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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1868158

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Radius of the inscribed circle, r is; r = 2 cm. Therefore; Shaded Region is; area sector with radius 2 cm and angle 120°- area triangle with two equal lengths 2 cm and the angle they form, 120° + Area quarter circle with radius 2 cm - 2(area sector with radius 2 cm and angle 60°) + Area equilateral triangle with sides 2 cm each. = ⅓(*4π)-½(4sin120)+(¼*4π)-2((60π*4)/360)+(½*4sin60) = ⅓(4π)-√(3)+π-⅓(4π)+√(3) = π cm² Or Shaded Region is; Quarter the area of the inscribed...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1868088

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Calculating shaded area considering semi circle PQ. Diameter of semi circle PQ will be 10cm, therefore radius is 5cm. Shaded area will be; Area semi circle PQ of radius 5cm and angle 180° - 2(area sector of radius 5cm and 60° - area equilateral triangle of equal side lengths 5cm each and angle 60° each angle) - 2(area sector of radius 5cm and 60° - area equilateral triangle of equal side lengths 5cm each and angle 60° each angle). =(5²π)/2 - (2(5²π*60)/360 - (5²sin60)/2) - (2(5²π*6...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1867952

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Let angle CAD = a Let angle BAD = b Let AD = c tana = 3/c ----- (1). tanb = 2/c ----- (2). Adding (1) and (2). Therefore; tana+tanb = 5/c ----- (3). Notice; tan(a+b) = tana+tanb/(1-tanatanb) tana+tanb = tan(a+b)(1-tanatanb) And, a+b = 45° It implies; tana+tanb = tan(45)(1-tanatanb) tan45 = 1 Therefore; tana+tanb = (1-tanatanb) ----- (4). Substituting (1), (2) and (3) in (4) to get c (height of triangle ABC) 5/c = 1-(3/c)(2/c) 5/c = 1-(6/c²) (5/c)+(6/c²) = 1 (5c+6)/c² = 1 c²-5c-6...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1867311

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Let the radius of one the the two congruent inscribed circles be a. Therefore; a = 2 cm. tanb = (2/4) b = atan(½)° c = (180-2atan(½))° c = 126.86989764584° Shaded Region is; Half area rectangle with length 4 cm and width 2 cm - Area square with side 2 cm + Area quarter circle with radius 2 cm - Area sector with angle 126.86989764584° and radius 2 cm + Area triangle with two equal lengths 2 cm and angle they form, 126.86989764584° = ½(4*2)-(2*2)+¼(2²π)-(126.86989764584π*2²)360+(0.5*2*2si...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1867163

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AC = √(12²+12²) AC = 12√(2) cm. AE = √(12²+6²) AE = 6√(5) cm. Angle BAE = atan(½)° Angle CAE = 45-atan(1/2) Angle CAE = 18.43494882292° cos18.43494882292 = (AF)/12√(2) AF = 16.099689438 cm. EF = AF-AE EF = 16.099689438-6√(5) EF = 2.683281573 cm. Angle BCG = 90-45-18.43494882292 Angle BCG = 26.56505117708° tan26.56505117708 = (BG)/12 BG = 6 cm. CF = √(6²-2.683281573²) CF = 5.366563146 cm. (BF)² = 16.099689438²+12²-24*16.099689438cos(atan(1/2)) BF = 7.58946638441 cm. (7.58946638441/sin...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1866286

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Let AB = a. Let BE = b. It implies; 2a+b = 392 b = 392-2a BC = ½(b-70) BC = ½(392-2a-70) BC = ½(322-2a) Therefore; BC = DE = 161-a c = ½(70)+161-a c = 196-a Where c is half BE. d² = a²-(196-a)² d² = a²-38416+392a-a² d² = 392a-38416 d = √(392a-38416) Where d is the height of the triangle ABE. e² = √(392a-38416)²+35² e² = 392a-38416+1225 e² = 392a-37191  e = √(392a-37191) Where e is AC = AD. It implies; Calculating a (AB = AE) AB+BC = AD+CD a+161-a = √(392a-37191)+70 161-70 = √(392a-3719...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1866128

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a² = 6²+12² a = 6√(5) cm. Where a is CE = EF. b² = 2(a²) b² = 2(6√(5))² b² = 360 b = 6√(10) cm. Where b is CF. Angle DCE = atan(½)° Angle ECF = 45° Angle FCM = 90-45-atan(½) Angle FCM = 18.43494882292°  tan18.43494882292 = c/12 c = 4 cm. Angle AEF = 90-atan2 Angle AEF = atan(½)° (AF)² = 36+180-2*6*6√(5)cos(atan(½)) AF = 6√(2) cm. (6/sind) = (6√(2)/sin(atan(0.5))) d = 18.43494882292° Where d is angle AFE. Angle CFM = 180-45-18.43494882292 Angle CGM = 116.56505117708° Angle CMF = 180-1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1865815

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Let CE = 1 unit. (1/sin20) = (a/sin40) a = 1.87938524157 units. Where a is AE. Notice; b = a-1 b = 1.87938524157-1 b = 0.87938524157 units. Where b is AD. Angle AEC = 180-20+40 Angle AEC = 120° (c/sin120) = (1/sin20) c = 2.53208888624 units. Where c is AC = BC. BE = 2.53208888624-1 BE = 1.53208888624 units. Angle BED = 180-120 = 60° (BD)² = 1.53208888624²+1-2*2.53208888624cos60 BD = 1.34729635534 units. Calculating Angle BDE (alpha, the required angle)....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1865498

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a = ½(180-150) a = 15° b = 60-15 b = 45° c² = 2(6²) c = 6√(2) cm. d² = 72+12²-144√(2)cos135 d= 6√(10) cm. (6√(10)/sin135) = (12/sine) e = 26.56505117708° f = 180-135-26.56505117708  f = 18.43494882292° g = 90-18.43494882292 g = 71.56505117708° tan71.56505117708 = 12/h h = 4 cm. tan71.56505117708 = j/2 j = 6 cm. It implies the height and base of the shaded region is 6 cm and 4 cm respectively. Therefore Area Shaded Region (triangle) is; ½*4*6 = ½(24) = 12 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1864230

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(6/sin(180-45-atan(2))) = (a/sin45) a = 4.472135955 cm. b = √(12²+6²)+a b = 17.88854382 cm. c² = 4.472135955²+6²-12*4.472135955cos(180-atan2) c = 8.94427191 cm. Where c is CF. (8.94427191/sin(180-atan2)) = (6/Sind) d = 36.86989764584° d is angle CFD. Angle BCF is; 180-(180-atan2)-36.86989764584° = atan2-36.86989764584° = 26.56505117708° Angle ECF is; 90-26.56505117708  = 63.43494882292° Notice; atan(½)+63.43494882292 = 90° It implies; sin(atan(½)) = e/12 e = 5.366563146 cm. cos(ata...
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