Question

The diagonals of a rhombus are 5: 12. If its perimeter is 104 cm, find the length of the sides and the diagonals

Answer 1

Solutions :-

Given :
Perimeter of Rhombus = 104 cm
The diagonals of a rhombus are 5: 12

Find the side of the rhombus :-

Side of Rhombus = Perimeter / 4
= 104 / 4 cm
= 26 cm

Let the diagonal BD and AC be 5x and 12x respectively.

∴ BO = 5x/2 = 2.5x & AO = 12x/2 = 6x

Find the length of diagonals :-

We know that the diagonal of rhombus bisect each other at 90°.

In right triangle AOB, By using Pythagoras theorem :-

(AO)² + (BO)² = (AB)²
=> (6x)² + (2.5x)² = (26)²
=> 36x² + 6.25x² = (26)²
=> 42.25x² = 676
=> x² = 676 / 42.25
=> x = √16 = 4

Therefore,
BD = 5x = 5 × 4 = 20 cm
AC = 12x = 12 × 4 = 48 cm

Hence,
Length of side of the rhombus = 26 cm
And its diagonals are 20 cm and 48 cm.

Answer 2

$\underline{\underline{\huge\mathfrak{Solution ;}}}$

Stated that ;-
• Perimeter of Rhombus = 104 cm
• Ratios of their diagonals = 5:12

We have to find the Lenght and sides of the given diagonals , So let's find out ;-

Firstly , finding the sides ;-

• Side of Rhombus = Perimeter ( P ) / 4
$= > \frac{104}{4} = 26 \: cm$
Assuming here the diagonals as BD and AC with and summing their values as 5y and 12y respectively.

Hence ,
$= > \frac{5y}{2} = 2.5 \: y$
and

$= > \frac{12y}{2} = 6 \: y$

Now ,
We are well know known with it that the diagonals of a Rhombus bisects eachother at 90°.

Now, from the above attachment , by using the Pythagoras theorem , let's find the value ;-

$= >( 6y)^{2} +( 2.5y)^{2} = (26)^{2}$

$= > 36y^{2} + 6.25y^{2} = (26)^{2}$

$= > 42.25y^{2} = 676$

$= > y^{2} = \frac{676}{42.25} = 16$

$= > y = \sqrt{16} = 4$

Hence ,
• AC = 12y = ( 12 × 4 ) = 48 cm
• BD = 5y = ( 5 × 4 ) = 20 cm

Therefore ,
$<b>Lenght if Sides = 26 cm and Diagonals = 20 cm and 48 cm.</b>$