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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1916258

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Calculating R. (5+R)² = (10-R)²+(5-R)² 25+10R+R² = 100-20R+R²+25-10R+R² 20R = 100-20R+R² R²-40R+100 = 0 Therefore; R ≠ 20+10√(3) units. R = 20-10√(3) units. R = 2.67949192431 units. Calculating r. (5-r)² = r²+a² 25-10r+r² = r²+a² a = √(25-10r) units. Therefore; (5+r)² = (5-r)²+ (5+√(25-10r))² 25+10r+r² = 25-10r+r²+25+10√(25-10r)+25-10r 20r = 25+10√(25-10r)+25-10r 30r-50 = 10√(25-10r) (3r-5)² = 25-10r 9r²-30r+25 = 25-10r 9r² = 20r 9r = 20 r = ⅑(20) units. r = 2.22 units. It implies; r ÷...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1916187

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b² = 12²+12² b² = 144+144 b² =288 b = 12√(2) cm. Where b is AE = AC. c² = 12²+6² c = 6√(5) cm. Where c is AM. d = b-c d = (12√(2)-6√(5)) d = 3.55415488348 cm. Where d is EM. e = 45-atan(1/2) e = 18.43494882292° Where e is angle MAC. f = 45+18.43494882292  f = 63.43494882292° Where f is angle EAB. sin63.43494882292 = g/(12√(2)) g = 15.17893276881 cm. cos63.43494882292 = h/(12√(2)) h = 7.5894663844 cm. i = 12-7.5894663844  i = 4.4105336156 cm. j = atan(15.17893276881/4.4105336156) j = 73...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1916055

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Let the AC = AB = a (diameter of the semi circle). Therefore; sin22.5 = AD/AC AD = asin22.5 units. cos22.5 =CD/AC CD = acos22.5 units. Calculating a. 84+25+½(acos22.5)(asin22.5) = ½(a²) 109+⅛√(2)a² = ½(a²) 109 = ½(a²)-⅛√(2)a² 109 = ⅛(4-√(2))a² 872 = (4-√(2))a² a = √(872/(4-√(2))) a = 18.36377300632 units. It implies; Green Area is; 0.5a²-84-25 = 0.5*18.36377300632²-84-25 = 59.61407951382 square units. Or = 0.5*18.36377300632²*sin22.5*cos22.5 = 59.6140...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1915865

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Let a be the radius of the inscribed circle. Calculating a. tan30 = a/6 a = 2√(3) cm. b = inscribed circle diameter. b = 4√(3) cm. 12² = c²+6² c = 144-36 c = √(108) c = 6√(3) cm. d = 6√(3)-4√(3) d = 2√(3) cm. Where d is AD. e² = 12²+(2√(3))²-2*12*2√(3)cos30 e = 2√(21) cm. Where e is BD. (2√(21)/sin30) = (12/sinf) f = 139.10660535087° Where f is angle BDA. g = 180-30-139.10660535087  g = 10.89339464913° Where g is angle ABD. h = 60-10.89339464913  h = 49.10660535087° Where h is angle FB...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1915750

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The single interior angle of the ascribed regular hexagon is; ⅙(180*4) = 120° a = 360-90-120-(120-53) a = 150-120+53 a = 83° b = 360-120-53-53 b = 240-106 b = 134° It implies; The required angle ? is; 360-46-134-83 = 360-180-83 = 180-83 = 97°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1914803

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Let the small inscribed half circle be r. Big inscribed half circle's radius is 0.5 cm. Calculating r. (0.5+r)² = 0.5²+(1-r)² 0.25+r+r² = 0.25+1-2r+r² 3r = 1 r = ⅓ cm. 1² = 2a² a = 1/√(2) a = 0.70710678119 cm. b = 0.70710678119-(1/3) b = 0.37377344785 cm. b is the height of triangle ABC. c = 0.5-(1/3) c = (1/6) cm. d² = (1/2)²-(1/6)² d = ⅓√(2) cm. d = 0.47140452079 cm. e = 2(0.70710678119-0.47140452079) e = 0.4714045208 cm. e is the base of triangle ABC. It implies; Area Triangle ABC...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1914507

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18² = 14²+16²-2*14*16cosa 448cosa = 14²+16²-18² a = acos((14²+16²-18²)/448) a = 73.39845040098° (18/sin73.39845040098) = (14/sinb) b = 48.18968510422° c = 180-48.18968510422-73.39845040098 c = 58.4118644948° cotd = cota+cotb+cotc cotd = 1.80748828181  d = acot(1.80748828181) d = 28.95373454979° e = 58.4118644948-d e = 29.45812994501° f = 180-28.95373454979-73.39845040098 f = 77.64781504923° (14/sin77.64781504923) = (g/sin73.39845040098)  g = 13.73434221224 units. It implies; Area Blue...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1914466

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Total green sector angle is; atan(3)+atan(⅓) = 90° It implies; 210 = ¼(πr²) Where π is (22/7). Calculating r, radius of the green sectors. 210*7*2 = 11r² r = √(2940/11) units. Where r is also the diameter of the inscribed yellow half circle. Radius x, of the inscribed yellow half circle is; x = ½√(2940/11) It implies; Area yellow Inscribed half circle is; ½*π(½√(2940/11))² = ⅛*(22/7)*(2940/11) = ¼*(2940/7) = ¼(420) = 105 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1913898

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NP = 3 cm. AN = √(7²-3²) AN = √(49-9) AN = 2√(10) cm. MN = ½(AN) MN = √(10) cm. Angle APN = atan(2√(10)/3) = 64.62306647485° Angle CPN is; (180-atan(2√(10)/3))° Angle CNP is; ½(180-(180-atan(2√(10)/2))) = ½(atan(2√(10)/3))° Angle BNM is;  57.68846676258 ° = 57.68846676258° Therefore x ( length BM) is; tan(57.68846676258) = x/MN And MN is √(10) cm. x = √(10)tan(57.68846676258) x = 5 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1911927

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tan15 = a/4 a = 1.07179676972 cm. Where a is AD. b² = 1.07179676972²+4² b = 4.14110472164 cm. Where b is DF. c = 4-1.07179676972 c = 2.92820323028 cm. Where c is CD. sin60 = d/2.92820323028  d = 2.53589838487 cm. Where d is DE = BE. e² = 16+16-2*16cos150 e = 7.72740661031 cm. Where e is BF. f² = 7.72740661031²+2.53589838487²-2*2.53589838487*7.72740661031cos45 f = 6.19925645689 cm. Where f is EF. (6.19925645689/sin45) = (2.53589838487/sing) g = 16.81321456806° Where g is angle BFE. h = 1...
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