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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3099190

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Area Black will be; Area triangle BCD - area sector of radius 4m and angle 45° + area sector of radius 4m and angle 45° - quarter circle of radius 2m - 2(right-angled triangle of height and base 2m each) + quarter circle of radius 2m. = ½(16) - (16π)/8 + (16π)/8 - ¼(4π) - 2(½(4)) + ¼(4π) = 8 - 2π + 2π - π - 4 + π = 8 - 4 = 4 m²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3096525

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Let the radius of the ascribed quarter circle be 2x. Therefore the radius of the inscribed yellow half circle is x. Calculating x. ¼(π(2x)²)-½(πx²) = 17 πx²-½(πx²) = 17 2πx²-πx² = 34 πx² = 34 x² = 34/π x = √(34/π) cm. Therefore, shaded yellow area is; ½(πx²) And x = √(34/π) cm. = ½(π√(34/π)²) = 17 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3095137

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Let R be radius of the quarter circle. Let r be radius of the inscribed circle. R will be; √(3²+4²) R=√(25) R= 5 units. Therefore r will be; (5-r)²=(3+r)²+r² 25-10r+r²=9+6r+r²+r² r²+16r-16=0 Using quadratic equation app to resolve the quadratic equation above. Therefore; r = 0.94427 units.

Mathematics Question and Solution.

https://onlineedumath.com/update/mathematics-question-and-solution/3093436

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Sir Mike Ambrose is the author of the question. Exact Area R in its simplest from will be; Area semi circle of radius 2 units - Area right-angled triangle of height 2√(2+√(2)) units and width 2√(2-√(2)) units - Area sector of radius 2 units and angle 45° + Area isosceles triangle of two equal length, 2 units each and angle 45° - Area quarter circle of radius 2 units + Area isosceles right-angled triangle of height and width 2 units respectively - Area sector of radius 2 units and angle 22....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3091998

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Blue Area is; Area square of single side 4cm - 2(area circle of radius 1cm) - 8(area square of single side 1cm) + 8(area quarter circle of radius 1cm). = 16 - 2π - 8 + ¼(8π) = 8 - 2π + 2π = 8 cm².

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3086817

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Shaded/green area will be; Area semi circle of radius 9unit - 2(area sector of radius 9unit and angle 60°) + area equilateral triangle of single side length 9unit - area sector of radius 9unit and angle 50° - area quarter circle of radius 9unit + 2(area sector of radius 9unit and angle 60°) - area equilateral triangle of single side length 9unit = ½(81π) - ⅓(81π) + ¼(81√(3)) - ¼(45π) - ¼(81π) + ⅓(81π) - ¼(81√(3)) = ½(81π) - ¼(45π) - ¼(81π)  = ¼(162π - 45π - 81π) = ¼(162π - 126π...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3077883

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Area Area Blue is; Area semi circle with radius √(2) units - Area quarter circle with radius 2 units + Area triangle with height and base 2 units respectively. = ½(√(2)²)π - ¼(2²)π + ½(2²) = π - π + 2 = 2 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2990381

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Sir Mike Ambrose is the author of the question. Let the base of the parallelogram be 2 units. Therefore the area of the parallelogram is; 2(area triangle with height (sin70/sin35) unit and base 2sin75 unit). = 2*½*(sin70/sin35)*2sin75 = 3.16496046094 square units. Area shaded is; Area triangle with height 0.55803775975 unit and base 1.17917770336sin55 unit. = ½*0.55803775975*1.17917770336sin55 = 0.26951154209 square unit. It implies; The percentage of the shaded part of the parallelo...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2982438

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Area blue is; Area square with side 10 cm - Area quarter circle with radius 10 cm - Area square with side 5 cm + Area quarter circle with radius 5 cm. = (10*10) - (¼*10*10*π) - (5*5) + (¼*5*5*π) = 100 - 25π - 25 + ¼(25π) = 75 - 25π + ¼(25π) = ¼(300 - 100π + 25π) = ¼(300-75π) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2980854

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Area pink is; 4(area semicircle with radius 4 cm - area quarter circle with radius 4√(2) cm + area triangle with height and base 4√(2) cm) respectively. It implies; 4(½(π(4²)) - ¼(π(4√(2))²) + ½(4√(2))²) = 4(8π - 8π + 16) = 4*16  = 64 cm²
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