Question

A metal wire, when bent in the form of an equilateral triangle of largest area, encloses an area of 484√3 sq. cm. If the same wire is bent in the form of a circle of largest area, find the area of this circle.

Answer 1

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Answer 2

Hey user

Here is your answer :-

The area of equilateral triangle formed by bending a metal wire is " 484√3 sq.cm "

Let the side of equilateral triangle be " x cm "

We know that

$area \: of \: equilateral \: triangle \: = \frac{ \sqrt{3} }{4} \times {x}^{2} \\ \\ 484 \sqrt{3} = \frac{ \sqrt{3} }{4} \times {x}^{2} \\ \\ {x}^{2} = \frac{484 \sqrt{3} \times 4}{ \sqrt{3} } \\ \\ {x}^{2} = 484 \times 2 \\ \\ taking \: square \: root \: \\ \\ x = \sqrt{484 \times 4} \\ \\ x = 22 \times 2 \\ \\ x = 44$

Therefore the side of equilateral triangle is " 44 cm "

Perimeter of equilateral triangle = 3 × side

= 3 × 44

= 132 cm

So the perimeter of equilateral triangle is " 132 cm "

Perimeter of equilateral triangle is nothing but total lenght of the wire .

Hence the total lenght of the wire is " 132 cm ".

If the circle is formed its circumference is 132 cm

We know that

$circumference \: of \: circle \: = 2\pi \: r \\ \\ 132 = 2 \times \frac{22}{7} \times r \\ \\ r \: = \frac{132 \times 7}{22 \times 2} \\ \\ r = 3 \times 7 \\ \\ r = 21$

So the radius of circle formed is " 21 cm " .

Area of the circle = πr²

= 22/7 × 21 × 21

= 22 × 3 × 21

= 1386 sq.cm

Thank you.