A wire, in the shape of an equilateral triangle,encloses an area 'A' cm2 if the same wire - Maths - Areas Related to Circles

Question

A wire, in the shape of an equilateral triangle,encloses an area
'A' cm2 if the same wire is bent to form circle then the
area of the circle will be:
(A) a2 /22/7 (B) 2A2 /22/7 (C) 3A/22/7 (D)
3Aroot3/22/7

Ajit Singh · Accepted Answer

Dear Student!

Here is the answer to your query.

Let the side of the equilateral triangle be a.

we know that area of equilateral triangle $= \frac{\sqrt{3}}{4} a^{2}$

$\Rightarrow A= \frac{\sqrt{3}}{4} a^{2} \Rightarrow a^{2} = \frac{4 A}{\sqrt{3}} ....... (1)$

Now as the equilateral triangle is bent to form a circle of radius r (say).

⇒ Circumference of circle = perimeter of triangle

⇒ 2πr = 3a

$\Rightarrow r = \frac{3 a}{2 π} ...... (2)$

Now area of circle = πr²

$= π (\frac{3 a}{2 π})^{2} (from (2)) = π \times \frac{9 a^{2}}{4 π^{2}} = \frac{9 a^{2}}{4 π} = \frac{9 \times \frac{4 A}{\sqrt{3}}}{4 π} (from (1)) = \frac{3 \sqrt{3} A}{π} = \frac{3 A \sqrt{3}}{\frac{22}{7}}$

Cheers!

A wire, in the shape of an equilateral triangle,encloses an area 'A' cm2 . if the same wire is bent to form circle then the area of the circle will be: (A) a2 /22/7 (B) 2A2 /22/7 (C) 3A/22/7 (D) 3Aroot3/22/7

A wire, in the shape of an equilateral triangle,encloses an area 'A' cm² . if the same wire is bent to form circle then the area of the circle will be:

(A) a² /22/7 (B) 2A² /22/7 (C) 3A/22/7 (D) 3Aroot3/22/7