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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3253085

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Sir Mike Ambrose is the author of the question. Let the side length of the square be 10 units. Area square is; 10² = 100 square units. Therefore; Area R is; Area semi circle with radius 5 units - Area sector with radius 5 units and angle (180-2atan(2))° - Area kite with diagonal lengths √(10²+5²) units and √(200-200cos(2atan(½))) units + Area sector with radius 10 units and angle 2atan(½)° - 2(area quarter circle with radius 5 units - area triangle with height and base 5 un...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3241835

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Calculating Green Inscribed Triangle. tana = 12/10 a = atan(6/5)° b = (180-atan(6/5)) b = (90+atan(⅚))° c² = 10²+12² c = √(244) c = 2√(61) cm. d = (2√(61)-x) units. It implies; 2(2√(61)-x)sin(90+atan(⅚)) = ½*12xsin(atan(6/5)) 24-1.53644255919x = 4.60932767758x 24 = 6.14577023677x x = 3.90512483796 cm. It implies; 2A = 0.5*3.90512483796*12sin(atan(6/5)) Where 2A is area green. 2A = 18 cm² It implies; A = 9 cm² Green area is; ½(2+12)*12-18-9...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3241154

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Sir Mike Ambrose is the author of the question. Calculating Shaded Area ÷ Green Area to 2 decimal places. Let the side of the regular pentagon be 1 unit. a = 108-90 a = 18° b = 180-108-18 b = 54° (1/sin54) = (c/sin108) c = 1.17557050458 units. (1/sin54) = (d/sin18) d = 0.38196601125 unit. e = 1-d e = 0.61803398875 unit. f² = 0.61803398875²+1²-2*0.61803398875cos108 f = 1.3281310261 units. (1.3281310261/sin108) = (1/sing) g = 45.73230144788° h = 180-108-45.73...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3237960

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Calculating the required angle, let it be x. Let a be the side length of the equilateral triangle. It implies; 6² = a²+10²-2*10*a cosy 36 = a²+100-20acosy 20acosy = a²+64 cosy = (a²+64)/(20a) --- (1). 8² = a²+10²-2*10*acos(60-y) 64 = a²+100-20a(½*cosy+½√(3)siny) 64 = a²+100-10acosy-10√(3)asiny 10acosy+10√(3)asiny = a²+36 --- (2). At (1). cosy = (a²+64)/(20a) Calculating siny. siny = opp/hyp opp = √((20a)²-(a²+64)²) opp = √(400a²-(a⁴+128a²+4096)) opp = √(-a⁴...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3235654

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Sir Mike Ambrose is the author of the question. Calculating the exact hatched area. Let x be the radius of the inscribed circle. 2x² = a² a = √(2)x cm. b = x+a b = (x+√(2)x) cm. c = 16+b c = (16+x+√(2)x) cm. c is the diagonal of the ascribed square. Again. c² = 2(16)² c = 16√(2) cm. Therefore, calculating x. Equating the both c. (16+x+√(2)x) = 16√(2) (1+√(2))x = (16√(2)-16) x = (48-32√(2)) cm. x = 16(3-2√(2)) cm. x = 2.74516600406 cm. Again, x is the radi...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3231294

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Let Alpha be x. Calculating tanx. Angle ABD = Angle ACD = y. Let AB = AC = a. Let AD = b. b² = a²+10²-2a*10cosy b² = a²+100-20acosy --- (1). b² = a²+6²-2a*6cosy b² = a²+36-12acosy --- (2). cosy = a/10 --- (3). Equating (1) and (2). a²+100-20acosy = a²+36-12acosy 64 = 8acosy 8 = acosy cosy = 8/a --- (4). Substituting (3) in (4). 8/a = a/10 a² = 80 a = √(80) a = 4√(5) units. cosc = 4√(5)/10 c = acos(⅕(2√(5)))° Calculating alpha (x). 180 = 2acos...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3228162

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Calculating the red shaded area. Notice. 10 cm is the radius of each of the four equal inscribed circles. tana = 10/(4*10) a = atan(¼)° a = 14.0362434679° b = 3*10 b = 30 cm. (10/sin14.0362434679) = (30/sinc) c = 133.313856658° d = 180-a-c d = 180-14.0362434679-133.313856658 d = 32.6498998741° e = 90-d e = 90-32.6498998741 e = 57.3501001259° f = 180-2a f = 180-2*14.0362434679 f = 151.927513064° Therefore, the required red shaded area is; Area rectangle...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3226397

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Sir Mike Ambrose is the author of the question. Calculating Exactly Angle Alpha - Angle Beta. Let Alpha be x. Let Beta be y. (6+6√(2))² = 8²+10²-2*8*10cosa 36+72√(2)+72 = 164-160cosa  160cosa = 164-36-72-72√(2) 160cosa = 56-72√(2) 20cosa = 7-9√(2) cosa = (7-9√(2))/20 a = acos((7-9√(2))/20)° a = 106.642318459° 8² = 10²+(6+6√(2))²-2*10(6+6√(2))cosb 64 = 100+36+72√(2)+72-(120+120√(2))cosb (120+120√(2))cosb = 144+72√(2) cosb = (144+72√(2))/(120+120√(2)) cosb = ⅕(3√(2)) b...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3215499

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Calculating the required angle. Notice. The ascribed plane shape is a square, let it's side length be x. Let the longest side length of the inscribed red lengths triangle be 1 unit. a = 180-45-57 a = 180-102 a = 78° (1/sin78) = (b/sin45) b = 0.72290396731 units. (1/sin78) = (c/sin57) c = 0.85740696733 units. d²+x² = 0.85740696733² d = √(0.85740696733²-x²) units. cosy = x/0.85740696733 --- (1). siny = √(0.85740696733²-x²)/0.85740696733 --- (2). cos(45-y) = x/...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3214346

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Notice. The ascribed quadrilateral is a square. Calculating x, the side length of the square. a²+x² = 8² a = √(64-x²) units. b²+x² = 10² b = √(100-x²) units. It implies;. c = (x-√(64-x²)) units. d = (x-√(100-x²)) units. Notice again. The inscribed triangle is a right-angle triangle. Therefore; √(x-√(100-x²) ~ √(64-x²) 6 ~ 8 Cross Multiply. 8(x-√(100-x²)) = 6√(64-x²) 4(x-√(100-x²)) = 3√(64-x²) 4x-4√(100-x²) = 3√(64-x²) 16x²-32x√(100-x²)+1600-16x² = 57...
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