Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2692326

Area Orange exactly in fraction is; Area triangle with height. 7.87908624144 units and base 4.37727013143 units + Area triangle with side (38/9) units and 9.01335086564 units, and angle 53.0170930749° = ½(7.87908624144*4.37727013143) + ½((38/9)*9.01335086564)sin53.0170930749 = 17.2444444 + 15.2 = (776/45) + (76/5) = 1460/45 = 292/9 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2691721

Let MP be 2 units. a² = 1²+1² a = √(2) units. 2b² = 1² b² = ½ b = ½√(2) units. c = a+b c = ½(3√(2)) units. d² = 1²+(½√(2))²-2*1*½√(2)cos135 d = 1.5811388301 units. Therefore, the required angle, x is; (1.5811388301/sin135) = (0.5√(2)/sinx) x = 18.4349488229°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2691662

Calculating yellow area. a² = 5²+5²-2*5*5cos120 120° is the single interior angle of the regular hexagon. a = 5√(3) units. a = 8.6602540378 units. tanb = a/(2.5) b = atan(5√(3)/(5/2)) b = atan(2√(3))° b = 73.897886248° b is the angle of the yellow incomplete sector. c² = 2.5²+(5√(3))² c = 8.6746757864 units. d = ½(c) d = 4.3373378932 units. d is the radius of the yellow incomplete sector. e = 90-atan(2.5/(5√(3))) e = 73.897886248° f² = 4.33733789322+5²-2*5*4.33733...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2691653

Let the side length of the three congruent regular hexagon be 1 unit each. Notice. The inscribed green triangle is equilateral. a² = 2-2cos120 120° is the single interior angle of one of the congruent regular hexagon. a = √(3) units. b = 2a b = 2√(3) units. c² = (2√(3))²+1² c² = 13 c = √(13) units. tand = 2√(3)/1 d = atan(2√(3))° e = atan(1/(2√(3)))° f = 120-d f = 46.102113752° It implies; (1/sin60) = (g/sin(atan(2√(3))) g = 1.1094003925 units. (1/sin6...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2691650

Calculating angle x. tana = 1/3 a = atan(1/3)° b² = 3²+1² b = √(10) units. 1 - c √(10) - 6 Cross Multiply. √(10)c = 6 c = ⅗√(10) units. c = 1.8973665961 units. d² = 3²+1.8973665961²-2*3*1.8973665961cos(atan(⅓)) d = 1.3416407865 units. (1.3416407865/sin(atan(⅓))) = (3/sine) e = 45° Therefore, the required angle, x is; x = e x = 45°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2691646

Let a be the side length of the inscribed red equilateral triangle. It implies; (5/sin120) = (2.5/sinb) b = 25.6589062733° c = 60-b c = 34.3410937267° Calculating a. a² = 2.5²+5²-2*2.5*5cos34.3410937267 a = 3.2569390943 units. Therefore, area red inscribed equilateral triangle is; 0.5*3.2569390943²sin60 = 4.5932481676 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2690133

Sir Mike Ambrose is the author of the question. Let BC be 1 unit. Therefore; AB is 2 units. Area small square is; (⅓√(13))² = (13/9) square units. Area blue is; Area trapezium with two parallel side √(13) units and ⅓√(13) units and height ⅓√(13) units - area triangle with height ⅓√(13) units and base √(13)/6 units - area triangle with side √(13) units and 3.25764407172 units, and angle atan(1/3)° = ½(√(13)+⅓√(13))⅓√(13) - ½(⅓√(13)*√(13)/6) - ½(√(13)*3.25764407172)sin(a...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2689776

Let a be the base of the blue inscribed triangle. b = 4-1-a b = (3-a) units. c² = 2(1)² c = √(2) units. d² = 2(4)² d = 4√(2) units. e = d-c e = 3√(2) units. f² = a²+4² f = √(a²+16) units. g²+1² = (3√(2))² g² = 18-1 g = √(17) units. h = g+b h = (√(17)+(3-a)) units. Therefore, equating f and h to get a, the base of the blue inscribed triangle. √(a²+16) = (√(17)+(3-a)) a²+16 = (√(17)+(3-a))² a²+16 = ((√(17)+3)-a)² a²+16 = (7.1231056256-a)² 16 = 50.738633753...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2689761

Calculating blue inscribed half circle area. Let the radius of the blue inscribed half circle be a. 20² = 2b² b² = 200 b = √(200) b = 10√(2) units. c²+(0.5b)² = a² c² = a²-(5√(2))² c = √(a²-50) units. d = a-c d = (a-√(a²-50)) units. d is OC. Calculating a. 5√(2) - (a-√(a²-50)) 10√(2) - 5√(2) It implies; ½ = (a-√(a²-50))/(5√(2)) Cross Multiply. 2(a-√(a²-50)) = 5√(2) 2a-2√(a²-50) = 5√(2) 2a-5√(2) = 2√(a²-50) (2a-5√(2))² = 4(a²-50) 4a²-20√(2)a+50 = 4a²-2...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2689755

x = 180-2(70) x = 180-140 x = 40°

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