Question

If the diagonals of a rhombus are 12 cm and 5 cm, find the perimeter of the
rhombus.​

Answer 1

Answer:

the diagonals of a rhombus bisect perpendicular ly.

here 4 equal right angle triangle .

each triangle have base 12/cm=6 cm

and other side is 5/2 cm

so hypotenuse= √(6^2+5/2^2) cm

=√(36+25/4) cm

=√(144+25)/4 cm

=√169/4 cm

=13/2 cm

it is the side of the rhombus

so perimeter of the rhombus is 4*13/2 cm

=26 cm

Answer 2

Given:

• AC = 12 cm
• DB = 5 cm

To find:

⟶ Perimeter of rhombus = ?

Method:

First let's understand the concept:

Concept used:

→ Diagonals of a rhombus bisect each other

→ All sides in a rhombus are equal

∴ Perimeter of rhombus = 4 × Side

→ Pythagoras theorem

We know that AC = 12 cm

$\Rightarrow \sf{AO + OC = 12\:cm}\\\\\rm{Since\:AO =OC,}\\\\\sf{AO + AO = 12\:cm}\\\\\Rightarrow \sf{2AO = 12\:cm}\\\\\\\Rightarrow \sf{AO = \dfrac{12}{2}}\\\\\\\Rightarrow \bf{AO = 6\:cm}$

$\bf{Also, BD = 5 \:cm}\\\\\Rightarrow \sf{BO+OD = 5\:cm}\\\\\rm{Since\:BO=OD,}\\\\\sf{BO + BO = 5\:cm}\\\\\Rightarrow \sf{2BO = 5\: cm}\\\\\\\Rightarrow \sf{BO=\dfrac{5}{2}}\\\\\\\Rightarrow \bf{BO = 2.5 \:cm}$

$\sf{In \: \triangle AOB,}\\\\\sf{By\: Pythagoras\: theorem,}\\\\\sf{AO^{2} + BO^{2} = AB^{2}}\\\\\Rightarrow \sf{AB^{2} = 6^{2}+ 2.5^{2}}\\\\\Rightarrow \sf{AB^{2} = 36 + 6.25}\\\\\Rightarrow \sf{AB^{2} = 42.25}\\\\\Rightarrow \sf{AB = \sqrt{42.25} }\\\\\Rightarrow \bf{\underline{AB = 6.5 \: cm}}$

$\mathfrak{Now\: we\: have\: got\: the \: measure\: of \: 1 \: side}$

$\sf{Perimeter = 4 \times Side}\\\\\\\Rightarrow \sf{Perimeter = 4\times 6.5}\\\\\\\therefore \large{\boxed{\boxed{\bf{Perimeter = 26 \:cm}}}}$