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A piece of wire is bent in shape of an equilateral triangle of each side $6.6\ cm$. It is rebent to form circular ring. What is the diameter of ring?
Given: A piece of wire is bent in shape of an equilateral triangle of each side $6.6\ cm$. It is rebent to form circular ring.
To do: To find the diameter of ring.
Solution:
As given, side of the triangle$=6.6\ cm$
$\therefore$ Perimeter of the equilateral triangle$=6.6+6.6+6.6$
$=19.8\ cm$
Now, let's $r$ be the radius of the ring.
Then, circumference of the ring$=$perimeter of the equilateral triangle
$\Rightarrow 2\pi r=19.8$
$\Rightarrow 2\times\frac{22}{7}\times r=19.8$
$\Rightarrow r=\frac{19.8\times7}{2\times22}$
$\Rightarrow r=3.15\ cm$
$\therefore$ Diameter of the ring$=2\times r$
$=2\times3.15$
$=6.30\ cm$
Thus, the diameter of the ring is $6.30\ cm$.
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