Question

ABCD is a rhombus with diagonals AC = 18 cm and BD = 24 cm. Find the perimeter of ABCD.
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Answer 1

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Answer 2

The perimeter of the given rhombus ABCD is 60 cm.

Given :

Length of diagonal AC of rhombus = 18 cm

Length of diagonal BD of rhombus = 24 cm

To Find :

The perimeter of the rhombus ABCD

Solution :

Let the diagonals AC and BD intersect at point O. According to a property of Rhombus, the diagonals perpendicularly bisect each other.

Thus, AOB would be a right-angled triangle at O and measures being :

AO = Half of diagonal AC

      $= 18/2$

       $= 9 cm$

BO = Half of diagonal BD

      $= 24/2$

       $= 12 cm$

Applying Pythagoras' theorem,

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

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

                 $(AB)^{2} = 81 + 144$

                 $(AB)^{2} = 225$

                   $AB = \sqrt{225}$

                   $AB = 15$

Thus, the side of the rhombus ABCD is 15cm. As all the sides of the rhombus are the same,

Perimeter or Rhombus ABCD = $4 * 15$

                                                  $= 60$

Hence, the perimeter of the Rhombus is 60cm.

To learn more about Rhombus, visit

https://brainly.com/question/88523

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