Question

If area of the rhombus is 24 cm square and one of its diagonal us 4 cm find the perimeter of the rhombus

Answer 1

Area of rhombus is = $\frac{product of diagonals }{2}$


24 = $\frac{4(bd)}{2}$

24 = 2bd

12 = bd 

as the diagonals of a bisector bisect each other AO = OC = 2

and BO= OD = 6 

$(AB)^{2}= (OB)^{2} + (OD)^{2}$

$(AB)^{2}= 6^{2} + 2^{2}$

$AB = \sqrt{40}$

$AB= 2 \sqrt{10}$

as the sides are equal in rhombus i.e AB=BC=CD=DA 

the perimeter of a rhombus is 4 x sides 

4 ( $2 \sqrt{10}$ )

perimeter $= 8\sqrt{10}$

Answer 2

$\large\underline\blue{Diagram:-}$

$\large\underline\pink{Given:-}$

• Area of the Rhombus is 24 cm²
• And one of its diagonal is 4 cm.

$\large\underline\pink{To find:-}$

• Find the perimeter of the Rhombus ....?

$\large\underline\pink{Solutions:-}$

• Let the 1st diagonal be x cm.
• Let the 2nd diagonal be 4 cm.

$\: \: \: \: \: \therefore \: \: Area \: \: of \: \: Rhombus \: \: = \: \: \frac{1}{2} \: \times {d_1} \: \times \: {d_2}$

$\: \: \: \: \: \leadsto \: \: {24} \: \: = \: \: \frac{1}{2} \: \times {x} \: \times \: {4}$

$\: \: \: \: \: \leadsto \: \: {24} \: \times \: {2} \: \: = \: \: {x} \: \times \: {4}$

$\: \: \: \: \: \leadsto \: \: {48} \: \: = \: \: {4x}$

$\: \: \: \: \: \leadsto \: \: {x} \: \: = \: \: \frac{48}{4}$

$\: \: \: \: \: \leadsto \: \: {x} \: \: = \: \: {12} \: cm$

$\: \: \: \: \: \: \: Now, \\ \therefore \: \: In \: \ \triangle \: AOB$

$\: \: \: \: \: \: \: \underline\red{ Using \: \: Pythagoras \: \: Theorem.}$

$\: \: \: \: \: \: \: \leadsto \: \: {AB}^{2} \: \: = \: \: {AO}^{2} \: + \: {OB}^{2}$

$\: \: \: \: \: \: \: \leadsto \: \: {AB}^{2} \: \: = \: \: {(6)}^{2} \: + \: {(2)}^{2}$

$\: \: \: \: \: \: \: \leadsto \: \: {AB}^{2} \: \: = \: \: {36} \: + \: {4}$

$\: \: \: \: \: \: \: \leadsto \: \: {AB}^{2} \: \: = \: \: {40}$

$\: \: \: \: \: \: \: \leadsto \: \: {AB} \: \: = \: \: {\sqrt{40}}$

$\: \: \: \: \: \: \: \leadsto \: \: {AB} \: \: = \: \: {2} \: \sqrt{10}$

$\: \: \: \: \: \: \: \therefore \: \: Perimeter \: \: of \: \: the \: \: Rhombus \: \: = \: \: {4} \: {(side)}$

$\: \: \: \: \: \: \: \leadsto \: \: {4} \: {({2} \: \sqrt{10})}$

$\: \: \: \: \: \: \: \leadsto \: \: {{8} \: \sqrt{10}} \: {cm}^{2}$

$\: \: \: \: \: \: \: Hence, \: \: Perimeter \: \: of \: \: the \: \: Rhombus \: \: is \: \: {{8} \: \sqrt{10}} \: {cm}^{2}.$

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