Question

A piece of wire is bent in shape of an equilateral triangle of each side being 7.2cms . It is rebent to form of circular ring. Find diameter of the ring ​

Answer 1

AnsweR :

$\bf{\Large{\underline{\sf{Given\::}}}}$

A piece of wire is bent in shape of an equilateral triangle of each side being 7.2 cm. It is  rebent to form of circular ring.

$\bf{\Large{\underline{\sf{To\:find\::}}}}$

The diameter of the ring.

$\bf{\Large{\underline{\rm{\blue{Explanation\::}}}}}$

$\bigstar\it{Reference\:of\:image\:is\:in\:the\:diagram\::}$

$\setlength{\unitlength}{1cm}\begin{picture}(6,8)\linethickness{0.075mm}\put(1,.5){\line(2,1){3}}\put(4,2){\line(-2,1){2}}\put(2,3){\line(-2,-5){1}}\put(.7,.3){ A }\put(4.05,1.9){ B }\put(1.7,2.95){ C }\put(0.6,1.7){ 7.2cm }\end{picture}\\\\\\\bigstar\it{\underline{Reference\:in\:image\:rebent\:to\:form\:\:circular\:rinG\::}}$

$\setlength{\unitlength}{1cm}\thicklines\begin{picture}(8,10)\put(5,10){\line(1,0){0.7}}\put(5.1,10.1){\scriptsize{ R }}\put(5,10){\circle{12}}\end{picture}$

According to the question :

$\leadsto\sf{\red{Perimeter\:of\:triangle\:=\:Circumference\:of\:circle}}$

So,

$|\implies\sf{3*side\:=\:\pi d}\\\\\\\\|\implies\sf{3*7.2\:cm\:=\:\dfrac{22}{7} *d}\\\\\\\\|\implies\sf{21.6\:cm\:=\:\dfrac{22}{7} *d}\\\\\\\\|\implies\sf{d\:=\:\dfrac{21.6*7}{22} }\\\\\\\\|\implies\sf{d\:=\:\cancel{\dfrac{151.2\:cm}{22} }}\\\\\\\\|\implies\sf{\pink{d\:=\:6.87\:cm}}$

Thus,

$\bigstar$ The diameter of the ring is 6.87 cm.

Answer 2

Step-by-step explanation:

$\huge\underline\purple{\sf Answer :-}$

Given:- Side of equilateral triangle is 7.2

• Then it is rebent to form circular ring

To find:- The diameter of ring

A/q

$\large\implies{\sf }$ Perimeter of circle= Circumference of circle

$\large\implies{\sf }$3 X Side = πd

$\large\implies{\sf }$(3 X 7.2)cm= $\large{\sf {\frac{</strong><strong>2</strong><strong>2</strong><strong>}{</strong><strong>7</strong><strong>}}}$x D

$\large\implies{\sf }$21.6cm= $\large{\sf {\frac{</strong><strong>2</strong><strong>2</strong><strong>}{</strong><strong>7</strong><strong>}}}$x D

$\large\implies{\sf }$D= $\large{\sf {\frac{</strong><strong>2</strong><strong>1</strong><strong>.</strong><strong>6</strong><strong> </strong><strong>x</strong><strong> </strong><strong>7</strong><strong>}{</strong><strong>2</strong><strong>2</strong><strong>}}}$cm

$\large\implies{\sf }$D= $\large{\</strong><strong>t</strong><strong>t</strong><strong> {\frac{</strong><strong>1</strong><strong>5</strong><strong>1</strong><strong>.</strong><strong>2</strong><strong>}{</strong><strong>2</strong><strong>2</strong><strong>}}}$Cm

$\large\implies{\sf }$D= 6.87 Cm.

$\large\red{\boxed{\sf }}$Hence the diameter of ring is 6.87cm