Question

Find the area of a rhombus one side of which measure 20cm and one of whose diagonals is 24 cm

Answer 1

Answer:

384 cm ^2

Step-by-step explanation:

see attachment

Hope it helps you

Answer 2

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 $\bf 384 \: cm^2$.