Question

Q58 The diagonals of a rhombus measure
16cm and 30cm. Find its perimeter.​

Answer 1

Answer :-

68cm

Explanation :-

Given :

Diagonals  of the rhombus = 16cm and 30cm

To Find :

Perimeter of the rhombus

Solution :

Let ABCD be a rhombus and its diagonals, AC and BD, are intersecting each other at point O.

We know diagonals of a rhombus are perpendicular therefore they bisect each other at 90°

So,

$\implies \sf{}AO=\dfrac{AC}{2}$

$\sf{}\implies \dfrac{16}{2}$

$\sf{}\therefore 8$

$\implies \sf{}BO=\dfrac{BD}{2}$

$\sf{}\implies \dfrac{30}{2}$

$\sf{}\therefore 15$

By applying Pythagoras theorem in ΔAOB,

$\sf{}\implies OA^2+OB^2=AB^2$

$\sf{}\implies 8^2+15^2=AB^2$

$\sf{}\implies64+225=AB^2$

$\sf{}\implies 289=AB^2$

$\sf{}\implies AB=\sqrt{289}$

$\sf{}\implies AB=17$

Therefore, the length of the side of rhombus is 17 cm.

We know,

All sides of the rhombus are equal.

Perimeter of rhombus,

= 4 × Side of the rhombus

= (4 × 17)cm

= 68 cm

Therefore,the perimeter of the given rhombus is equal to to 68cm

Answer 2

Diagram :-

$\setlength{\unitlength}{2mm}\begin{picture}(0,0)\thicklines\put(0,0){\line(3,4){1cm}}\put(0,0){\line(3,-4){1cm}}\put(5.1,-6.6){\line(3,4){1cm}}\put(10,0){\line(-3,4){1cm}}\put(5,-6.6){\line(0,3){2.7cm}}\put(0,0){\line(3,0){2cm}}\put(5.5,1){\sf\footnotesize 30cm}\put(1.5,-1.5){\sf\footnotesize 16cm}\put(4.6,7){\textbf{\textsf{A}}}\put(-2.5,-0.5){\textbf{\textsf{ B}}}\put(10.5,-0.5){\textbf{\textsf{C}}}\put(4.5,-8.5){\textbf{\textsf{D}}}\put(5,1){\line(-3,0){2mm}}\put(4,0){\line(0,3){2mm}}\put(5.4,-1.5){\textbf{\textsf{O}}}\put(5.1,0){\circle*{0.5}}\end{picture}$

Solution :-

Taking a rhombus ABCD with centre O

As we know that

• Diagonals of rhombus bisect perpendicularly.

Now, taking ∆AOB from the rhombus it has a right angle at O.

Using Pythagoras theorem .

Hypotenuse² = Base² + Height²

Here ,

• Hypotenuse = AB = side of rhombus
• Base = BO = 16/2 = 8
• Height = AO = 30/2 = 15

Substituting we have ,

⇒ Hypotenuse² = 8² + 15²

⇒ Hypotenuse² = 64 + 225

⇒ Hypotenuse² = 289

⇒ Hypotenuse = $\sqrt{289}$

⇒ Hypotenuse = 17 = side of rhombus

Now finding perimeter of the rhombus.

As we know that ,

Perimeter of rhombus = 4 ( side )

$\sf\longrightarrow Perimeter_{\sf rhombus} = 4 ( 17 )$

$\sf\longrightarrow Perimeter _{rhombus}=\textbf{\textsf{68cm}}$

Hence , Perimeter of rhombus = 68cm