Question

Q.the diagonals of a rhombus are in the ratio 5:12.If it's perimeter is 104 cm,find the lengths of the sides and the diagonals.​

Answer 1

$\mathfrak{\underline{Answer:Each\:side\:=26\:cm;Diagonals\:=20\:cm\:&\:48\:cm,respectively.}}$

$\rule{100}2$

$\textbf{\underline{\underline{Step-By-Step\:Explanation:-}}}$

In a rhombus, all four sides are of equal measure.

Let it's unknown side be ' P ' cm.

Now,

→ Perimeter of rhombus = sum of all four sides or 4 × side.

$\sf P + P + P + P = 104 \:(\: Perimeter\:is\: 104\: cm\:) \\\\ \implies 4P = 104 \\\\ \implies P =\dfrac{104}{4} \\\\ \therefore P = 26$

Hence,

All the 4 unknown sides of the given rhombus are of 26 cm.

Also,

→ Diagonals of a rhombus are in the ratio 5 : 12

Let the common factor in their ratios be  ' x '

By Pythagoras Theorem,

( Hypotenuse )² = ( Perpendicular )² + ( Base )²

$\sf ( 26 )^2 = ( 5x )^2 + ( 12x )^2 \\\\ \implies {676} = {25}x^2 + {144}x^2 \\\\ \implies {676} = {169} x^2 \\\\ \implies {x}^2 = \cancel{\dfrac{676}{169}} \\\\ \implies {x}^2 = {4} \\\\ \therefore x = 2$

Hence,

Diagonals are :-

• 5x = 5 × 2 = 10 cm; for it's length, multiply it by 2 ; 10 × 2 = 20 cm
• 12x = 12 × 2 = 24, now, 24 × 2 = 48 cm

$\rule{200}2$

Answer 2

Step-by-step explanation:

we know that, perimeter of rhombus = 4 × side.

so, side = 104÷4 = 26 cm.

now let diagonals be 5x and 12x respectively

By Pythagoras Theorem,

( Hypotenuse )² = ( Perpendicular )² + ( Base )²

\begin{lgathered}\sf ( 26 )^2 = ( 5x )^2 + ( 12x )^2 \\\\ \implies {676} = {25}x^2 + {144}x^2 \\\\ \implies {676} = {169} x^2 \\\\ \implies {x}^2 = \cancel{\dfrac{676}{169}} \\\\ \implies {x}^2 = {4} \\\\ \therefore x = 2\end{lgathered}

(26)

2

=(5x)

2

+(12x)

2

⟹676=25x

2

+144x

2

⟹676=169x

2

⟹x

2

=

169

676

⟹x

2

=4

∴x=2

Hence,

Diagonals are :-

5x = 5 × 2 = 10 cm; for it's length, multiply it by 2 ; 10 × 2 = 20 cm

12x = 12 × 2 = 24, now, 24 × 2 = 48 cm