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If two consecutive sides of a rhombus are represented by 3x-6 and x+14. Find the perimeter of the rhombus.

please help!!....__​


Sagot :

Two adjacent sides are given by 3x-6 and x+14.
Now, because its a rhombus all sides of it are
equal.
By the problem,
3x-6=X+14
2x=20
X=10
Perimeter of rhombus- 4a (a=side of rhombus)
4*10=40 units

Given that:

The consecutive sides of a rhombus are 3x-6 and x+14 .

We know that

all sides are equal in a rhombus

⇛ 3x-6 = x+14

⇛ 3x-x = 14+6

⇛ 2x = 20

⇛ x = 20/2

⇛ x = 10

Now,

3x-6 = 3(10)-6 = 30-6 = 24

Length of the side = 24 units

Now,

Perimeter of a rhombus = 4×side

⇛ P = 4×24 units

P = 96 units

Answer:-The perimeter of the given rhombus is 96 units.

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