 1600 = b²

81 = b²

√81 = b

9 = b

b = 9 m

i.e Breadth of the rectangle = 9 m

We know that

Perimeter of the rectangle = 2(l + b)

= 2(40 + 9)

= 2(49)

= 98 m

Perimeter of the rectangle = 98 m

Area of the rectangle = lb

= 40(9)

= 360 m²

Area of the rectangle = 360 m²