Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2443447

Let the radius of the inscribed circle be x. a = (3-x) units. b = 7-3 b = 4 units. c = (7-x) units. d = a+c d = (3-x)+(7-x) d = (10-2x) units. Therefore, x (radius of the inscribed circle) is; (2x)²+b² = d² (2x)²+4² = (10-2x)² 4x²+16= 100-40x+4x² 16 = 100-40x 40x = 100-16 40x = 84 10x = 21 x = 2.1 units. Again, x is the radius of the inscribed circle.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2443407

Notice! The gray inscribed triangle is equilateral. Calculating length x. 4² = 2a²-2a²cos120 16 = 3a² a = √(16/3) a = ⅓(4√(3)) units. a = 2.3094010768 units. a is the radius of the circle. b² = 2(⅓(4√(3)))²-2(⅓(4√(3)))²cos60 b = 2.3094010768 units. tanc = 1/2.3094010768 c = atan(1/2.3094010768) c = 23.413224446° sin23.413224446 = 1/d d = 2.5166114785 units. e = 60-c e = 60-23.413224446 e = 36.586775554° cos36.586775554 = 2.3094010768/f f = 2.876127404 units. It implies, x, the required...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2442925

Calculating length x. 5² = 8²+7²-2*8*7cosa 112cosa = 64+49-25 cosa = 88/112 a = acos(88/112) a = 38.2132107017° a is the angle opposite 5 units. (5/sin38.2132107017) = (7/sinb) b = 60° b is the angle opposite 7 units. c = 2a c = 2*38.2132107017 c = 76.4264214034° sin38.2132107017 = (0.5*5)/d d = 4.0414518843 units. d is the radius of the circle. e = ½(180-76.4264214034) e = 51.7867892983° f = e-0.5b f = 51.7867892983-0.5*60 f = 21.7867892983° cos21.7867892983 = g/4.0414518843 g = 3.7527...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2441905

Let the side length of the bigger regular pentagon be 1 unit. a = ⅕*180(5-2) a = 108° a is the single interior angle of the regular pentagon. b² = 2-2cos108 b = 1.6180339887 units. c = 108-½(180-108) c = 108-36 c = 72° sin(0.5*108) = (0.5*1)/d d = 0.6180339887 units. d is the side length of the smaller regular pentagon. Area Blue is; 0.5*1.6180339887*1sin72 - 0.5*1.6180339887*0.6180339887sin36 = 0.7694208843-0.2938926261 = 0.4755282582 square units. e = 108+36 e = 144° Are Red is; 0.5...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2441227

Let the bigger white semi circle's radius be a. b = ½(a) units. b is the radius of the smaller white circle. c = a+b c = a+½(a) c = ½(3a) units. c is the radius of the bigger white and green semi circle. Therefore; Calculating a. (½(3a))² = 2²+(a/2)² (9a²/4) = 4+(a²/4) 2a² = 4 a² = 2 a = √(2) units. Again, a is the radius of the bigger white semi circle. b = ½(a) And a = √(2) units. b = ½√(2) units. b = 0.5√(2) units. Again, b is the radius of the smaller white s...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2440596

Let the radius of the ascribed quarter circle be a. It implies; a² = 5²+b² b = √(a²-25) units. 6² = b²+c² c² = 36-b² c = √(36-√(a²-25)²) c = √(36-(a²-25)) c = √(61-a²) units. d = 5+c d = (5+√(61-a²)) units. Calculating a. (5+√(61-a²))² = a²+6² 25+10√(61-a²)+61-a² = a²+36 10√(61-a²)-a² = a²+36-86 100(61-a²) = (2a²-50)² 100(61-a²) = 4a⁴-200a²+2500 6100-100a² = 4a⁴-200a²+2500 4a⁴-100a²-3600 = 0 a⁴-25a²-900 = 0 Let a² = p. p²-25p-900 = 0 Resolving the above quadratic equation via completing...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2440569

Calculating the required angle, x. a = (√(3)-1) units. Observing similar plane shape (right-angled triangle) side length ratios. √(3) - 1 b - (√(3)-1) Calculating b. Cross Multiply. 3-√(3) = b b = (3-√(3)) units. c = √(3)-b c = √(3)-(3-√(3)) c = (2√(3)-3) units. tand = c/√(3) tand = (2√(3)-3)/√(3) d = atan(⅓(6-3√(3))) d = atan(2-√(3))° Therefore, x, the required angle is; x = 90-d x = 90-atan(2-√(3)) x = 90-15 x = 75°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2440551

a² = 12²+6² a = 6√(5) cm. a = AE. b = atan(1/2)° b = angle DAE. c = 180-atan(2) c = 116.56505117708° c = angle AFC. d = 180-116.56505117708-atan(1/2) d = 36.86989764584° (6/sin36.86989764584) = (e/sin116.56505117708) e = 8.94427191 cm. f = 6√(5)-8.94427191 f = 4.472135955 cm. g = 180-2atan(2) g = 53.13010235416° g = angle AEB. cos53.13010235416 = h/4.472135955 h = 2.683281573 cm. It implies; Area S1 is; 0.5*2.683281573*4.472135955sin53.13010235416 = 4.8 cm² Calculating S2 Area. j =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2439598

Calculating Grey Area. a² = 5²+2² a = √(29) m. tanb = 5/2 b = atan(5/2)° c = (90+b)° c = (90+atan(5/2))° d² =√(29)²+4²-2*4√(29)cos(90+atan(5/2)) d = 9.2195444573 m. d is the diagonal of the rectangle. e = x m. f = (2-e) f = (2-x) m. Observing similar plane shape (right-angled triangle) side length ratios. 5 - 4 e - f It implies; 5 - 4 x - (2-x) Cross Multiply. 5(2-x) = 4x 10-5x = 4x 10 = 9x x = (10/9) m. f = (2-x) And x = (10/9) m. f = 2-(10/9) f = ⅑(18-10) f = (8/9) m. 9.21954445...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2439228

Let a be the side length of the purple inscribed square. b² = 4 b = 2 cm. b is the side length of the green inscribed square. c² = 1 c = 1 cm. c is the side length of the yellow ascribed square. d = a-c d = (a-1) cm. e = a-b e = (a-2) cm Calculating a, side length of the inscribed purple square. Considering similar plane shapes (right-angled triangle) side length ratios. d - 2 1 - e It implies; (a-1) - 2 1 - (a-2) Cross Multiply. (a-1)(a-2) = 2 a²-3a+2 = 2 a²-3a = 0 a = 3 cm. Ther...

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