Question

A piece of wire is bent in the shape of an equilateral triangle each of whose sides measure 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
                             
${=\:3}$ × Side of the equilateral triangle

$= 3 \times 8.8$

$= 26.4 \: cm$

Let the wire be bent into the form of a circle of radius r cm.

$\bold{Now,}$

Circumference of the circle$= 26.4 \: cm$

$2\pi \: r = 26.4$

$2 \times \frac{22}{7} \times r = 26.4$

$r = ( \frac{26.4 \times 7}{2 \times 22} )$

$r = 4.2 \: cm$

$\bold{Therefore,}$

Diameter

$= 2r$

$= 2 \times 4.2$

$= 8.4 \: cm$

Hence, the diameter of the ring is $8.4 \: cm$