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

length of wire = The perimeter of the equilateral triangle
=3*side =3*6.6=198m
Let the wire be bent to form circular ring of radius r cm
Then,
circumference=19.8 cm
=2*22/7*r = 19.8 cm
r= 19.8*7/22*2=3.15cm
so the diameter of ring =2*3.15=6.30cm

Answer 2

Length of one side of triangle = 6.6 cm
Perimeter of triangle = Length of the wire
Perimeter of the triangle = 3 * side
Perimeter of the triangle = 3 * 6.6 cm
Perimeter of the triangle = 19.8 cm
Length of wire = 19.8 cm
It is bent to form a circle, so
Circumference = Length of wire
Circumference = 19.8 cm
also Circumference = 2πr
so, 2πr = 19.8 cm ( π = 22 / 7 )
2 * 22 / 7 * r = 19.8 cm
44 / 7 * r = 19.8 cm
as radius = r = diameter / 2 = d / 2
44 / 7 * d / 2 = 19.8 cm
44d / 14 = 19.8 cm
d = 19.8 * 14 / 44
d = 6.3 cm
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Hope it helps