Question

if the area of a rhombus is 24cm^ on of its diagonal one be 4cm find the perimeter of rhombus​

Answer 1

$\Large {\underline { \sf \orange{Clarification :}}}$

Here, we are given that the area of the rhombus is 24 cm² and one of its diagonals is 4 cm. We have to find the perimeter of the rhombus.

In order to find the perimeter of the rhombus, we need its side. Also, in order to calculate its side we need the length of its another diagonal. Then, by using pythagoras property we'll find its side. After that, by using the formula for the perimeter of the rhombus,we'll find its perimeter.

$\Large {\underline { \sf \orange{Explication \: of \: steps :}}}$

Let ABCD be the rhombus here. [Kindly refer to the attachment.]

• AC and BD are its diagonals respectively.
• Let O be the point where diagonals are bisecting each other.

As we know that,

★ Diagonals of a rhombus bisect each other at 90°.

So, we'll use here pythagoras property as there are 4 right angles forming in the rhombus.

In ∆ ABO :

→ AO² + BO² = AB²

→ √(AO² + BO²) = AB (Side)

Here,

• AO = $\sf {\dfrac{D_1}{2}}$

• BO = $\sf {\dfrac{D_2}{2}}$

So, we can say that :

$\bigstar \: \boxed{\sf { Side =\sqrt{ {\Bigg \lgroup \dfrac{D_1}{2} \Bigg \rgroup }^{2} + {\Bigg \lgroup \dfrac{D_2}{2} \Bigg \rgroup }^{2} }}}$

$\underline{\small \sf {\maltese \; \; \; Finding \: its \: other \: diagonal \: (D_2) : \; \; \; }}$

We know that,

$\bigstar \: \boxed{\sf { Area_{(Rhombus)} = \dfrac{D_1 \times D_2}{2} }}$

• $\sf { D_1 = 4 \: cm}$

$\longrightarrow \sf { 24 \: {cm}^{2}= \dfrac{\cancel{4} \: cm \times D_2}{\cancel{2} }}$

$\longrightarrow \sf { 24 \: {cm}^{2}= 2 \: cm \times D_2 }$

$\longrightarrow \sf { \dfrac{24 \: {cm}^{2}}{2\: cm}= D_2 }$

$\longrightarrow \boxed{ \sf { 12 \: cm = D_2 }}$

Now,

$\underline{\small \sf {\maltese \; \; \; Finding \: its \: side : \; \; \; }}$

$\bigstar \: \boxed{\sf { Side =\sqrt{ {\Bigg \lgroup \dfrac{D_1}{2} \Bigg \rgroup }^{2} + {\Bigg \lgroup \dfrac{D_2}{2} \Bigg \rgroup }^{2} }}}$

$\longrightarrow \sf {Side =\sqrt{ {\Bigg \lgroup \cancel{\dfrac{4}{2}} \Bigg \rgroup }^{2} \: cm + {\Bigg \lgroup \cancel{\dfrac{12}{2}} \Bigg \rgroup }^{2} \: cm } }$

$\longrightarrow \sf {Side =\sqrt{ (2)^2 \: cm + (6)^2 \: cm}}$

$\longrightarrow \sf {Side = \sqrt{4 \: cm + 36 \: cm}}$

$\longrightarrow \sf {Side = \sqrt{40 \: cm}}$

$\longrightarrow \boxed {\sf {Side = 2\sqrt{10} \: cm}}$

$\underline{\small \sf {\maltese \; \; \; Finding \: its \: perimeter : \; \; \; }}$

We know that,

$\bigstar \: \boxed{\sf { Perimeter_{(Rhombus)} = 4 \times Side }}$

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

$\longrightarrow$ ❝ $\boxed{ \sf \orange {Perimeter_{(Rhombus)} = 8\sqrt{10} \: cm }}$ ❞

Therefore, perimeter of the rhombus is $\pmb { \mathfrak \gray { 8\sqrt{10} \: cm }}$.