A circle is inscribed in an equilateral triangle of side 8 m. The approximate area of the unoccupied space inside the triangle is
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The center of the inscribed circle = O = incenter = centroid of the triangle ABC, as it is equilateral. Given AB = 8m
Draw the median = perpendicular from A to BC to meet BC at D. Draw a perpendicular from O to AB to meet AB at E. OE is the radius of incircle.
Altitude = AD = √3/2 * AB
AO = 2/3 * AD = AB/√3
AE = AB/2
In the Right angle triangle AOE, AO^2 = AE^2 + OE^2
OE^2 = AB^2 /3 - AB^2/4 = AB^2 /12
Area of Incircle = π OE^2 = π AB^2/12
Area of Equilateral Triangle = 1/2 * √3/2 * AB^2
Unoccupied space inside the triangle = AB^2 [ √3/4 - π/12 ]
= 10.957 m^2
Draw the median = perpendicular from A to BC to meet BC at D. Draw a perpendicular from O to AB to meet AB at E. OE is the radius of incircle.
Altitude = AD = √3/2 * AB
AO = 2/3 * AD = AB/√3
AE = AB/2
In the Right angle triangle AOE, AO^2 = AE^2 + OE^2
OE^2 = AB^2 /3 - AB^2/4 = AB^2 /12
Area of Incircle = π OE^2 = π AB^2/12
Area of Equilateral Triangle = 1/2 * √3/2 * AB^2
Unoccupied space inside the triangle = AB^2 [ √3/4 - π/12 ]
= 10.957 m^2
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yes he is right......
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