Question

A piece of wire is bent in the shape of an equilateral triangle of each side 6.6 cm. it is re-bent to form a circular ring. What is the diameter of the ring?

Answer 1

We have:
Length of the wire = The perimeter of the equilateral triangle

= 3 x side = 3 x 6.6 = 19.8 cm.

Let the wire be bent to form a circular ring of radius r cm.Then,

Circumference = 19.8 cm

$$\begin{gathered} \Rightarrow 2\mathrm{\pi r}=19.8\mathrm{cm} \\ \Rightarrow 2\times \frac{22}{7}\times r=19.8\mathrm{cm} \\ \Rightarrow r=\frac{19.8\times 7}{22\times 2}=\frac{138.6}{44}\mathrm{cm}=3.15\mathrm{cm}. \end{gathered}$$

So, the diameter of the ring = 2 x 3.15 = 6.30 cm.