Question

Each side of a rhombus is of length 20 cm. what is the area of this rhombus if one of its diagonals equals to 240 cm ?




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Answer 1

Correct Question :-

Each side of a rhombus is of length 20 cm. What is the area of this rhombus if one of its diagonals equals to 24 cm ?

Answer :-

Area of the rhombus is 384 cm².

Explanation :-

With given information draw a rhombus ABCD.

[ Refer to the attachment ]

Consider ΔAOD

→ AB = 20 cm

We know that diagonals in a rhombus bisect each other perpendicularly.

→ ∠AOD = 90°

→ OD = BD / 2 = 24 / 2 = 12 cm

As ΔAOD is right angled at O it is a Right angled triangle.

By pythagoras theorem

⇒ AO² + OD² = AD²

⇒ AO² + 12² = 20²

⇒ AO² + 144 = 400

⇒ AO² = 200 - 144

⇒ AO² = 256

⇒ AO = √256 = 16 cm

We know that diagonals in a rhombus bisect each other perpendicularly.

→ AC = 2AO = 2( 16 ) = 32 cm

So here

One of the diangonal ( d₁ ) = BD = 24 cm

Another diagonal ( d₂ ) = AC = 32 cm

Area of the rhombus ABCD = ( 1 / 2 ) * d₁ * d₂

= ( 1 / 2 ) * BD * AC

= ( 1 / 2 ) * 24 * 32

= ( 1 / 2 ) * 768

= 768 / 2

= 384 cm²

Hence, Area of the rhombus is 384 cm².

Answer 2

Correct Question :

Each side of a rhombus is of length 20 cm. What is the area of this rhombus if one of its diagonals equals to 24 cm ?

AnswEr :

Let's make a Rombous PQRS with Midpoint O. [ Refer to Attachment ]

⋆ In ∆POQ, where ∠O = 90°

➛ OQ = SQ ( Diagonal ) /2

➛ OQ = 24 /2

➛ OQ = 12 cm

⋆ Using Pythagoras Theorem :

➛ PQ² = PO² + OQ²

➛ (20)² = PO² + (12)²

➛ 400 = PO² + 144

➛ 400 - 144 = PO²

➛ PO² = 256

• Square Root to Both Side

➛ PO = 16 cm

⋆ Diagonals of rhombus bisects each other perpendicularly. Therefore :

➛ PR = 2 × PO

➛ PR = 2 × 16 cm

➛ PR = 32 cm

◑ Area of Rhombus PQRS

⇒ 1 /2 × Diagonal₁ × Diagonal₂

⇒ 1 /2 × PR × SQ

⇒ 1 /2 × 32 cm × 24 cm

⇒ 16 cm × 24 cm

⇒ 384 cm²

჻ Area of the rhombus PQRS is 384 cm².