Question

area of a rhombus is 24 cm square and one of its diagonal is 8 and its perimeter is​

Answer 1

Answer:

20cm

Step-by-step explanation:

Area of rhombus = $\frac{1}{2}$ x d1 x d2

24 = $\frac{1}{2}$ x 8 x d2 = 4 x d2

d2 = $\frac{24}{4}$ = 6cm

half of the diagonals = 4cm and 3 cm

So, applying PYTHAGORAS THEOREM, we get the equation

3² + 4² = S²

9 + 16 = S²

√25 = S = 5cm

So, side of the rhombus is 5cm

P(rhombus) = 4s

= 4 x 5

= 20cm

PLEASE MARK AS BRAINLIEST

Answer 2

Given :-

➠ Area of the rhombus = 24 cm²

➠ One of it's diagonal = 8 cm

To Find :-

➠ Perimeter of this rhombus

Solution :-

➠ As we know that

$\sf \rightarrow Area\;of\;rhombus\;=\;\dfrac{1}{2} \times \sf D_1 \times \sf D_2$

$\begin{gathered}\sf Where,\\\\D_1 = 1^{st} Diagonal \\\\D_2 = 2^{nd} Diagonal \\\end{gathered}$

$\sf \implies 24\;cm^{2} = \dfrac{1}{2} \times 8 \times D_2$

$\sf \implies D_2 = \dfrac{24}{4} = 6\;cm$

➠ As we know that ,

Diagonals bisect each other

∴  $\sf AO = \dfrac{8}{2} = 4\;cm$

∴ $\sf BO = \dfrac{6}{2} = 3\;cm$

➠ We need to find the side length now

According to the Pythagoras theorem

⇢ AB² = AO² + OB²

⇢ AB² = 3² + 4²

⇢ AB = √3²+4²

⇢ AB = √25

⇢ AB = 5 cm

➠ Now , we need to find the perimeter :

Perimeter of a rhombus = 4 × Side

⇒   4 × 5

⇒   20 cm

∴ Perimeter of this rhombus is 20 cm