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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3095856

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Each of the triangles is equilateral with single side length; (√(24)÷3^(1/4)) units. Total yellow area will be; (√(24)÷3^(1/4))²x(3/4)x0.5sin60 + (√(24)÷3^(1/4))²x(2/3)x(0.5)²sin60 + (√(24)÷3^(1/4))²x(1/4)(1/3)x0.5sin60 = 4.5 + 2 + 0.5 = 7 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3095852

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Diameter of circle Z will be; ½(Diameter of the ascribed circle - length of the inscribed square) = ½(√(16²+16²) - 16) = ½(16√(2)-16) = (8√(2)-8) cm The radius of circle Z will be; ½ the diameter of circle Z. = ½(8√(2)-8) = (4√(2)-4) cm. Radius of circle X will be; 2(area triangle with height r cm and base 16 cm) + Area triangle with height r cm and base 16√(2) cm = Area triangle with height and base 16 cm respectively. 2*½*16r + ½*16√(2)r = ½*16² 16r + 8√(...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3095850

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The required angle is; 180° - (11.25°+33.75°) = 180° - 45° = 135°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3095849

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Let the single side length of the regular pentagon be 4 unit. Therefore angle x degree will be; Sinx = (2/4) Sinx = ½ Therefore; x = sin–¹(½) x = 30°

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/3090099

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Blue Area will be; Area square of single side 4cm - 3(Area circle of radius 1cm) It will be; (4x4) - 3(1²(π)) It is; 16 - 3π Answer is; (16-3π) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3088475

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The radius of the circle with the inscribed square will be; ½(√(4²+4²)) = ½(√(32)) = ½(4√(2)) = 2√(2) cm. Therefore Blue Area will be; 2(area circle of radius 2cm) - area circle of radius 2√(2)cm + area square of single side length 4cm. It will be; 2(2²π) - ((2√(2))²π) +(4x4) = 8π - 8π + 16 = 16 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...
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