Question

A wire when bent in the form of an equilateral triangle encloses an area of `121sqrt(3)" cm"^2`. If the same wire is bent into the form of a circle, what will be the area of the circle?

Answer 1

Let a cm be the side of the equilateral triangle.

Now,

Area of the equilateral triangle`=sqrt(3)/4"a"^2`

We have:

`sqrt(3)/4"a"^2 = 121sqrt(3)`

`=> a^2/4 = 121`

⇒ a2 = 484

⇒ a = 22

Perimeter of the triangle = Circumference of the circle
Perimeter of the triangle = (22 + 22 + 22) cm
                                        = 66 cm
Now, let r cm be the radius of the circle.
We know:
Circumference of the circle = 2πr

`=> 2xx22/7xx"r" = 66`

`=> "r"=(66xx7/44)"cm"`

`=> "r" = 21/2 "cm"`

Also,

Area of the circle = πr² 

`=(22/7xx21/2xx21/2)  "cm"^2`

`=693/2 "cm"^2`

= 346.5 cm²