Question

In the given figure, ABCD is a rhombus. If AC = 24 cm and BD = 10 cm, then the perimeter of the quadrilateral ABCD is 
13
52
68
104​

Answer 1

Answer:

perimeter = 52 cm

Step-by-step explanation:

Let the diagonals intersect at point O

Given :

ABCD is a rhombus

AC=24 cm

BD = 10cm

=> the diagonals will bisect each other at 90°

In ∆ DOC,

angle DOC = 90°

DO= 5 cm ; OC= 12 cm

Therefore by Pythagoras theorem,

${dc}^{2} = {do}^{2} + {oc}^{2}$

${dc}^{2} = {5}^{2} + {12}^{2}$

DC^2 = 25+144= 169

=> DC = 13 cm

Since the sides of a rhombus are equal,

=> AB=BC=CD=DC= 13cm

Therefore, perimeter = 13 x 4

= 52 cm

Hope it helps......!