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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2552895

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Sir Mike Ambrose is the author of the question. The base/length of the rectangle with height or width 4 units is 3√(3) units. Therefore calculating red area (right-angled triangle with hypotenuse 4 units). The right-angled triangle's height, x is; Cos27.45707609594 = x/4 x = (4cos27.45707609594) units. It implies; Area Red exactly in decimal is; Area triangle with height (4cos27.45707609594) units and base (4sin27.45707609594) units. = ½*(4cos27.45707609594)*(4sin27.4570...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2552194

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a = (4-x) units. It implies; 4² = (4-x)²+2² 12 = 16-8x+x² x²-8x+4 = 0 (x-4)² = -4+(-4)² (x-4)² = -4+16 (x-4)² = 12 x = 4±√(12) x = 4±2√(3) It implies; x = (4-2√(3)) units. a = (4-x) And x = (4-2√(3)) units. a = 4-(4-2√(3)) a = 2√(3) units. b = 2+a b = (2+2√(3)) units. b is the radius of the big green quarter circle. c = b-4 c = (2+2√(3))-4 c = (2√(3)-2) units. c is the radius of the small green quarter circle. It implies, area green is; ¼*π(2+2√(3))²+¼*π(2√...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2551706

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a = (b+1) units. a is the diameter of the circle. c = ½(a) c = ½(b+1) units. c is the radius of the circle. d² = 1+1 d = √(2) units. d is the diagonal of a square. e = d-c e = √(2)-½(b+1) e = ½(2√(2)-1-b) units. Calculating b. (½(2√(2)-1-b))² = 2(½(b+1))² ½(2√(2)-1-b) = √(2)(½(b+1)) 2√(2)-1-b = √(2)(b+1) 2√(2)-1-b = √(2)b+√(2) √(2)-1 = √(2)b+b (1+√(2))b = (√(2)-1) b = (√(2)-1)/(√(2)+1) b = (2-2√(2)+1)/(1) b = (3-2√(2)) units. b = 0.1715728753 units. c = ½(b+1)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2551427

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tana = 4/7 a = atan(7/4)° b = 180-a b = (90+atan(4/7))° c² = 7²+4² c = √(49+16) c = √(65) units. Therefore, Blue Area is; Area triangle with height √(65) units and base 4sin(90+atan(4/7)) units. = 0.5*√(65)*4sin(90+atan(4/7)) = 14 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2551415

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a = 180-44 a = 136° b = ½(180+44) b = ½(224) b = 112= c = ½(a) c = 68° d = ½(b) d = 56 e = 180-c-d e = 180-68-56 e = 180-124 e = 56° Where e is the required angle, angle CBO. It implies triangle OBC is isosceles.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2551398

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7² = 5²+6²-2*5*6cosa 60cosa = 25+36-49 cosa = (61-49)/60 a = acos(12/60) a = 78.4630409672° Calculating r, radius of the inscribed half circle. 5r+6r = 5*6sin78.4630409672 11r = 30sin78.4630409672 r = (30sin78.4630409672)/11 r = 2.6721706285 units. Again, r is the radius of half the inscribed circle.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2550654

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tan30 = a/h a = htan30 units. tan60 = b/h b = htan60 units. It implies; a+b = 100 htan30+htan60 = 100 ⅓(√(3)h)+√(3)h = 100 ⅓(4√(3)h) = 100 4√(3)h = 300 h = 300/(4√(3)) h = ¼*⅓*300√(3) h = 25√(3) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2550595

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a²+R² = 7² a = √(49-R²) units. b = (3+R) units. c²+R² = 3² c = √(9-R²) units. d = a+c d = √(49-R²)+√(9-R²) units. R is; (√(49-R²)+√(9-R²))² = (3+R)²+√(49-R²)² 49-R²+2√((49-R²)(9-R²))+9-R² = 9+6R+R²+49-R² 58-2R²+2√((49-R²)(9-R²)) = 58+6R 2√((49-R²)(9-R²)) = 2R²+6R √((49-R²)(9-R²)) = R²+3R (49-R²)(9-R²) = (R²+3R)² 441-58R²+R⁴ = R⁴+6R³++9R² 441-67R²-6R³ = 0 6R³+67R²-441 = 0 It implies; R = ⅓(7) units. R = 2⅓ units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2549647

mathematics-question-and-solution_1716874878O23bLw.jpeg
0.5*a*6sin75 = 9 a = 9/(3sin75) a = 3/sin75 a = 3.1058285412 units. a is AB. b² = 3.1058285412²+6²-2*3.1058285412*6cos75 b = 6 units. b is AC. It implies, triangle ABC is isosceles. Therefore, the required angle alpha, let it be c is; c = 180-2*75 c = 180-150 c = 30° Again, c is the required angle, alpha.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2549160

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Calculating length CD. Notice! The composite plane shape is not drawn to scale. tan60 = 8/a a = 8/√(3) a = ⅓(8√(3)) units. a is OE. (⅓(8√(3)))²+8² = b² b² = ⅓(64)+64 b² = ⅓(64+192) b² = ⅓(256) b = ⅓(16√(3)) units. b is CE. Or sin60 = 8/c c = 8/sin60 c = 16/√(3) c = ⅓(16√(3)) units. c is CE. (8/sin120) = (⅓(8√(3))/sind) d = 30° d is angle ODE. It implies that DE = OE = ⅓(8√(3)) units. Calculating CD. Let it be e. e² = (⅓(8√(3)))²+(⅓(16√(3)))²-2*⅓(8√(3))*⅓...
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