Solution for 29
The figure one
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area of shaded portion is 12.56
Ben111:
incorrect answer
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see diagram. Your diagram is not very clear related to the shaded regions.
I will calculate the various regions. Please find the appropriate region yourself.
r = 2 cm
Side of the equilateral triangle = a = 2+2 = 4 cm
Angle ∠A = ∠B = ∠C = 60°
Area B1 = B2 = B3 = π 2² * (60°/360°) = 2π/3 cm²
Rose color shaded region = 3 * 2π/3 = 2π cm²
Area of ΔABC: √3/4 a² = √3/4 * 4² = 4√3 cm²
Violet colored region enclosed among the circles = (4√3 - 2π) cm²
In an equilateral triangle, O is the centroid. AO= BO= CO = 2/3 * altitude
AO = 2/3 * (√3/2 * a) = a/√3 = 4/√3 cm
So radius of the biggest circle: R = AO + r = 2 + 4/√3 cm
Grey shaded region: π [(2+4/√3)]² - 3 * π * 2² cm²
= 8 π [ 2√3 - 1 ]/3 cm²
Blue shaded region = 3 * π 2² * (300°/360°) = 10π cm²
I will calculate the various regions. Please find the appropriate region yourself.
r = 2 cm
Side of the equilateral triangle = a = 2+2 = 4 cm
Angle ∠A = ∠B = ∠C = 60°
Area B1 = B2 = B3 = π 2² * (60°/360°) = 2π/3 cm²
Rose color shaded region = 3 * 2π/3 = 2π cm²
Area of ΔABC: √3/4 a² = √3/4 * 4² = 4√3 cm²
Violet colored region enclosed among the circles = (4√3 - 2π) cm²
In an equilateral triangle, O is the centroid. AO= BO= CO = 2/3 * altitude
AO = 2/3 * (√3/2 * a) = a/√3 = 4/√3 cm
So radius of the biggest circle: R = AO + r = 2 + 4/√3 cm
Grey shaded region: π [(2+4/√3)]² - 3 * π * 2² cm²
= 8 π [ 2√3 - 1 ]/3 cm²
Blue shaded region = 3 * π 2² * (300°/360°) = 10π cm²
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