Question

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

Please explain elaborately​

Answer 1

$\large\underline{\underline{\sf Given:}}$

• Side of equilateral triangle (a) = 8.8cm

$\large\underline{\underline{\sf To\:Find:}}$

• Diameter of the ring (D) = ?

$\large\underline{\underline{\sf Solution:}}$

We know :-

${\boxed{\sf Perimeter\:Of\: Triangle=a+b+c }}$

$\large\implies{\sf 8.8+8.8+8.8 }$

$\large\implies{\sf 26.4cm}$

And ,

$\large{\boxed{\sf Circumcircle\:Of\: Circle=2πr }}$

$\large\implies{\sf 2×\frac{22}{7}×R}$

____________________________________

$\large\implies{\sf 26.4=2×\frac{22}{7}×R }$

$\large\implies{\sf R=\frac{26.4×7}{2×22}}$

$\large\implies{\sf R=\frac{184.8}{44}}$

$\large\implies{\sf R=4.2cm}$

We know that ,

$\large{\boxed{\sf Diameter (D)=2R}}$

$\large\implies{\sf D=2×4.2}$

$\large\implies{\sf D=8.4cm}$

$\Large\underline{\underline{\sf Answer:}}$

•°•Diameter of ring is 8.4cm