Question

if area of a rhombus be 24cm² and one of its diagonal is 4cm .find the perimeter of the rhombus.​

Answer 1

Given:

• Area of the rhombus = 24 cm²

• Measure of one of its diagonals = 4 cm

To calculate:

• Perimeter of the rhombus.

Calculation:

In order to calculate the perimeter of the rhombus, we need to find its side. We'll calculate its side by forming an algebraic equation and by solving that equation & by using pythagoras property.

As we know that,

${\underline {\boxed {\bf {{Area}_{(Rhombus)} = \dfrac{Product \: of \: diagonals }{2} } }}}$

$\longrightarrow \sf {{Area}_{(Rhombus)} = \dfrac{d_1 \times d_2 }{2} }$

• d₁ = 4 cm

• d₂ = ?

$\longrightarrow \sf {24 = \dfrac{\cancel{4}\times d_2 }{\cancel{2}} }$

$\longrightarrow \sf {24 = 2 \times d_2 }$

$\longrightarrow \sf {\dfrac{24}{2} = d_2 }$

$\longrightarrow \sf {12 \: cm = d_2 }$

• Measure of the 2nd diagonal of the rhombus is 12 cm.

Now, we'll make a diagram. [Refer to the attachment].

» The rhombus has 4 right angles. The side of the rhombus is the hypotenuse (x) of each angle here. Base and perpendicular become half of its diagonals as the diagonals of the rhombus bisect each other (at 90°).

Remember : All the sides of the rhombus are equal.

Now, by using pythagoras property:

→ H² = B² + P²

→ x² = 6² + 2²

→ x² = 36 + 4

→ x² = 40

→ x = √40

→ x = 2√10

• Side of the rhombus = 2√10 cm

Now, as we know that :

${\underline {\boxed {\bf { {Perimeter}_{(Rhombus)} = 4 \times Side} }}}$

$\longrightarrow \sf {{Perimeter}_{(Rhombus)} = 4 \times x \: cm }$

$\longrightarrow \sf {{Perimeter}_{(Rhombus)} = 4 \times 2 \times \sqrt{10} \: cm}$

$\longrightarrow \sf {{Perimeter}_{(Rhombus)} = 4 \times 2 \times \sqrt{10} \: cm}$

$\longrightarrow \sf {{Perimeter}_{(Rhombus)} = 8 \times \sqrt{10} \: cm}$

$\longrightarrow\boxed{ \sf \red{{Perimeter}_{(Rhombus)} = 8 \sqrt{10} \: cm}}$

Therefore, perimeter of the rhombus is 8√10 cm.