Question

the perimeter of a rhombus is 68 square cm and one of it's diagonal is 30 cm.Find the length of it's other diagonal ​

Answer 1

• The length of the other diagonal (BD) = 16 cm.

Given :

The perimeter of a rhombus = 68 cm².

The one diagonal (AC) = 30 cm.

To Find :

The length of the other diagonal (BD).

Solution :

First, we need to find the side of the rhombus.

Given,

Perimeter the rhombus = 68 cm²

We know that,

Perimeter of a rhombus = 4a

• a = side.

That means,

$\bf :\implies 4a = 68 \: cm \\ \\ \bf :\implies a = \dfrac{68}{4} \: cm \\ \\ \bf :\implies a = 17 \: cm$

Hence, the side of the rhombus is 17 cm.

Given,

The one diagonal (AC) is 30 cm.

A rhombus divided into four right angled triangle and O is the mid-point of the rhombus.

So,

• OA = 15 cm.

• OC = 15 cm.

In ∆OCD,

By the Pythagoras theorem,

$\underline{ \boxed{ \bf{ {(H)}^{2} = {(B)}^{2} + {(P)}^{2} }}}$

Where,

• H = Hypotenuse (CD).
• B = Base (OC).
• P = Perpendicular (OD).

We have to find the perpendicular.

We have,

• Hypotense = CD = 17 cm.

• Base = OC = 15 cm.

$\bf :\implies {CD}^{2} = {OC}^{2} + {OD}^{2} \\ \\ \bf :\implies {(17 \: cm)}^{2} = {(15 \: cm)}^{2} + {OD}^{2} \\ \\ \bf :\implies 289 \: {cm}^{2} = 225 \: {cm}^{2} + {OD}^{2} \\ \\ \bf :\implies 289 \: {cm}^{2} - 225 \: {cm}^{2} = {OD}^{2} \\ \\ \bf :\implies 64 \: {cm}^{2} = {OD}^{2} \\ \\ \bf :\implies \sqrt{64 \: {cm}^{2} } = OD \\ \\ \bf :\implies 8 \: cm = OD \\ \\ \: \bf \: \therefore \: OD = 8 \: cm$

The other diagonal = OD + OB

O is the mid-point of the rhombus.

So,

• OB = OD

• OB = 8 cm.

$\bf :\implies 8 \: cm + 8 \: cm \\ \\ \bf :\implies 16 \: cm$

Hence,

The length of the other diagonal (BD) is 16 cm.