Question

find the perimeter of a the rhombus ABCD if ac=24cm and BD=10cm pls do fast​

Answer 1

Answer:

52 cm

Step-by-step explanation:

AC and BD are diagonals.

    Diagonals of rhombus bisect each other at 90.

So, using Pythagoras theorem:

⇒ (1/2 of AC)² + (1/2 of BD)² = side²

⇒ (1/2 of 24)² + (1/2 of 10)² = side²

⇒ 12² + 5² = side²

⇒144 + 25 = 169 = side²

⇒ 13 = side

    Hence, perimeter is

         4 * side = 4*13 cm = 52 cm

Answer 2

Answer:-

Need to know:

properties of Rhombus:

• All sides of rhombus are equal.

• Diagonals of rhombus bisect each other.

According to question,

• AC = 24 cm

• BD = 10 cm

To find the Perimeter,

AE = EC = 12 cm and BE = ED = 5cm (according to property 2)

now,

In ∆AED

By Pythagoras theorem:

AD ² = AE²+ED²

AD²=(12)²+(5)²

AD²= 169

AD=√169

AD=13 cm

Perimeter= AB + BC + CD + DA

Perimeter=13+13+13+13

Perimeter=52 cm ( According to property 1)

Perimeter of Rhombus is 52 cm!!!

sanjushri Reddy'