Question

If the diagonals of a rhombus are
24 cm and 10 cm, what is its perimeter?​

Answer 1

Given that :

Let diagonal 1 = 10 cm.

Let diagonal 2 = 24 cm.

Find out :

The perimeter of the Rhombus = ?

Solution :

Using Formula :-

P = 4a

a = $\frac{√p²+q²}{2}$

So,

Let p = 24 and q = 10.

Solving for p :

→ P = 2√p²+q² = 2.√24²+10² = 52 cm.

Answer 2

Answer:

Here, AC = 24 cm and BD = 10 cm, therefore, AO = 12 cm and BO = 5 cm.

Now to find the perimeter, we need the length of AB.

Let’s find AB using the pythagoras theorem in ABC,

AB2 = AO2 + OB2

AB2 = 122 + 52

AB2 = 144 + 25

AB2 = 169

AB = √169

AB = 13 cm

Since the length of one side of the rhombus is AB = 13 cm, perimeter of the rhombus, P = 4 × side of a rhombus

P = 4 × AB

P = 4 ×13 = 52 cm

Thus, the perimeter of the rhombus is 52 cm