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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2674711

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Notice. 3x = 15 It implies; x = 5 m. x is the width of the inscribed green rectangle. The length is 15 m. Therefore, area inscribed green rectangle is  5 *15 = 75 m²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2674127

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Let the square side be a. a² = πr²  And r = 1 unit, π = 22/7 a² = (22/7) a = √(22/7) units. b = ½(a) b = ½√(22/7) units. c² = 2a² c² = 2√(22/7)² c = √(44/7) c = ⅐(2√(77)) units. c is the diagonal of the square. cosd = ½√(22/7)/1 d = acos(½√(22/7))° d = 27.5750477105° e = 45-d e = 17.4249522895° f = 3e f = 34.849904579° f is the required angle, phi. g² = 2-2cos34.849904579 g = 0.5989126703 units. 2h² = 0.5989126703² h = 0.4234952105 units. Shaded Area...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2673610

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Calculating MN, diameter of the yellow semi circle. Let the radius of the yellow semi circle be a. b = ½(3+3+2) b = 4 units. b is the radius of the bigger semi circle. c = b-3 c = 1 unit. Therefore. a² = 1.5²+d² d = √(a²-2.25) units. e = 3+1.5-b e = 4.5-4 e = 0.5 units. f² = 0.5²+√(a²-2.25)² f² = 0.25+a²-2.25 f = √(a²-2) units. It implies; 4² = √(a²-2)²+a² 16 = a²-2+a² 2a² = 18 a² = 9 a = 6 units. Again, a is the radius of the yellow semi circle. The...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2673604

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sin30 = 1/a ½ = 1/a a = 2 units. cos30 = ½(a)/b ½√(3) = 1/b √(3)b = 2 b = ⅓(2√(3)) b = ⅓(2√(3)) units. b = 1.1547005384 units. b is the radius of the ascribed semi circle. Calculating c, radius of the inscribed green semi circle. d = 2b-c d = (⅓(4√(3))-c) units. sin30 = c/d ½ = c/(⅓(4√(3))-c) 2c = ⅓(4√(3))-c 9c = 4√(3) c = ⅑(4√(3)) units. Again, c is the radius of the inscribed green semi circle. Area inscribed green semi circle is; ½(c)² = ½*⅑(4√(3))*⅑(4√(3))...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2673119

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Let CN = a b = (9+a) m. b is the side length of the square. c = ½(b) c = ½(9+a) m. c = CM = DM. Calculating a. (9+a) - ½(9+a) ½(9+a) - a Cross Multiply. a(9+a) = ¼(9+a)² a = ¼(9+a) 4a = 9+a a = 3 m. Recall. b = (9+a) And a = 3 m. b = 9+3 b = 12 m. Again, b is the side length of the square. c = ½(9+a) And a = 3 m. c = ½(9+3) c = 6 m. Again, c = CM = DM. Shaded Area is; (12*12)-½(12*6)-½(12*9) = 144-36-54 = 144-90 = 54 m²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2673096

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Calculating the perimeter of the rectangle (QRSP). Let MR = NS = a b = (2+a) b is the width of the rectangle (RS). Let PS = QR = c, the length of the rectangle. It implies; ½*ac = A ac = 2A c = (2A/a) m. d = 3A d is the area of the ascribed rectangle (QRSP). Therefore; c*b = d (2A/a)*(2+a) = 3A 4A+2Aa = 3Aa 4A = Aa a = 4 m. Again, a is MR = NS. Recall. b = (2+a) And a = 4 m. b = 2+4 b = 6 m. b is RS, the width of the rectangle. Calculating c, leng...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2672364

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Please, move the above question left/right one time to review the solution. Thank you. Radius, r of the inscribed yellow circle is; 4.192325 cm

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2671882

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Sir Mike Ambrose is the author of the question. Area yellow is; ½(0.63245553204*0.75)sin(atan⅓) = 0.075 square units. Area ABC is; ½(3*3.75)sin(2tan⅓) = 3.375 square units. Area yellow ÷ Area ABC is; 0.075 ÷ 3.375 = 75/3375 = 1/45

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2671875

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Calculating green inscribed triangle area. Notice. CM = DM a = 10-6 a = 4 m² Therefore, green inscribed triangle area is; (6+6)-a = 12-4 = 8 m²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2671866

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Calculating angle x  Let AC = CD = a. b² = a²+10² b = √(a²+100) units. It implies, observing similar plane shape (right-angled) side length ratios to get a. a - √(a²+100) 5 - 10 Cross Multiply. 10a = 5√(a²+100) 2a = √(a²+100) 4a² = a²+100 3a² = 100 a = √(100/3) a = 10/√(3) a = ⅓(10√(3)) units. The required angle, x is; tanx = a/10 tanx = ⅓(10√(3))/(10) tanx = ⅓√(3) x = atan(⅓√(3)) x = 30°
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