Question

A piece of thin wire in the form of an equilateral triangle of side 8.8 DM is bent into a circle with no loss of wire. Find the diameter of the wire.

Answer 1

Perimeter of triangle=perimeter of circle.
1 dm=10cm
P of ∆=8.8dm+8.8dm+8.8dm
=26.4dm=264cm
P of∆=p of circle.
P of circle =2πr=264cm
2πr = 264
2r = 264÷π
2r = 264 × 7/22
2r = 84 cm = 8.4 dm
Hence, the diameter of the wire is 8.4dm .

Answer 2

Answer:

1dm=10cm

p of triangle=8.8dm+8.8dm+8.8dm

=26.4dm=264cm

p of triangle=p of circle

p of circle=2 pi.r=264

2pi.r=264

2r=264/r

2r=264×7/22

2r=84cm=8.4dm