Question

how we can solve the the diagonals of a rhombus are in the ratio 3 ratio 4 and the area is 54 CM square find the side of the Rhombus​

Answer 1

AnswEr :

• Diagonal₁ : Diagonal₂ = 3 : 4
• Area of Rhombous = 54 cm²
• Find the Side of Rhombous.

Let the Diagonal₁ and, Diagonal₂ of the Rhombous will be 3x and, 4x.

• According to the Question Now :

$\longrightarrow \tt Area \:of \:Rhombous = \dfrac{D_1 \times D_2}{2} \\ \\\longrightarrow \tt 54 = \dfrac{3x\times4x}{2} \\ \\\longrightarrow \tt 54 = \dfrac{12{x}^{2}}{2} \\ \\\longrightarrow \tt54= {6}{x}^{2} \\ \\\longrightarrow \tt \cancel\dfrac{54}{6}= {x}^{2} \\ \\\longrightarrow \tt9 = {x}^{2} \\ \\\longrightarrow \tt \sqrt{9} = x \\ \\\longrightarrow \green{\tt x = 3}$

$\rule{300}{1}$

• D I A G O N A L S :

◗ Diagonal₁ = 3x = 3(3) = 9 cm

◗ Diagonal₂ = 4x = 4(3) = 12 cm

$\rule{300}{2}$

• We know that Diagonal of Rhombous Bisect Each Other.
• So Diagonal AC and BD Bisect Each Other at Point O, such that BO = 9 /2 cm, and CO = 6 cm.
• We Know that Each Angle at Point O will be Equal to 90°.

⋆ Refrence of Image is in the Diagram :

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\thicklines\put(8.6,3){\large{A}}\put(9,1.3){\mathsf{9/2 cm}}\put(10,1.3){\mathsf{6 cm}}\put(7.7,0.9){\large{B}}\put(11.1,0.9){\large{C}}\put(9.9,2.1){\large{O}}\put(8,1){\line(1,0){3}}\put(11,1){\line(1,2){1}}\put(9,3){\line(3,0){3}}\put(11,1){\line(-1,1){2}}\put(8,1){\line(2,1){4}}\put(8,1){\line(1,2){1}}\put(12.1,3){\large{D}}\end{picture}$

• In ∆ BOC ; Using Pythagoras theorem :

$\implies \tt BC^{2} = BO^{2} + CO^{2} \\ \\\implies \tt BC^{2} = \bigg(\frac{9}{2}\bigg)^{2} + (6)^{2} \\ \\\implies \tt BC^{2} = \dfrac{81}{4} + 36 \\ \\\implies \tt BC^{2} = \dfrac{81+144}{4} \\ \\\implies \tt BC^{2} = \dfrac{225}{4} \\ \\ \implies \tt BC= \sqrt{ \dfrac{225}{4} } \\ \\\implies \tt BC= \sqrt{ \dfrac{15 \times 15}{2 \times 2}} \\ \\\implies \tt BC= \dfrac{15}{2} \\ \\\implies \large \boxed{ \red{\tt BC= 7.5 \: cm}}$

⠀

∴ Side of the Rhombous will be 7.5 cm.

#answerwithquality #BAL

Answer 2

Answer:

7.5 cm

Step-by-step explanation:

GIVEN THAT :

Diagonals of a rhombus are in the ratio = 3 : 4

Area of rhombus = 54 cm²

TO FIND :

Length of side of the rhombus

SOLUTION :

Let D1 and D2 be the two diagonals of the rhombus.

Let D1 = 3x cm

Let D1 = 3x cmLet D2 = 4x cm

We know,

Area of rhombus = 1/2 × D1 × D2

Putting the values in the above equation, we get :

=> 54 = 1/2 × 3x × 4x

=> 6x² = 54

=> x² = 54/6

=> x² = 9 cm

=> x = √9 = 3 cm

Therefore,

D1 = 3x = 9 cm

D1 = 3x = 9 cmD2 = 4x = 12 cm

Now,

Diagonals of a rhombus bisect each other at right angles.

So, by Pythagoras theorem, we can write :

=> (1/2 D1 )² + (1/D2 )² = ( Side of rhombus )²

Let each side of the rhombus be a cm

=> Putting the values in the equation :

=> a² = ( 1/2 × 9 )² + ( 1/2 × 12 )²

=> a² = ( 81/4 ) + ( 144/4 )

=> a² = 225/4

=> a = √225/√4

=> a = 15/2

=> a = 7.5 cm

ANS ) Length of each side of the rhombus is 7.5 cm

Thanks !

#answerwithquality

#BAL