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03 Nov 2020, 01:40
Kudos
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The Area Described is (1/3rd) of the Area OUTSIDE the Inscribed Circle, but INSIDE the Equilateral Triangle
(1st) Area of Equilateral Triangle:
Side of Equilateral Triangle = 6
Area of Equilateral Triangle = (6)^2 * sqrt(3) * (1/4) = 9 * sqrt(3)
(2nd) Area of Inscribed Circle
The InRadius of an Inscribed Circle inside an Equilateral Triangle is equal to = (1/3) * (Altitude of Equilateral Triangle)
Altitude of Equilateral Triangle = (side of 6) * sqrt(3) * (1/2) ------> 3 * sqrt(3)
InRadius of Inscribed Circle = (1/3) * (3) * sqrt(3) = sqrt(3)
Area of Inscribed Circle = (InRadius)^2 * (pi) -------> (sqrt(3))^2 * (pi) = 3 * (pi)
(3rd) Area of Region = (1/3) * (Area of Equilateral Triangle - Area of Inscribed Circle)
Area of Region = (1/3) * [ (9 * sqrt(3) ) - 3(pi) ]
Area of Region = 3 * sqrt(3) - (pi)
-B-
(1st) Area of Equilateral Triangle:
Side of Equilateral Triangle = 6
Area of Equilateral Triangle = (6)^2 * sqrt(3) * (1/4) = 9 * sqrt(3)
(2nd) Area of Inscribed Circle
The InRadius of an Inscribed Circle inside an Equilateral Triangle is equal to = (1/3) * (Altitude of Equilateral Triangle)
Altitude of Equilateral Triangle = (side of 6) * sqrt(3) * (1/2) ------> 3 * sqrt(3)
InRadius of Inscribed Circle = (1/3) * (3) * sqrt(3) = sqrt(3)
Area of Inscribed Circle = (InRadius)^2 * (pi) -------> (sqrt(3))^2 * (pi) = 3 * (pi)
(3rd) Area of Region = (1/3) * (Area of Equilateral Triangle - Area of Inscribed Circle)
Area of Region = (1/3) * [ (9 * sqrt(3) ) - 3(pi) ]
Area of Region = 3 * sqrt(3) - (pi)
-B-
21 Apr 2024, 22:17
Kudos
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