Question

The diagonals of a Rhombus are in ratio 3:4 . If its perimeter is 40 cm,find the lengths of sides and diagonals of Rhombus.

Answer 1

$\mathfrak{\huge{\underline{Answer:}}}$

Given ABCD is a Rhombus

$\therefore \: \sf AB = BC = CD = AD$

$\therefore \sf \: Side \: Of \: Rhombus = \dfrac{1}{4} \times 40$

$\sf \: Side = 10cm$

Let BD = 3x and OC = 4x

Now,Triangle DOC is a right-angled triangle.

$\therefore \sf\: ({OD) }^{2} + {(OC) }^{2} = ( {CD )}^{2}$

$\sf \implies \: (\dfrac{3}{2} x) ^{2} + (\dfrac{4}{2} x) ^{2} = 10$

$\sf \implies \: \dfrac{9 {x}^{2} }{4} + \dfrac{16 {x}^{2} }{4} = 100$

$\sf \implies25 {x}^{2} = 100 \times 4$

$\sf \implies \: {x}^{2} = \dfrac{400}{25} = 16$

$\sf \: x = \sqrt{16} = 4$

$\therefore$ The diagonals BD and AC of Rhombus are 12cm and 16cm respectively and each side is 10cm.

Answer 2

Answer:

The lengths of all the diagonal is 10 cm.

Step-by-step explanation:

Given,

In a rhombus ABCD

As we know that all the sides of the rhombus are equal. [Property of Rhombus]

.°. AB = BC = CD = AD

So, the sides of the rhombus = 1/4 × 40 = 10

.°. Sides of the rhombus = 10 cm

Now,

Let us assume that,

BD = 3x

OC = 4x

Now,

In ∆DOC, ∆DOC is a right angled triangle.

By using the Pythagoras theorem,

.°. (OD)² + (OC)² = (CD)²

=> (3/2x)² + (4/2x)² = 10²

=> 9x²/4 + 16x²/4 = 100

=> 25x² = 100 × 4

=> x² = 400/25 = 16

.°. x = √16 = 4

Therefore, the diagonal BD and AC of the rhombus are 12 cm and 16 cm respectively. The each sides of the rhombus are equal to the 10 cm.