Question

one diagonal of a rhombus is 6cm if the area of the rhombus is 24sq.cm then what is the length of its side? ( step by step explanation)​

Answer 1

Answer:

9000

Step-by-step explanation:

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Answer 2

$\large{\sf \underline{Solution-}}$

Here it is given that,

• Length of one diagonal (AC) = 6 cm
• Area of rhombus = 24 cm²

⟶ We have to find the length of its side.

We know,

$\sf \implies \boxed{ \sf Area \: of \: rhombus = \dfrac{1}{2} \times d_{1}\times d_{2}}$

$\sf \implies \dfrac{1}{2} \times AC \times BD = 24$

$\sf \implies \dfrac{1}{2} \times 6 \times BD = 24$

$\sf \implies BD = \dfrac{24 \times 2}{6}$

$\sf \implies BD = 8 \: cm$

So,

➢ The other diagonal of rhombus is 8 cm.

Now,

We know that diagonals of rhombus bisect each other at 90°

• Here, OB = 4 cm and AO = 3 cm

⟶ By using Pythagoras theorem :

$\bigstar \: {\boxed{\sf{(Hypotenuse)^{2} = (Base)^{2} + (Height)^{2}}}}$

$\implies \: \sf{(AB )^{2} = (AO )^{2} + (OB)^{2}}$

$\implies \: \sf{(AB )^{2} = (3)^{2} + (4)^{2}}$

$\implies \: \sf{(AB )^{2} = 9 + 16}$

$\implies \: \sf{(AB )^{2} = 25}$

$\implies \: \sf{AB = \sqrt{25} }$

$\implies \: \bf{AB = 5 \: cm}$

Therefore,

➢ The length of the side of rhombus is 5 cm.