Question

find the perimeter,area and length of diagonal of a rectangle,having length=3.2cm, breadth=2.4cm​

Answer 1

Step-by-step explanation:

Length ( l ) = 3.2 cm

Breadth ( b ) = 2.4 cm

Perimeter = 2 ( l + b )

= 2 ( 3.2 + 2.4 )

= 2 ( 5.6 )

= 11.2 cm

Area = l × b

= 2.4 × 3.2

= 7.68 cm^2

$lenght \: of \: diagonal \: = \sqrt{ {l}^{2} + {b}^{2} } \\ \\ \\ = \sqrt{ {3.2}^{2} + {2.4}^{2} } \\ \\ \\ = \sqrt{10.24 + 5.76} \\ \\ \\ = \sqrt{16} \\ \\ \\ = 4 \: cm$

THANKS

Answer 2

Question -:

Find the perimeter,area and length of diagonal of a rectangle having :

length = 3.2cm, breadth = 2.4cm

Explanation -:

Given :

• Length of a rectangle = 3.2 cm
• Breadth of the rectangle = 2.4 cm

Need to find :

• Perimeter
• Area
• Diagonal

Solution :

First we will calculate the Perimeter

$\small \boxed{\sf{ Perimeter = 2(L + B)} }$

Substituting the value we get

→ 2 (3.2 + 2.4)

→ 2 × 5.6

→ 11.2 cm

• Perimeter = 11.2 cm

$\rule{40mm}{2pt}$

Now we will calculate the Area

$\small \boxed{\sf{ Area = L × B}}$

Substituting the value we get

→ (3.2 × 2.4) ²

→ 7.68 cm ²

• Area =7.68 cm ²

$\rule{41mm}{2pt}$

Now we will calculate the diagonal

$\small \boxed{\sf Diagonal = \sqrt{ L² + B²}}$

Substituring the value we get

$\small \sqrt{({3.2)}^{2} +{( 2.4)}^{2} }$

$\small \sqrt{10.24 + 5.76 }$

$\small \: \sqrt{16} = 4cm$

• Diagonal = 4 cm

$\rule{71mm}{4pt}$