Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1954045

Let the square side be 2 units. Notice; a+b = c --- (1). 2√(2) = c+a 2√(2)-a = c --- (2). a² = 2d² d = (a/√(2)) units. e = 2-d e = (2-(a/√(2))) units. b² = (2-(a/√(2)))²+(a/√(2))² b² = 4-(4a/√(2))+½(a²)+½(a²) b = √(4-(4a/√(2))+a²) --- (3). At (1). a+b = c Substituting (3) in (1). a+√(4-(4a/√(2))+a²) = c --- (4). Substituting (2) in (4). a+√(4-(4a/√(2))+a²) = 2√(2)-a √(4-(4a/√(2))+a²) = 2√(2)-2a 4-(4a/√(2))+a² = 8-8√(2)a+4a² 3a²-8√(2)a+2√(2)a+4 = 0 3a²-6√(2)a+4 = 0 3a²-8.48528137424a+4...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1952625

Let the square side, x be 1 unit. sin67.5 = a/1 a = 0.92387953251 units. cos67.5 = b/1 b = 0.38268343237 units. c = 2b c = 0.76536686473 units. d² = 1²+1² d = √(2) units. sin67.5 = e/√(2) e = 1.30656296488 units. e is the radius of the half circle. f = e-c f = 1.30656296488-0.76536686473 f = 0.54119610015 units. g = r-b g = (r-0.38268343237) units. r² = (r-0.38268343237)²+h² h² = r²-r²+0.76536686474r-0.14644660941 h² = 0.76536686474r-0.14644660941 h = √(0...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1952379

Let the side of the regular pentagon be 1 unit. Calculating Area Red. a² = 1²+1²-2*1*1cos108 a = 1.61803398875 units. b = 108-36-60 b = 12° (c/sin12) = (1.61803398875/sin96) c = 0.33826121272 unit. d = 1-c d = 0.66173878728 unit. e = 96-60 e = 36° tan36 = f/0.66173878728 f = 0.48078137139 unit. Area Red is; ½(df) = 0.5*0.48078137139*0.66173878728 = 0.15907584083 square units. Calculating Area Lilac. sin36 = 0.48078137139/g g = 0.81795412443 units. h = 1-f h = 0.51921862861 unit....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1952000

a² = 15²+12²-2*12*15cos120 a = 23.43074902772 cm. Where a is the side length of the equilateral triangle ABC. (23.43074902772/sin120) = (12/sinb) b = 26.32950349168° c = 60-b c = 33.67049650832° d² = 15²+15²-2*15*15cos120 d = 25.98076211353 cm. e = atan(7.5/25.98076211353) e = 16.10211375199° f = 90-e f = 73.89788624801° g = 180-f-c g = 72.43161724367° (15/sin72.43161724367) = (h/sin73.89788624801) h = 15.11661227595 cm. i = a-h i = 8.31413675177 cm. Let the radius of the green ins...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1951105

Let the side length of triangle ABC be 1 unit. a = 120-78 a = 42° b = 60-42 b = 18° (1/sin42) = (c/sin18) c = 0.4618186516 unit. tan60 = d/0.5 d = 0.86602540378 unit. Area Green is; 0.5*0.4618186516*0.86602540378 = 0.19997334211 square units. e = ⅓ unit. Where e is the side length of the red inscribed regular hexagon. Area Red is; ½*6*(⅓)²*(1/2tan(30)) = ⅓*(1/2tan(30)) = 0.28867513459 square units. f = 60+b f = 78° g = 180-2(78) g = 24° h = 1+c h = 1.4618186516 units. Where h i...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1950719

Let the inscribed square side be a. Radius of the half circle is 5 units. Considering Similar Triangle Ratios. 5 = a a = (5-a) Cross Multiply. 25-5a = a² a²+5a-25 = 0 (a+(5/2))² = 25+(5/2)² a+(5/2) = √(25+(25/4)) a+(5/2) = √(125/4) a = -(5/2)±½(5√(5)) a ≠ -½(5+5√(5)) units. a = ½(5√(5)-5) units. Therefore; Area of the inscribed square is; a² = (½(5√(5)-5))² = ¼(125-50√(5)+25) = ¼(150-50√(5)) square units. = 12.5(3-√(5)) square units. = 9.54915028125 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1949823

Let AB be 1 unit. (1/sin100) = (b/sin30) b = 0.50771330594 units. (1/sin110) = (d/sin30) d = 0.53208888624 units. d is CB. (0.50771330594/sin30) = (c/sin50) c = 0.77786191343 units. a² = 0.77786191343²+0.53208888624²-2*0.77786191343*0.53208888624cos10 a = 0.27014860749 units. Therefore, showing that; c = a+b Again; a = 0.27014860749 units. b = 0.50771330594 units. c = 0.77786191343 units. It implies; a+ b = 0.27014860749+0.50771330594 = 0.77786191343 units, which is c. Showed!

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1949559

Notice; The inscribed star is formed from a pentagram (a regular pentagon is ascribing the star). Let the side of the pentagram be 1 unit. tan54 = a/0.5 a = 0.68819096024 unit. cos54 = 0.5/b b = 0.85065080835 unit. c = b+a c = 1.53884176859 units. Let the side length of the ascribed regular hexagon be d. Calculating d. sin45 = e/d e =0.5√(2)d unit. It implies; e+d = c 0.5√(2)d+d = 1.53884176859 d = 0.90143263769 unit. f = 0.90143263769+2(0.90143263769sin45) f = 2.17625089948 units...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1949513

Notice; The figure is not drawn to scale. Let BC be 1 unit. (1/sin135) = (a/sin30) a = 0.70710678119 units. (1/sin135) = (b/sin15) b = 0.36602540378 units. Notice; MB = NC, let it be c. Therefore; MN = (1-2c) units. (AM)² = 0.36602540378²+c²-2*0.36602540378ccos30 (AM)² = 0.13397459621+c²-0.63397459621c --- (1) (AN)² = 0.70710678119²+c²-2*0.70710678119ccos15 (AN)² = 0.5+c²-1.36602540379c --- (2) Therefore; Calculating c. (MN)² = (AM)²+(AN)² (1-2c)² = 0.13...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1948290

Let the inscribed purple regular heptagon side be 1 unit. a = ⅐(180*5) a = ⅐(900)° b = 180-⅐(900)) b = ⅐(360)° c = 180-2(⅐(360)) c = ⅐(540)° (1/sin⅐(540)) = (d/sin⅐(360)) d = 0.8019377358 unit. e = 2(0.8019377358)+1 e = 2.60387547161 units. f = 90-½(⅐(540)) f = ⅐(360)° cos⅐(360) = g/2.60387547161 g = 1.62348980186 units. h = 2g h = 3.24697960372 units. Where h is the base of the ascribed quadrilateral. cos(180/7) = i/0.8019377358 i = 0.72252093395 unit. j = 2i j = 1.4450418679 units...

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