Math, asked by BlessedMess, 2 months ago

★Find the area of a rhombus each side of which measures 20cm and one of whose diagonals is 24cm.​

Answers

Answered by Clαrissα
33

Given :

  • Each side of the rhombus measures 20 cm.

  • And also one of its diagonals is 24 cm.

To Find :

  • Area of the rhombus.

Calculation :

Here,

  • AB = 20 cm
  • BO = 12 cm

We know that diagonals of a rhombus bisects each other at right angle [at  \tt \: 90° ]

In right \Delta  \bf{AOB}, by Pythagoras theorem.

 \sf{\longrightarrow AB^2  \: = \:  BO^2 \:  +  \: AO^2} \\  \\ \\  \sf{\longrightarrow (20)^2  \: = \:  (12)^2  \: +  \: AO^2 } \\  \\ \\   \sf \: \longrightarrow \: 400 \:  -  \: 144 \:  = AO^2 \:  \\  \\  \\ \longrightarrow \underline{\boxed{\bf{256 \:  =  \:  \: AO^2 }}} \:

Now, calculating,

 \sf{\longrightarrow AO \:  =  \:  \sqrt{256}  \: cm^2}  \\  \\  \\  \longrightarrow \underline{\boxed{\bf{AO \:  =  \: 16 \: cm }}} \:

Therefore,

 \sf{\longrightarrow BD \:  =  \: 16 + 16  \: cm}  \\  \\  \\  \longrightarrow \underline{\boxed{\bf{BD \:  =  \: 32 \: cm }}} \:

 \dag According to the Question,

We know that,

  • \boxed{\bf{ Area  \: of \:  rhombus \:  =  \dfrac{1}{2} \:  \times d_1 \:  \times d_2 }}

Where,

  • Diagonal 1 is 24 cm [Given]

  • Diagonal 2 is 32 cm

Putting the values,

 \sf{\longrightarrow  \dfrac{1}{2}  \: \times  \: d_1 \times \:  d_2}  \\  \\  \\  \sf \: {\longrightarrow  \dfrac{1}{2}  \: \times  \: 24  \: cm \: \times \:  32 \: cm}  \\  \\  \\ \longrightarrow \underline{\boxed{\bf{ 384 \: cm^2 }}} \: \purple{\bigstar}

 \therefore Hence, area of the rhombus is  \bf 384 \: cm^2 .

Attachments:
Answered by ppig2478
3

Answer:

Given :

Each side of the rhombus measures 20 cm.

And also one of its diagonals is 24 cm.

To Find :

Area of the rhombus.

Calculation :

Here,

AB = 20 cm

BO = 12 cm

We know that diagonals of a rhombus bisects each other at right angle [at \tt \: 90°90° ]

In right \Delta \bf{AOB}ΔAOB , by Pythagoras theorem.

\begin{gathered} \sf{\longrightarrow AB^2 \: = \: BO^2 \: + \: AO^2} \\ \\ \\ \sf{\longrightarrow (20)^2 \: = \: (12)^2 \: + \: AO^2 } \\ \\ \\ \sf \: \longrightarrow \: 400 \: - \: 144 \: = AO^2 \: \\ \\ \\ \longrightarrow \underline{\boxed{\bf{256 \: = \: \: AO^2 }}} \: \end{gathered}

⟶AB

2

=BO

2

+AO

2

⟶(20)

2

=(12)

2

+AO

2

⟶400−144=AO

2

256=AO

2

Now, calculating,

\begin{gathered} \sf{\longrightarrow AO \: = \: \sqrt{256} \: cm^2} \\ \\ \\ \longrightarrow \underline{\boxed{\bf{AO \: = \: 16 \: cm }}} \: \end{gathered}

⟶AO=

256

cm

2

AO=16cm

Therefore,

\begin{gathered} \sf{\longrightarrow BD \: = \: 16 + 16 \: cm} \\ \\ \\ \longrightarrow \underline{\boxed{\bf{BD \: = \: 32 \: cm }}} \: \end{gathered}

⟶BD=16+16cm

BD=32cm

\dag† According to the Question,

We know that,

\boxed{\bf{ Area \: of \: rhombus \: = \dfrac{1}{2} \: \times d_1 \: \times d_2 }}

Areaofrhombus=

2

1

×d

1

×d

2

Where,

Diagonal 1 is 24 cm [Given]

Diagonal 2 is 32 cm

Putting the values,

\begin{gathered} \sf{\longrightarrow \dfrac{1}{2} \: \times \: d_1 \times \: d_2} \\ \\ \\ \sf \: {\longrightarrow \dfrac{1}{2} \: \times \: 24 \: cm \: \times \: 32 \: cm} \\ \\ \\ \longrightarrow \underline{\boxed{\bf{ 384 \: cm^2 }}} \: \purple{\bigstar} \end{gathered}

2

1

×d

1

×d

2

2

1

×24cm×32cm

384cm

2

\therefore∴ Hence, area of the rhombus is \bf 384 \: cm^2384cm

2

.

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