Question

the perimeter of a rhombus is 100 cm one of its diagonal is 40 cm find the length of the Other diagonal​

Answer 1

Answer:

Perimeter of Rhombus: 100 cm

We know that Rhombus has 4 equal sides. Therefore, the length of each side is:  100 / 4 = 25 cm

Therefore length of each side of rhombus is 25 cm.

Also, the diagonals of a rhombus bisect each other at a point. In our case, the point is called O. We are given that, AC = 40 cm.

Therefore OA = 40 / 2 = 20 cm

Now consider Δ AOB.

Here if we apply Pythagoras Theorem we get,

⇒ OA² + OB² = AB²

⇒ 20² + OB² = 25²

⇒ OB² = 25² - 20²

⇒ OB² = 625 - 400

⇒ OB² = 225

⇒ OB = √ 225 = 15 units.

Now we need the length of BD. We know that OB is half of BD.

Therefore BD = 2 × 15 = 30 units

Therefore the length of other diagonal is 30 cm.

Answer 2

Answer:

The diagonals are :

• AC = 30 cm.
• BD = 40 cm.

Step-by-step explanation:

Given :

Perimeter of the rhombus = 100 cm.

Now,

Let, ABCD be the rhombus and AC and BD be the diagonals.

We will find the Perimeter first,

Therefore,

We know that :

Perimeter of the rhombus :

${\bigstar\color{aqua}{\boxed{\sf 4\times Sides.}}}$

So,

=> 100 = 4 × side

=> Side = 25

______________Now,

As given,

BD = 40 cm.

We also know that,

The diagonals of a rhombous bisect each other at right angles.

____________{ Property of Rhombous! }

This implies that,

AO = OC and OB = OD = 20cm.

Therefore,

∠AOB = ∠BOC = ∠COD = ∠AOD = 90°

______{ Making each triangle a right triangle! }

According to Pythagorean theorem,

We get,

In the triangle of AOB,

• OB = 20cm
• AB = 25 cm
• ∠O is 90°

So,

⇒ AB² = OB² + AO²

⇒ 25²  =  20² + AO²

⇒  25² - 20²  =  AO²

⇒ 625 - 400 =  AO²

⇒ 225 =  AO²

⇒  15²  =  AO²

⇒ AO = 15 cm.

Implies - The diagonal is AC = (AO + OC) = 30 cm.

∴  The diagonals are AC = 30 cm and BD = 40 cm.