Question

The diagonals of a Rhombus
are 48cm and 14cm. Find its perimeter​

Answer 1

$\large{\boxed{\boxed{\sf Let's \: Understand \: Question \: F1^{st}}}}$

It is a simple question. In which we have given two diagonals of a Rhombus and have to find it's perimeter.

$\large{\boxed{\boxed{\sf How \: to \: do \: it?}}}$

Here, f1st of all we will recall all the properties of rhombus which is that diagonals of a Rhombus Bisect each other at right angle and it's all sides are equal. Now, we will use Pythogoras Theorem in the Triangle made by Diagonals and will get the length of 1side. Now, we know that perimeter of rhombus will be 4times the side (4×side) and hence by substituting value we will get our required answer.

Let's Do It⚡

$\huge{\boxed{ \underline{\sf AnSwer}}}$

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Given:-

• Diagonals of a Rhombus = 48cm and 14cm

Find:-

• Perimeter of Rhombus.

Diagram:-

$\setlength{\unitlength}{1 cm}\begin{picture}(20,15)\thicklines\qbezier(1,1)(1,1)(6,1)\qbezier(1,1)(1,1)(1.6,4)\qbezier(1.6,4)(1.6,4)(6.6,4)\qbezier(6,1)(6,1)(6.6,4)\qbezier(6.6,4)(6.6,4)(1,1)\qbezier(1.6,4)(1.6,4)(6,1)\put(0.7,0.5){\sf A}\put(6,0.5){\sf B}\put(1.4,4.3){\sf D}\put(6.6,4.3){\sf C} \put(3.7,2.7){\sf O} \end{picture}$

Solution:-

we, know that in a rhombus diagonals Bisect each other at 90°

So,

OA = OC

→OA + OC = AC = 48cm

→OA + OA = 48

→2OA = 48

→OA = 48/2 = 24cm

Similarly,

OB = OD

→OB = OD = 7cm

and, $\angle$ AOB = 90°

Now, In ∆AOB

$\huge{\underline{\boxed{\sf H^2 = P^2 + B^2}} \quad\bigg \lgroup \sf Pythogoras \: Theorem \bigg \rgroup}$

$\implies\sf AB^2 = OA^2 + OB^2$

$\sf where \small{\begin{cases} \sf OA = 24cm\\ \sf OB = 7cm \end{cases}}$

$\red\bigstar$ Substituting these values:-

$\dashrightarrow\sf AB^2 = OA^2 + OB^2$

$\dashrightarrow\sf AB^2 = 24^2 + 7^2$

$\dashrightarrow\sf AB^2 = 576 + 49$

$\dashrightarrow\sf AB^2 = 625$

$\dashrightarrow\sf AB = \sqrt{625}$

$\dashrightarrow\sf AB =25cm$

Now, using

$\huge{\underline{\boxed{\sf Perimeter \: of \: Rhombus = 4 \times (side)}}}$

$\sf where \small{\begin{cases} \sf Side, AB = 25cm \end{cases}}$

$\pink\bigstar$ Substituting this value:-

$\implies\sf Perimeter = 4 \times (side)$

$\implies\sf Perimeter = 4 \times (25)$

$\implies\sf Perimeter = 100cm$

$\underline{\boxed{\therefore\sf Perimeter\:of\: rhombus = 100cm}}$

Answer 2

Given:

Diagonals of a Rhombus are 48cm and 14cm.

To Find:

It's Perimeter

Solution:

Before doing this question let us know about the properties of the rohmbus which will be used to solve this problem.

• All the sides of the rohmbus are equal
• It's diagonals bisect each other at 90°

Which means if ABCD is a Rhombus with diagonals AC and BD with centre point O where it's diagonals bisect each other then

• AB = BC = CD = DA
• AC and BD are diagonals
• AO = OC and BO = OD

______________________________________

Given:

AC = 48 and BD = 14

AO = OC = 48/2 = 24cm

BO = OD = 14/2 = 7cm

Perimeter of Rhombus = 4*Side

Now we need to find it's side

Using Pythagoras theorem in triangle AOD

AD² = AO² + OD²

AD² = 24² + 7²

AD² = 576 + 49

AD² = 625

AD = 25

Hence, side of Rhombus is 25cm.

Perimeter lf Rhombus = 4×Side

= 4×25

= 100cm

Hence, perimeter of Rhombus is 100cm.