Question

The diagonals of a rhombus measure 18 cm and 24 cm. Find its perimeter.​

Answer 1

Answer:

In a rhombus, the diagonals bisect each other at 90

0

Referring the diagram, if PR =18cm, SQ =24cm, then OP =9cm, OQ =12cm

Since, triangle POQ is a right angled triangle,

PQ

2

=OP

2

+OQ

2

=>9

2

+12

2

=225

PQ=

225

=15 cm

Hence, the length of the sides of the diagonal =15 cm.

Since the length of all sides of a rhombus are equal, Perimeter =4× length of one side =4×15=60cm

Answer 2

Answer:The perimeter of given rhombus is 60 cm.

Step-by-step explanation:

Given:The diagonals of a rhombus measure 18 cm and 24 cm.

To find: We have to find the perimeter of rhombus.

Explanation:

Step 1: We know that, the diagonals of a rhombus bisect each other at right angles.

∴ $AO=OC=\frac{24}{2}$

                    $= 12$ cm

and $BO=OD= \frac{18}{2}$

                        $= 9$

Step 2: Consider right angled triangle Δ AOB,

              From pythagoras theorem, we have

                      $(AB)^{2} = (AO)^{2} + (BO)^{2}$

                                 $= (12)^{2} + (9)^{2}$

                                 $=144+81$

                                 $= 225$

⇒                        $AB =\sqrt{225}$

                           $AB = 15$ cm

Hence the measure of each side og given rhombus is 15 cm.

Step 3: The perimeter of a rhombus is given by,

                          $P = 4a$

                          $P = 4$×$15$

                          $P = 60$ cm

Hence, the perimeter of given rhombus is 60 cm.

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