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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2593252

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Calculating x (length OD). a = (x+4) units. a is the radius of the quarter circle (length OA = length OB). It implies; (x+4)² = 8²+x² x²+8x+16 = 64+x² 8x = 48 x = 6 units. Again, x is required length OD.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2591938

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Area Coloured Region is; Area triangle with height and base 10 units - Area triangle with height and base 3√(2) units - Area triangle with height 5 units and base 7√(2)sin45 units = ½*10*10) - ½*3√(2)*3√(2) - ½*5*7√(2)sin45 = 50 - 9 - ½(35) = 41 - ½(35) = ½(82-35) = ½(47) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2591226

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Calculating the area of the inscribed blue circle. a = ½(14+6) a = 10 units. a is the radius of the ascribed semi circle. b = a-6 b = 4 units. c²+4² = d² c is the radius of the inscribed half circle. d = √(c²+16) units. 10² = √(c²+16)²+c² 100 = c²+16+c² 84 = 2c² c² = 42 c = √(42) units. Again, c is the radius of the inscribed half circle. e = (c+f) e = (√(42)+f) units. f is the radius of the inscribed blue circle. g = (10-f) units. Notice! d = √(c²+16) And c...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2591193

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a² = 30²+40²-2*30*40cos150 a = 67.6643256752 units. (67.6643256752/sin(150)) = (30/sinb) b = 12.807876266° c = 83-b c = 70.192123734° It implies, the required length x is; x² = 50²+67.6643256752²-2*50*67.6643256752cos70.192123734 x = 69.1775868761 units. x = 40√(3) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2591172

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c²+12² = b² c = √(b²-144) units. For b = 15 units. c = √(15²-144 c = 9 units. tana = 12/9 a = atan(4/3)° cos(atan(4/3)) = 12/d ⅗ = 12/d d = 20 units. e = 24-d e = 4 units. It implies; 12 - 4 9 - f Cross Multiply. 12f = 36 f = 3 units. Therefore, the required area of the blue triangle is; ½*e*f = ½*4*3 = 6 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2589559

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Let the single side length of the regular pentagon be 1 unit. Therefore; Area blue is; ½(sin108) square unit. Area red is; (1-cos36)sin108 square unit. Therefore; Area blue : Area red is; ½(sin108) ÷ (1-cos36)sin108 = 1 ÷ (2-2cos36) = 2.61803398875

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2589555

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Coloured region is; Area triangle with height 6√(2) units and base 4.5sin45 units + Area triangle with height 3 units and base 5.65685424949sin45 units + Area triangle with height 1.5 units and base 2.82842712475sin45 units = ½*6√(2)*4.5sin45 + ½*3*5.65685424949sin45 + ½*1.5*2.82842712475sin45 = 13.5 + 6 + 1.5 = 21 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2588840

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Sir Mike Ambrose is the author of the question. Area Red is; Area triangle with height 2 units and base 2sin(180-2ata(¾)) units. = ½*2*2sin(180-2ata(¾)) = 2sin(180-2ata(¾)) = 48/25 square units. Area Green is; Area triangle with height √(13) units and base 2sin(2atan(3/4)-atan(3/2)) units. = ½*√(13)*2sin(2atan(3/4)-atan(3/2)) = √(13)sin(2atan(3/4)-atan(3/2)) = 27/25 square units. It implies; Area Green ÷ Area Red exactly is; (27/25) ÷ (48/25) = 27/48 = 9/16

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2587649

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Calculating Shaded Black Area. a² = 9²+12² a = √(81+144) a = 15 units. a is the hypotenuse of the ascribed right-angled triangle. b = ½(a) b = 7.5 units. tanc = 9/12 c = atan(3/4)° cos(atan(3/4)) = 7.5/d 0.8 = 7.5/d d = ⅛(75) units. d = 9.375 units. e = 12-d e = ⅛(96-75) e = ⅛(21) units. e = 2.625 units. Therefore, shaded black area is; ½*9*⅛(21) = (189/16) square units. = 11.8125 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2587361

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a = ⅑*180(9-2) a = 140° a is the single interior angle of the regular nonagon. b = ½(180-140) b = 20° c = 140-20 c = 120° d = ½((180(5-2))-3*140) d = ½(540-420) d = ½(120) d = 60° e = 140-d e = 80° It implies, the required angle x is; x = ½(180-120) x = ½(60) x = 30°
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