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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1892765

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Let the radius of the ascribed quarter circle be a. a² = 2(a-1)² a² = 2(a²-2a+1) a² = 2a²-4a+2 a²-4a+2 = 0 (a-2)² = -2+4 a = 2±√(2) a ≠ 2-√(2) a = 2+√(2) units. a = 3.41421356237 units. Where a is AB = BC. BQ = AB-AQ BQ = 2+√(2)-2 BQ = √(2) units. Area Blue, Area Triangle CBQ is; ½*√(2)*(2+√(2)) = ½(2√(2)+2) = (√(2)+1) square units. = 2.41421356237 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1892231

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Let DE = BF = a. Let EG = GH = HF = b. (3b)² = 12²+(12-2a) 9b² = 144+144-48a+4a² b² = ⅑(288-48a+4a²) ----- (1). CF = √(144+a²) (CG)² = (12-a)²-b² CG = √((12-a)²-b²) Therefore; (144+a²) = 4b²+((12-a)²-b²) 144+a² = 4b²+144-24a+a²-b² 24a = 3b² b² = 8a ----- (2). Equating (1) and (2). ⅑(288-48a+4a²) = 8a 4a²-120a+288 = 0 a²-30a+72 = 0 (a-15)² = -72+225 a = 15±√(153) a ≠ 15+3√(17) a = 15-3√(17) cm. a = 2.63068312315 cm. AF = 12-2.63068312315 AF = 9.36931687685 cm. HF = √(8a) HF = √(8...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1890114

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Let r be the radius of the inscribed circle. Calculating r. tan30 = r/(6-r) √(3)r = 6-r r = 6/(√(3)+1) cm. r = 3(√(3)-1) cm. sin30 = r/a ½ = 3(√(3)-1)/a a = 6(√(3)-1) cm. Where a is BO, O is the center of the inscribed circle. BP = 6(√(3)-1)-r BP = 6√(3)-6-3√(3)+3 BP = 3√(3)-3 BP = 3(√(3)-1) cm. cos30 = 6/(BH) BH = 12/√(3) BH = 4√(3) cm. PH = BH-BP PH = 4√(3)-3√(3)+3 PH = (√(3)+3) cm. tan30 = (DH)/6 DH = 6/√(3) DH = 2√(3) cm  HR = 2√(3)-3√(3)+3 HR = (3-√(3)) cm. (PR)² = (√(3)+3)²+(3-√...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1886518

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tan(0.5atan(3/4))= a/24 a = 8 units. Where a is DF. b = 18-8 b = 10 units.  Where b is CF. tan(0.5atan(4/3))= c/18 c = 9 units. Where c is BE. d = 24-9 d = 15 units. Where d is CE. It implies; (EF)² = (CF)²+(CE)² (EF)² = 10²+15² EF = √(325) EF = 5√(13) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1886431

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(AM)² = 15²+8²-2*15*8cos60 AM = √(225+64-120) AM = √(105+64) AM = √(169) AM = 13 units. (13/sin60) = (8/sina) a = 32.20422750397° Where a is angle BAM. b = 90-32.20422750397  b = 57.79577249603° Where b is angle DAM. Therefore, calculating DM. (DM)² = 13²+11²-2*13*11cos57.79577249603 DM = 11.72943003449 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1886366

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Let AB = BC = CD = 1 unit. AC = √(1²+1²) AC = √(2) units. (AD)² = 3-2√(2)cos105 AD = 1.93185165258 units. Therefore; (1.93185165258/sin105) = (√(2)/sin?) ? = 45° Where ? is angle ADC.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1885846

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a² = 4²+4²-2*4*4cos150 a = 7.72740661031 cm. Where a is BE. b = angle ABE = ½(180-150) b = 15° c = angle AFE = 180-45-15 c = 120° (4/sin120) = (d/sin45) d = 3.26598632371 cm. Where d is EF. Notice; EF = CF Therefore; Length CF = 3.26598632371 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1885237

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a = asin(1/6) b = 2a b = 2asin(1/6) Where b is angle BAC cos(2asin(1/6)) = c/6 c = ⅓(17) units. Where c is AD. d = 6-c d = 6-⅓(17) d = ⅓ units Where d is CD. sin(2asin(1/6)) = e/6 e = 1.97202659437 units. Where e is BD. f = asin((1/3)/2) f = 9.59406822686° Where f is angle EBC = angle CAE. tan(9.59406822686) = g/(17/3) g = 0.95784148869 units. Or  g = 17√(35)/105 units. Where g is DE.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1883742

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Notice; Since angle BDC equals angle BAC which are the opposite inscribed angles of quadrilateral ABCD. And; CD+BD = 40 units. It implies; CD+BD = AC+AB It also implies that; Area Triangle BDC = Area Triangle BAC Therefore; AC+AB = 40 units. AC = AB = ½(40) AC = AB = 20 units. Area quadrilateral ABCD is; 2(½*20*20) = 400 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1882433

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9 = 8-8cosa 8cosa = -1 a = acos(-1/8) a = 97.18075578146° Therefore; a• = 180-a a• = 180-97.18075578146  a• = 82.81924421854° b = ½(180-82.81924421854) b = 48.59037789073° c = ½(48.59037789073) c = 24.29518894536° Where c is alpha. sin24.29518894536 = d/4 d = 1.64575131106 cm. cos24.29518894536 = e/4 e = 3.64575131106 cm. f = 90-24.29518894536 f = 65.70481105464° Red area is; 0.5*1*1.64575131106sin65.70481105464 = ¾ cm² Blue area is; 2(0.5*3.64575131106*1.64575131106)-(¾) = 6-(3/4)...
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