Q. 54.4( 9 Votes )

# A wire when

Answer :

In this question the wire is first bent in the shape of equilateral triangle and then same wire is bent to form a circle. The point to be noticed is that the same wire is used both the times which implies that the **perimeter of equilateral triangle and that of circle will be equal.**

Let the equilateral triangle be of side ‘a’ cm and radius of the circle be ‘r’.

Given: Area enclosed by equilateral triangle = 123√3 cm^{2}

Also, we know that Area of equilateral triangle

Where ‘a’ = side of equilateral triangle

⟹ a = √484

⇒** a = 22 cm**

Therefore, side of equilateral triangle, ‘a’ is 22 cm.

Also, circumference of the circle = Perimeter of equilateral triangle → eqn1

Perimeter of equilateral triangle = 3 × side

= 3 × 22

= 66 cm → eqn2

Also, we know Circumference of circle = 2πr → eqn3

Put values in equation 1 from equation 2 & 3, we get

2πr = 66

(put π = 22/7)

On rearranging,

⇒** r = 10.5 cm**

So, the radius ‘r’ of the circle is 10.5 cm.

Area of circle = πr^{2}

Where r = radius of the circle

⇒ Area of circle = π(10.5^{2})

= 346.5 cm^{2}

**Area of the circle is 346.5 cm ^{2}.**

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