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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3128818

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c² = 5³+12² c = √(169) c = 13 units. Calculating a, radius of the inscribed yellow circle. 5a+12a+13a = (12*5) 30a = 60 a = 2 units. Calculating b, radius of the blue inscribed circle. Let d be the radius of the big ascribed circle. d+x = 2b  b = ½(d+x) --- (1). d² = (½(13))²+x² d² = 42.25+x² x = √(d²-42.25) --- (2). e = (d-2) units. f = 6.5-3 f = 3.5 units. g = (x+2) units. It implies; e² = f²+g² (d-2)² = 3.5²+(x+2)² d²-4d+4 = 12.25+x²+4x+4 d²-4d...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3128816

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a²+m² = 12² a = √(144-m²) units. Calculating m. (4m)² = (3m)²+√(144-m²)² 16m² = 9m²+144-m² 8m² = 144 2m² = 36 m² = 18 m = 3√(2) units. Recall. a = √(144-m²) And m = 3√(2) units. a = √(144-(3√(2))²) a = √(144-18) a = √(126) a = 3√(14) units. b = 3(3√(2)) b = 9√(3) units. c = 4(3√(2)) c = 12√(2) units. cosd = 9√(2)/(12√(2)) d = acos(¾)° Calculating x. x² = 2(6√(2))²-2(6√(2))²cos(acos(¾)) x² = 144-144(3/4) x² = 144-108 x² = 36 x = √(36) x = 6 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3128813

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Notice. 4 units is the radius of the inscribed circle. a² = 2(4)² a = 4√(2) units. 8² = (4√(2))²+4²-2*4*4√(2)cosb 64 = 32+16-32√(2)cosb 32√(2)cosb = 48-64 cosb = -16/(32√(2)) b = acos(-1/(2√(2))) b = acos(-¼√(2)) b = 110.704811055° c = 180-b c = 69.2951889454° Calculating x. Let x = 2d. And; sin69.2951889454 = d/4 d = 4sin69.2951889454 d = 3.74165738677 units. Therefore; x = 2*3.74165738677 x = 7.48331477355 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3128811

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a²+9² = 15² a = √(15²-9²) a = 12 units. a is AF. c = (9-b) units. Therefore; 12 ~ b 9 ~ (9-b) Cross Multiply. 108-12b = 9b 21b = 108 b = ⅐(36) units. b is the radius of the inscribed half circle. Recall. c = (9-b) And b = ⅐(36) units. c = 9-⅐(36) c = ⅐(63-36) c = ⅐(27) units. tand = c/b d = atan((⅐(27))/(⅐(36))) d = atan(3/4)° e = 90-d e = 90-atan(3/4) e = atan(4/3)° Let f be the radius of the inscribed circle, the required length. Calculating f. g = b-f g = ((36/7)-f) units. Ther...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3128048

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Let a be the radius of the half circle. b = (a-1) units. c²+1² = b² c² = a²-2a+1-1 c = √(a²-2a) units. d = a+c  d = (a+√(a²-2a)) units. d is the required length. e = (a+1) units. Calculating a. (a+√(a²-2a))²+(a-1)² = (a+1)² a²+2a√(a²-2a)+a²-2a+a²-2a+1 = a²+2a+1 2a√(a²-2a)+2a² = 6a a√(a²-2a)+a² = 3a √(a²-2a)+a = 3 a²-2a = (3-a)² a²-2a = 9-6a+a² 4a = 9 a = ¼(9) units. a = 2.25 units. Recall. d = (a+√(a²-2a)) And a = ¼(9) units. d = ((9/4)+√((9/4)²-2(9/4))...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3128036

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Calculating yellow area. It implies; 2a² = 4² Where a is the height of the yellow area. Therefore; 2a² = 4² 2a² = 16 a² = 8 a = √(8) a = 2√(2) units. Area Yellow is; ½*a*6 = ½*2√(2)*6 = 6√(2) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3128035

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Calculating angle x. Let the two equal lengths be 1 unit each. (1/sin(180-24-30)) = (a/sin30) a = 0.61803398875 units. b = 1+a b = 1.61803398875 units. c = 24+30 c = 54° (1/sin(180-24-30)) = (d/sin24) d = 0.50275413978 units. e² = 1²+1.61803398875²-2*1.61803398875*1cos24 e = 0.81347328615 units. Therefore, the required angle, x is; (0.81347328615/sin54) = (0.50275413978/sinx) x = 30°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3128034

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Calculating the required angle, let it be x. Let a be the side length of the ascribed square. ½(ab) = 1 b = 2/a --- (1). 1+1+c = ½(a²) c = ½(a²)-2 c = ½(a²-4) square units. d = a-b d = a-(2/a) d = (a²-2)/a units. f²+d² = a² f² = a²-((a²-2)/a)² f² = a²-((a⁴-4a²+4)/a²) f² = (a⁴-a⁴+4a²-4)/a² f² = (4a²-4)/a² f = 2√(a²-1)/a units. g = a-f g = a-(2√(a²-1)/a) g = (a²-2√(a²-1))/a units. It implies; ½(ag) = c ½*a*(a²-2√(a²-1))/a = ½(a²-4) a²-2√(a²-1) = a²-4 2...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3127330

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Calculating x. Let a be the height of the inscribed isosceles triangle. b = ½(7)+7 b = 10.5 units. c² = 10.5²+a² c = √(a²+110.25) units. Let d be one of the two equal inscribed angles. cosd = 10.5/√(a²+110.25) --- (1). e² = 3.5²+a² e = √(a²+12.25) units. f = (10-√(a²+12.25)) units. It implies; (10-√(a²+12.25))² = √(a²+110.25)²+14²-2*14*√(a²+110.25)cosd 100-20√(a²+12.25)+a²+12.25 = a²+110.25+196-28√(a²+110.25)cosd 100-20√(a²+12.25) = 294-28√(a²+110.25)cosd 50-10√(a²+12.25) = 147-14√...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3127327

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Let a be the side length of the red inscribed square. b = (4-a) cm. Therefore; 4² = 2b² 4² = 2(4-a)² 16 = 2(16-8a+a²) 8 = 16-8a+a² a²-8a+8 = 0 (a-4)² = -8+(-4)² (a-4)² = 16-8 a = 4±√(8) a ≠ (4+2√(2)) cm. a = (4-2√(2)) cm. a = 1.17157287525 cm. Therefore, area red inscribed square is; a² = (4-2√(2))² = 16-16√(2)+8 = 24-16√(2) = 8(3-2√(2)) cm² = 1.37258300203 cm²
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