Question

perimeter of a rectangle whose length of one side is 20 cm and diagonal is 29 cm​

Answer 1

Given :

• One side = 20 cm
• Diagonal = 29 cm

To Find :

• Perimeter of rectangle

Solution :

• Using pythagoras theorem :

$\bold{\boxed{\sf{\red{Hypotenuse²\:=\: Perpendicular²\:+\:Base²}}}}$

$\leadsto \sf 29²\:=\: 20²\:+\:Side²$

$\leadsto \sf Side²\:=\:841\:-\:400$

$\leadsto \sf Side\:=\: \sqrt{441}$

$\leadsto \sf Side\:=\: 21\:cm$

• Perimeter of rectangle = 2( l + b )

$\leadsto \sf Perimeter\:=\: 2(20\:+\:21)$

$\leadsto \sf Perimeter\:=\: 2(41)$

$\leadsto \sf Perimeter\:=\:82\:cm$

• Perimeter of rectangle = 82 cm

_________________________

Answer 2

Answer :-

• Perimeter of the rectangle is 82cm.

Given :-

• Length and diagonal of a rectangle is 20cm and 29cm

To Find :-

• Perimeter of the rectangle.

Solution :-

Let ABCD be the rectangle in which

• AB = CD (length)
• AD = BC (breadth)
• AC = BD (diagonal)

Here

• AB (length) = 20cm
• AC (diagonal) = 29cm
• BC = ?

We've to find BC.

As we know that

• AC² = AB² + BC² [ Pythagoras theorem ]

⇒ (29)² = (20)² + (BC)²

⇒ 841 = 400 + BC²

⇒ 841 - 400 = BC²

⇒ 441 = BC²

⇒ BC = √441

⇒ BC = 21cm

→ BC (breadth) = 21cm

As we know that

• Perimeter of a rectangle is 2 (l + b)

Where

• l = length
• b = breadth

According to question :-

⇒ Perimeter of the rectangle = 2 (20 + 21)

⇒ Perimeter of the rectangle = 2 × 41

⇒ Perimeter of the rectangle = 82cm

Hence, the perimeter of the rectangle is 82cm.