Question

A wire when bent in the form of an equilateral triangle encloses an area of `121sqrt(3)  "cm"^2`. The same wire is bent to form a circle. Find the area enclosed by the circle.

Answer 1

Area of an equilateral triangle`=sqrt(3)/4xx("Side")^2` 

`=> 121sqrt(3) = sqrt(3)/4xx("Side")^2`

`=> 121xx4 = ("Side")^2`

`=> "Side" = 22  "cm" `

Perimeter of an equilateral triangle = 3 × Side

= 3 × 22

= 66 cm

Length of the wire = 66 cm

Now, let the radius of the circle be r cm.

We know:

Circumference of the circle = Length of the wire

2π = 66

`=> 2 xx 22/7xx"r" = 66`

`=> "r" = (66xx7)/(2xx22)`

`=> "r" = 21/2`

⇒ r = 10.5

Thus, we have :

Area of the circle = πr²

`=22/7xx10.5xx10.5`

=  346.5 sq.cm

Area enclosed by the circle = 346.5 cm²