Question

A wire when bent in the form of an equilateral triangle encloses an area of 121√3 cm². If the wire is bent in the form of a circle, find the area enclosed by the circle (Use π=22/7).

(Also show the workout of this question if it is possible)

Answer 1

Area of an equilateral ∆ = √3a²/4
Where a=side
Now, given that -
√3a²/4= 121√3
=>a²=121*4
=> a=11*2= 22
Therefore, length of wire = a+a+a=66
Now, this wire is bent into a circle, so circumference of the circle=2πr=66
=>2*(22/7)r=66
=> r=21/2
Now, area of this circle =πr²
=(22/7)*(21/2)²
=693/2
=346.5cm²

Answer 2

Let  the length of  wire  be l

to form it  in  a  shape  of  equilateral triangle, we  need  to divide  the wire  in  3  equal parts  of  l/3 as  shown in figure,

Now,Area  of  ΔABC with side l/3

$A=\frac{\sqrt{3}}{4}\times(side)^2$

Here,

A = 121√3 (as  given in question)

side = l/3

$A=\frac{\sqrt{3}}{4}\times(l/3)^2\\\;\\121\sqrt{3}=\frac{\sqrt{3}}{4}\times\frac{l^2}{9}\\\;\\121=\frac{l^2}{36}\\\;\\l^2=121\times36\\\;\\l=\sqrt{121\times36}\\\;\\l=11\times6\\\;\\l=66\;cm$

if  we bent  this  wire  of  length l in shape  of  circle, then parameter of  circle will be  equal to length of  wire.if the  radius  of  circle  is  r,

parameter  of  circle = l

$2\pi r=l\\\;\\2\pi r=66\\\;\\2\times\frac{22}{7}\times r=66\\\;\\2\times\frac{2}{7}\times r=6\\\;\\r=\frac{6\times7}{4}\\\;\\r=\frac{21}{2}\;cm$

$\text{area of circle}=\pi r^2\\\;\\=\frac{22}{7}\times(\frac{21}{2})^2\\\;\\=\frac{22}{7}\times\frac{21}{2}\times\frac{21}{2}\\\;\\=\frac{11\times3\times21}{2}\\\;\\=346.5\;cm^2$