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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2555576

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Area red is; Area semi circle with radius 5 cm - Area sector with radius 5 cm and angle (180-2atan(2))° - Area kite with diagonal lengths √(10²+5²) cm and √(200-200cos(2atan(½))) cm + Area sector with radius 10 cm and angle 2atan(½)° - 2(area quarter circle with radius 5 cm - area triangle with height and base 5 cm respectively). = 12.5π - 11.591190225 - 50 + 46.3647608999 - 14.2699081699 = 9.77357067487 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2555127

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Let the two equal lengths, AB and CD be 1 unit. a = 180-50-15 a = 115° a is angle ADB. b = 180-a b = 65° b is angle ADC. (1/sin115) = (c/sin50) c = 0.8452365235 units. c is AD. d² = 0.8452365235²+1²-2*0.8452365235cos65 d = 1 unit. d is AC. It implies, triangle ACD is isosceles. Therefore, the required angle theta, let it be e is; e+2b = 180 e+2(65) = 180 e = 190-130 e = 50° e is the regular angle, theta, angle ACB. Notice! Triangle ABC is isosceles.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2555052

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Calculating x, length CD. 4² = 6²+6²-2*6*6cosa 72cosa = 72-16 cosa = 56/72 a = acos(7/9) a = 38.942441269° a is angle ABC. b = ½(180-a) b = 70.5287793655° b is angle ADB = angle BAD. 2c = 180-b 2c = 109.4712206345 2c is 2 alpha, angle ADC. c = 54.7356103172° c is alpha, angle CAD. d = 180-2c-c d = 180-109.4712206345-54.7356103172 d = 15.7931690484° d is angle ACB. Therefore, x, length CD is; (x/sin54.7356103172) = (4/sin15.7931690484) x = 12 units. Again, x i...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2554627

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Notice! BD is the diameter of the circle. a = ½(BD)  a = ½(14) a = 7 cm. a is OB = OD. a is the radius of the circle. b = 2(45) b = 90° b is angle AOB. Calculating shaded area. It is; Area quarter circle with radius 7 cm - Area triangle with height and base 7 cm respectively. = ¼(7*7)π-½(7*7) = ¼(49π)-½(49) = ¼(49π-98) = ¼(49(π-2)) square units. = 13.9845100065 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2554584

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0.5*4*5sina = 8 Where a is the supplementary angle of angle ABC. It implies; 10sina = 8 a = asin(8/10) a = asin(4/5)° or a = acos(3/5)° or a = atan(4/3)° a = 53.1301023542° Therefore; Angle ABC is; Let it be b. It implies; b+a = 180 b = 180-a And a = atan(4/3)° b = 180-atan(4/3) b = (90+atan(¾))° b = 126.8698976458°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2554553

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Notice! The radius of the quarter circle is 4 units. 4² = 2²+a² 16-4 = a² a = √(12) a = 2√(3) units. It implies, area green is; Area triangle with height 2√(3) units and base 2 units. = ½*2√(3)*2 = 2√(3) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2554524

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Notice! The side length of the small inscribed square is twice the side length of the big inscribed square. Let the side length of the small inscribed square be a units  Therefore, the side length of the big inscribed square is 2a units. Radius of the ascribed semi circle is 1 unit. It implies; 1² = (2a)²+a² 1 = 5a² a = √(1/5) a = ⅕√(5) units. a is the side length of the small square. Let b be the radius of the inscribed shaded circle. c = (1-b) units. d = (⅕√(5...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2554168

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Area red is; Area rectangle with length 4 units and width 3 units - Area triangle with height 4 units and base 1 unit - Area rectangle with length 2 units and width 1 unit - Area triangle with height and base 3 units respectively. = (4*3)-½(4*1)-(2*1)-½(3*3) = 12-2-2-½(9) = ½(16-9) = ½(7) square units. = 3.5 square units. Or Area triangle with height and base 2 units respectively+ Area rectangle with length 4 units and width 1 unit - Area triangle with height 4 units and base...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2553292

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Let a be the radius of the half circle. a²+8² = b² b = √(a²+64) units. b is OC. Observing similar plane shape (right-angled) side length ratios. 8 - c √(a²+64) - a Cross Multiply. c = 8a/√(a²+64) units. Again; √(a²+64) - a a - d Cross Multiply. d = a²/√(a²+64) units. It implies; e = (√(a²+64)+a) units. e is AC. f = ((8a/√(a²+64))+8) units. f is CH. g = (a+(a²/√(a²+64))) g is AH. It implies; Observing Pythagoras Rule on Triangle AHC.  (√(a²+64)+a)² =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2553241

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a²+1.5² = 4² a² = 16-2.25 a = 3.7080992435 units. tanb = a/1.5 b = atan(3.7080992435/1.5) b = 67.9756871627° b is angle ODC. c = 90-b c = 22.0243128373° c is angle ODT. It implies; d² = 2²+4²-2*2*4cos22.0243128373 d = 2.2732362451 units. d is OT. Therefore, the required angle, alpha is; Let it be f. (2.2732362451/sin22.0243128373) = (4/sine) e = 41.2886037973° f = 180-e f = 138.7113962027° f is the required angle alpha. Or (2/sing) = (2.2732362451/sin...
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