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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

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3127322

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Calculating x. sina = 3.5/12.5 a = asin(3.5/12.5)° b = 2a b = 2asin(3.5/12.5)° sinc = 10/12.5 c = asin(10/12.5)° d = 2c d = 2asin(10/12.5)° e = 180-b-d e = 180-2asin(3.5/12.5)-2asin(10/12.5) e = 180-32.5204094166-106.260204708 e = 41.2193858754° It implies; x² = 2(12.5)²-2(12.5)²cos41.2193858754 x = 8.8 units. x = 88/10 units. x = ⅕(44) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3127317

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cosa = (2√(5))/(3√(5)) a = acos(⅔)° a is angle PRT. tana = 25/b b = 25/(tan(acos(⅔))) b = 22.360679775 units. b = 10√(5) units. Therefore, area triangle TRP is; ½*25b = ½*25*10√(5) = 125√(5) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3127294

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Let the radius of the ascribed half circle be 1 unit. a² = 2x² a = √(2)x units. b²+1² = a² b = √(2x²-1) units. c = 1-b c = (1-√(2x²-1)) units. d = ½(c) d = ½(1-√(2x²-1)) units. e²+x² = 2² e = √(4-x²) units. It implies; x ~ 2 ½(1-√(2x²-1)) ~ x Cross Multiply. (1-√(2x²-1)) = x² 1-x² = √(2x²-1) (1-x²)² = 2x²-1 1-2x²+x⁴ = 2x²-1 x⁴-4x²+2 = 0 Let x² = p. p²-4p = -2 (p-2)² = -2+(2)² (p-2)² = 2 p = 2±√(2) It implies; p = (2+√(2)) units. Or p = (2-√(2...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3126622

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Let a be the radius of the circle. b² = 2(2)² b = √(8) b = 2√(2) cm. c = (2√(2)-a) cm. It implies; 2a² = c² 2a² = (2√(2)-a)² 2a² = 8-4√(2)a+a² a²+4√(2)a-8 = 0 (a+2√(2))² = 8+(2√(2))² (a+2√(2))² = 16 a = -2√(2)±√(16) a = -2√(2)±4 Therefore; a = (4-2√(2)) cm. Area inscribed circle is; πa² = π(4-2√(2))² = π(16-16√(2)+8) = 8π(3-2√(2)) cm² = 4.31209667562 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3126411

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Let the bigger inscribed semi circle radius be 2 units. Let the big inscribed semi circle radius be r. Let the inscribed circle radius be q. Calculating r. (2+r)² = 2²+(4-r)² 4+4r+r² = 4+16-8r+r² 4r = 16-8r 12r=16 3r=4 r=4/3 unit. Calculating q. (2+(4/3))²=(2+q)²+((4/3)+q)² (10/3)²=4+4q+q²+(16/9)+(8q/3)+q² (100/9)-(16/9)-4=2q²+4q+(8q/3) (84/9)-4=2q²+4q+(8q/3) 84-36=18q²+36q+24q 48=18q²+60q 6q²+20q-16=0 3q²+10q-8=0 Therefore; q≠-4...
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