Question

If the area of a rhombus is 68 cm2
and one of its diagonals is 8 cm, then find the perimeter of the rhombus​

Answer 1

Step-by-step explanation:

let one diagonal be 'x'

area of rhombus =

$\frac{1}{2} \times diagonal1 \times diagonal2$

68= 1/2×x×8

x=68×2/8

x=68/4

x=17 cm

Answer 2

The perimeter of the rhombus​ is 37.6 cm .

Explanation:

Formula :

Area of rhombus = 0.5 x Diagonal 1 x Diagonal 2

Given : Area of rhombus = 68 cm²

Length of one diagonal =  8 cm

Put corresponding values in formula , we get

68 =  0.5 x 8 x Diagonal 2

⇒ 68 = 4 x Diagonal 2

⇒Diagonal 2 = 17 cm  [divide both sides by 4]

Let s be the side of rhombus.

Since diagonals of rhombus are perpendicular bisector of each other.

So by Pythagoras theorem , we have

$s^2= (\dfrac{\text{Diagonal} _1}{2})^2+(\dfrac{\text{Diagonal}_2}{2})^2$

$s^2=(\dfrac{8}{2})^2+(\dfrac{17}{2})^2\\\\ s^2= 88.25\\\\ s\approx 9.4\ cm$

Perimeter of  rhombus = 4s = 4(9.4) = 37.6 cm

Hence , the perimeter of the rhombus​ is 37.6 cm .

# Learn more :

If the area of the rhombus is 24 cm2, one of the diagonalis 8 cm .find it perimeter

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