Question

A wire bent in the form of equilatcral triangle has side 8.8 m. If same wire is bent in the form of circle, what would be area enclosed by it?​

Answer 1

Answer:

$\huge\tt\orange{Answer}$

Step-by-step explanation:

Given,

Side of a triangle = 8.8 m

Length of a wire is =the perimeter of the wire

$= 3 \times 8.8 = 26.4 m$

So, now let's move on the next step

The wire is converted into circular ring ( circle)

$circumference \: of \: circle \\ \: \: \: \: \: \: \: \: \: \: \: \: = 2\pi r \: \\ \\ 26.4 = 2\pi r \\ 26.4 = 2 \times \frac{22}{7} \times r \\ 26.4 \times 7 = 2 \times 22 \times r \\ r = \frac{26.4 \times 7}{44} \\ r = \frac{184.8}{44} \\ r = 46.2$

Therefore the diameter of the circle will be

$diameter = 2 \times 46.2 = 92.4$

$radius \: of \: the \: cirle \: is \: 46.2$

_______

Area of the circle = πr²

$= \pi {r}^{2} \\ = 3.14 \times {46.2}^{2} \\ = 3.14 \times 2134.44 \\ = 6702.1406$

Hence the area will be 6702.1406 meter