Question

A piece of wire is bent in the shape of an equilateral triangle each of whose sides measures 8.8 cm. This wire is rebent to form a circular ring. What is the diameter of the ring?

Answer 1

Length of the wire = Perimeter of the equilateral triangle
= 3 $\times$ Side of the equilateral triangle = (3 $\times$ 8.8) cm = 26.4 cm
Let the wire be bent into the form of a circle of radius r cm.
Circumference of the circle = 26.4 cm
⇒ $2\mathrm{\pi r}=26.4$
⇒ $2\times \frac{22}{7}\times r=26.4$
⇒ r = $\left(\frac{26.4\times 7}{2\times 22}\right)$ cm = 4.2 cm

∴ Diameter = 2r = (2 × 4.2) cm = 8.4 cm
Hence, the diameter of the ring is 8.4 cm.