Question

3. The perimeter of a rhombus is 180 cm and one of its diagonals is 72 cm. Find the length
of the other diagonal and the area of the rhombus.​

Answer 1

Answer:

$\large\bold\red{d_{2}=54\:cm}\\\\\large \bold\red{ Area = 1944 \: {cm}^{2} }$

Step-by-step explanation:

Given,

A Rhombus ABCD .

Note:- Refer to attachment for Diagram.

Now,

It's given that,

Perimeter of Rhombus = 180 cm

But,

We know that,

All the sides of Rhombus are equal.

Now,

Let's the side of Rhombus be 'a' cm

Therefore,

We get,

$= > 4a = 180 \\ \\ = > a = \frac{180}{4} \\ \\ = > a = 45 \: cm$

Also,

It's given that,

One of it's diagonal is 72 cm

But,

We know that,

Diagonals of a rhombus bisect each other at 90°.

Therefore,

In ∆OBC,

• OC = 36 cm
• BC = 45 cm

Now,

By Pythagoras Theorem,

We have,

$= > {OC}^{2} + {OB}^{2} = {BC}^{2} \\ \\ = > {OB}^{2} = {BC}^{2} - {OC}^{2}$

Now,

Putting the respective values,

We get,

$= > {OB}^{2} = {(45)}^{2} - {(36)}^{2} \\ \\ = > {OB}^{2} = (45 + 36)(45 - 36) \\ \\ = > {OB}^{2} = 81 \times 9 \\ \\ = > OB = \sqrt{81 \times 9} = \sqrt{729} \\ \\ = > OB = 27 \: cm$

Therefore,

Other diagonal is (27×2) i.e., 54 cm

Now,

We know that,

Area of Rhombus is equals to

$\large \boxed { \purple{\frac{1}{2} \times d_{1} \times d_{2}}}$

Therefore,

Putting the respect values,

We get,

$= > A = \frac{1}{2} \times 72 \times 54 \\ \\ = > A = 36 \times 54 \\ \\ = > \large \bold{ A = 1944 \: {cm}^{2} }$

Answer 2

Answer:

Each side of the rhombus is 180/4 = 45 cm.

One diagonal is 72 cm, so half the diagonal = 36 cm

Half of the other diagonal will be,

==> [45^2 - 36^2]^(1/2)

==> [729]^(1/2) = 27 cm.

So the other diagonal = 2 * 27 = 54 cm.

Thus, One diagonal is 72 cm the other diagonal is 54 cm.

The area of the rhombus = (72 * 54)/2 = 1944 sq cm.