Question

If the diagonals of a rhombus are 4.8 cm and 1.4 cm, then what is the perimeter of the rhombus?

Answer 1

As we know the formula,

Perimeterof Rhombus=
$2 \sqrt{(d1)^2+ (d2)^2}$
So,
$2 \sqrt{(4.8)^2 + (1.4)^2}$
$2 \sqrt{23.04+1.96}$
$2 \sqrt{25}$
$2 \times 5$
$10cm$

Answer 2

Answer:

$\large{\underline{\boxed{\sf Perimeter = 10 \: cm}}}$

Step-by-step explanation:

Given that ;

Diagonals of a rhombus are 12 cm and 16 cm.

Let AC = 4.8 cm and BD = 1.4 cm

We know that,

The diagonals of a rhombus bisect each other at right angles, i.e., 90°

Therefore,

AO = OC = 2.4 cm

BO = OD = 0.7 cm

Now, In ΔBOC,

BO = 2.4 cm

OC = 0.7 cm

∠BOC = 90°

Using Pythagoras theorem,

BC² = BO² + OC²

⇒ BC² = (2.4)² + (0.7)²

⇒ BC² = 5.76 + 0.49

⇒ BC² = 6.25

⇒ BC = √6.25

⇒ BC = 2.5 cm

Also, we know that each side of rhombus is equal.

Thus, the length of each side of rhombus is 2.5 cm.

We know that,

Perimeter of rhombus = 4 * each side

⇒ Perimeter = 4 * 2.5

⇒ Perimeter = 10 cm

Hence, the perimeter of rhombus is 10 cm.