Question

$\huge\fbox\pink{Answer : }$

The length of a side of a square field is 4 m. What will be the altitude of the rhombus, if the area of the rhombus is equal to the square field and one of its diagonal is 2 m?

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Answer 1

Answer:

• 16/√65 cm is the required altitude of Rhombus

Step-by-step explanation:

According to the Question

It is given that,

• Length of a side of square field is 4m.
• Area of Rhombus = Area of Square
• Diagonal ,d = 2m

We have to find the length of altitude of the rhombus.

Since it is given Area of Square is equal to Area of Rhombus .

• Area of Square = Area of Rhombus

On substituting the value we get

↠ 4 × 4 = ½ × 2 × d'

↠ 16 = d'

Length of another diagonal of Rhombus is 16cm .

Now, calculating the side of Rhombus

• Side of Rhombus = ½ √d²+d'²

On substituting the value we get

↠ Side of Rhombus = ½ √2²+16²

↠ Side of Rhombus = ½ √4+256

↠ Side of Rhombus = ½ √260

↠ Side of Rhombus = ½ × 2√65

↠ Side of Rhombus = √65 cm

Also,Area of Rhombus is calculated by

• Area of Rhombus = Side × Altitude

On substituting the value we get

↠ 16 = √65 × Altitude

↠ 16/√65 = Altitude

↠ Altitude = 16/√65 cm

• Hence, the altitude of rhombus is 16/√65

Answer 2

♦ Concept

Your question is based on the concept of Area and Perimeter firstly let's understand your question so in your question it is given that The length of a side of a square field is 4 m. and now they are asking us to find out the altitude of the rhombus , if the area of the rhombus is equal to the square field and one of its diagonal is 2 m

let's proceed calculation!!

$\underline{ \rule{190pt}{2pt}}$

♦ Formula Used

$\sf \color{pink}{Area \: of \: Rhombus = Area \: of \: square }$

$\sf \color{pink} \: Altitude \: = \frac{area \: of \: rhombus}{side}$

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♦ Solution

$\sf \rightarrow \:Area \: of \: Rhombus = Area \: of \: square$

$\sf \rightarrow \: (4 {)}^{2} \: cm$

$\sf \rightarrow \: 16 \: sq.cm$

$\sf \rightarrow \: Area \: of \: Rhombus = \frac{1}{2} \times d _{1} \times d _{2}$

$\sf \rightarrow \: 16 \: sq.cm = \frac{1}{2} \times 2cm \times d _{2}$

$\sf \rightarrow \: d _{2} = 16cm$

We know that Area of Rhombus = side × attitude

$\sf \rightarrow \: of \: side = \frac{1}{2} \sqrt{ {d}^{2}_{2} + {d}^{2}_{2} }$

$\sf \rightarrow \: \frac{1}{2} \sqrt{ 4 +256 }$

$\sf \rightarrow \: \frac{1}{2} \sqrt{260}$

$\sf \: altitude \: = \frac{area \: of \: rhombus}{side}$

$\sf \rightarrow \: \frac{16 \: sq.cm}{ \frac{1}{2} \sqrt{260m} }$

$\sf \rightarrow \: \frac{16}{ \sqrt{65} } cm$

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♦ Answer

$\sf \color{pink} \rightarrow \: \frac{16}{ \sqrt{65} } cm$

$\: \underline{ \rule{190pt}{2pt}}$

♦ Additional Information

$\begin{gathered} \begin{gathered}\begin{gathered}\boxed{\begin{array}{c} \\ \underline{ { \textbf {\textsf \red{ \dag \: \: More \: Formulas \: \: \dag}}}} \\ \\ \\ \footnotesize \bigstar \: \bf{Area \: _{Square} = Side \times Side} \\ \\ \\ \footnotesize\bigstar \: \bf{Area \: _{Rectangle} = Lenght \times Breadth} \\ \\ \\ \footnotesize \bigstar \: \bf{Area \: _{Triangle} = \frac{1}{2} \times Base \times Height } \\ \\ \\ \footnotesize \bigstar \: \bf{Area \: _{Parallelogram} = Base \times Height} \\ \\ \\ \footnotesize \bigstar \: \bf{Area \: _{Trapezium} = \frac{1}{2} \times [ \: A + B \: ] \times Height } \\ \\ \\ \footnotesize \bigstar \: \bf {Area \: _{Rhombus} = \frac{1}{2} \times Diagonal \: 1 \times Diagonal \: 2}\end{array}}\end{gathered}\end{gathered} \end{gathered}$