Question

. If the perimeter of a rhombus is 100 cm and one of the diagonals is 40 cm, then the area of the rhombus is in square centimetres is​

Answer 1

$\\ \\ \large\underline{ \underline{ \sf{ \red{given:} }}}$

• Perimeter of rhombus = 100cm.

• Diagonal(1) = 40cm.

$\\ \\ \large\underline{ \underline{ \sf{ \red{to \: find:} }}}$

• Area of rhombus.

$\\ \\ \large\underline{ \underline{ \sf{ \red{solution:} }}}$

All sides of rhombus are equal.

So,

• Perimeter = 4 × side

↪ 100 = 4 side

↪ Side = 25cm

Side of rhombus is 25cm.

⋆⋆⋆⋆ Diagonal bisect each other at right angle.

From figure ,

➛ 25² = 20² + x²

➛ 625 = 400 + x²

➛ x² = 625 - 400

➛ x² = 225

➛ x = 15cm

Second diagonal d(2) = 2x = 30cm

$\\ \small \bigstar \boxed{ \bf \: area \: of \: rhombus = \frac{product \: of \: two \: diagonals}{2} }$

$\\ \sf \: area = \frac{40 \times 30}{2} \\ \\ \sf \: area = \frac{1200}{2} \\ \\ \boxed{\sf \: \pink{area = 600 {cm}^{2} }}$

Therefore , area of rhombus is 600cm².

Answer 2

Given:

• Perimeter = 100 cm
• Diagonal₍₁₎ = 40 cm

To find:

☞ Area of rhombus = ?

Method:

Perimeter = 4 × side

⇒ 100 = 4 × side

⇒ Side = 100 ÷ 4

⇒ Side = 25 cm

So AB = BC = CD = DA = 25 cm

Length of diagonal = 40 cm

⇒ AC = 40 cm

⇒ AO + OC = 40 cm

⇒ AO + AO = 40 cm           [∵ AO = OC]

⇒ 2AO = 40 cm

⇒ AO = 40 ÷ 2

⇒ AO = 20 cm

In ΔAOD,

By Pythagoras theorem,

(AD)² = (AO)² + (OD)²

⇒ (25)² = (20)² + (OD)²

⇒ 625 = 400 + (OD)²

⇒ (OD)² = 625 - 400

⇒ (OD)² = 225

⇒ (OD) = √225

⇒ OD =  ±15

Side cannot be negative.

∴ OD = 15 cm

Diagonal₍₂₎ = OD + OB

⇒ Diagonal₍₂₎ = 2(OD)

⇒ Diagonal₍₂₎ = 2 × 15

⇒ Diagonal₍₂₎ = 30 cm

Area of Rhombus = (Diagonal₍₁₎ × Diagonal₍₂₎) ÷ 2

⇒ Area = (40 × 30) ÷ 2

∴ Area = 600 cm²