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In the given figure, ABC is an equilateral triangle. If the area of bigger circle is 1386 cm2, then what is the value of four times the area (in cm2) of smaller circle if the distance left between vertex A and point m on smaller circle can be assumed to be equal to the radius of small circle?

In the given figure, ABC is an equilateral triangle. If the area of bigger circle is 1386 cm2, then what is the value of four times the area (in cm2) of smaller circle if the distance left between vertex A and point m on smaller circle can be assumed to be equal to the radius of small circle? Correct Answer 616

Area of bigger circle = πr2

⇒ Area of bigger circle = πr2 = 1386

⇒ Radius of the bigger circle = 21 cm

We know that the radius of the inscribed circle in the equilateral triangle is 1/3rd times the altitude of that triangle.

Altitude of the equilateral triangle = 3 × 21 = 63 cm

Let radius of the smaller circle be x.

3x + (2 × 21) = 63

x = 7 cm

⇒ Four times of area of smaller circle = 4 × π × (7)2 = 4 × 154 = 616 cm2

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