Question

The length of diagonal of a rhombus are in the ratio 3 : 4. If its perimeter is 40 cm, find the

length of its sides and diagonals.​

Answer 1

Answer:

Let the sides of rhombus be a

Let the sides of rhombus be aand diagonals be d

Let the sides of rhombus be aand diagonals be d

Let the sides of rhombus be aand diagonals be d perimeter =40cm

Let the sides of rhombus be aand diagonals be d perimeter =40cm4a=40cm

Let the sides of rhombus be aand diagonals be d perimeter =40cm4a=40cma=10cm

$\frac{ {d}^{1} }{ {d}^{2} } = \frac{3}{4}$

ATQ

${h}^{2} = {p}^{2} + {b}^{2} \\ {10}^{2} = ( \frac{3x}{2 })^{2} + { (\frac{4x}{2} )}^{2} \\ 100 ={ \frac{9x}{4} }^{2} + { \frac{16x}{4} }^{2} \\ 100 = \frac{9 {x}^{2} + 16 {x}^{2} }{4} \\ 100 \times 4 = 25 {x}^{2} \\ \frac{100 \times 4}{25} = {x}^{2} \\ x = \sqrt{16} \\ x = 4$

D1=3x=3*4=12

D1=4x=4*4=16

Answer 2

Given :

The length of diagonal of a rhombus are in the ratio 3 : 4. Perimeter is 40 cm

To find :

Length of its sides and diagonals.​

Solution :

∵ Perimeter of rhombus = 4 × side

⇒ 40 = 4 × Side

⇒ Side = 40/4

⇒ Side = 10 cm.

Let the diagonal of rhombus be BD = 3m and AC = 4m respectively.

Now, from attachment :

⇒ BE = BD/2 = 3m/2

⇒ AE = AC/2 = 4m/2 = 2m

As diagonals of rhombus are perpendicular, so diagonals divide a rhombus into 4 right angled triangle.

AB = BC = CD = AD = 10 cm [Sides of rhombus are equal]

Now using pythagoras theorem in ΔABE :

⇒ AB² = AE² + BE²

⇒ 10² = (2m)² + (3m/2)²

⇒ 100 = 4m² + (9m²/4)

⇒ 100 = (16m² + 9m²)/4

⇒ 400 = 25m²

⇒ m² = 400/25

⇒ m² = 16

⇒ m = √16

⇒ m = 4

∴ Length of diagonal₁ = 3m = 3 * 4 = 12 cm

∴ Length of diagonal₂ = 4m = 4 * 4 = 16 cm

∴ Length of side of rhombus = 10 cm and length of diagonals are 12 cm and 16 cm respectively.