Question

diagonal of rhombus12 centimetre and 5 centimetre number find its perimeter​

Answer 1

$\huge\underline\bold{GIVEN}$

• Diagonals of rhombus are 12cm and 5cm

$\huge\underline\bold{Note}$

• In rhombus, adjacent sides are equal

AB = BC = CD = DA

• Diagonals of rhombus bisect each other at 90°

AC = OC = 6cm

BO = OD = 2.5cm

$\huge\underline\bold{TO\:FIND}$

• Perimeter of rhombus

$\huge\underline\bold{SOLUTION}$

★$\textbf{Applying Pythagoras theorem}$★

(Hypotenuse)²=(perpendicular)²+(base)²

$\textbf{In right triangle AOB}$

$\implies\sf AB^2=AO^2+BO^2$

$\implies\sf AB^2=(6)^2+(2.5)^2$

$\implies\sf AB^2=36+6.25$

$\implies\sf AB^2=42.25$

$\implies\sf AB=\sqrt{42.25}=6.5cm$

AB = BC = 6.5cm

★$\bold{Similarly\:in\:triangle\:DOC}$★

DC = 6.5cm

DC = AD = 6.5cm

★$\textbf{Perimeter of rhombus}$★

AB+BC+CD+AD

= 6.5 × 4

= 26cm²

Answer 2

Answer:

$\huge\mathrm\blue{Solution}</p><p>$

Let d1 be 1st diagonal

And, d2 be 2nd diagonal

Then,

a^2 = d1^2 + d2^2

Side of Rhombus = a^2 = 6^2 + 2.5^2 = 36 + 6.25 = 42.25

a = √42.25 = 6.5

Perimeter of Rhombus = 4 × 6.25 = 26cm

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