Answer:
The perimeter of the Rhombus is 100 cm
Step-by-step explanation:
First of all, we will need to find the length of the other diagonal.
let’s call the diagonals p and q
Mathematically, the area of the Rhombus is;
pq/2 = Area of Rhombus.
Let’s call the missing diagonal p
So;
(p * 14)/2 = 336
14p = 672
p = 672/14
p = 48 cm
Now, we can find the perimeter of the Rhombus using these diagonals.
Mathematically;
P = 2 √(p^2 + q^2)
Substituting these values, we have;
P = 2 √(14)^2 + (48^2)
P = 2 √(2500)
P = 2 * 50
P = 100 cm
The perimeter of the rhombus is the sum of its side lengths
The perimeter of the Rhombus is 100 cm
The length of one of its diagonal is given as:
[tex]p= 14[/tex]
And the area is given as:
[tex]A = 336[/tex]
Assume the other diagonal is q.
The area of the rhombus is represented as:
[tex]A = \frac{pq}2[/tex]
So, we have:
[tex]336 = \frac{14q}2[/tex]
This gives
[tex]336 = 7q[/tex]
Divide both sides by 7
[tex]48 = q[/tex]
Rewrite as:
[tex]q = 48[/tex]
The perimeter (P) of the rhombus is calculated as:
[tex]P =2\sqrt{p^2 + q^2[/tex]
So, we have:
[tex]P =2\sqrt{48^2 + 14^2[/tex]
Evaluate the squares
[tex]P =2\sqrt{2500[/tex]
Take positive root of 2500
[tex]P =2 \times 50[/tex]
[tex]P =100[/tex]
Hence, the perimeter of the Rhombus is 100 cm
Read more about areas and perimeters at:
https://brainly.com/question/14137384
